Đáp án:

\(\begin{array}{l}
1,\\
a,\\
x = \dfrac{1}{7}\\
b,\\
\left[ \begin{array}{l}
x = \dfrac{{\sqrt {105}  - 9}}{6}\\
x = \dfrac{{ - \sqrt {105}  - 9}}{6}
\end{array} \right.\\
c,\\
x = 8\\
d,\\
x = \dfrac{5}{2}\\
e,\\
Phương\,\,trình\,\,có\,\,vô\,\,số\,\,nghiệm\\
f,\\
x = 23\\
2,\\
a,\\
\left[ \begin{array}{l}
x =  - \dfrac{1}{3}\\
x =  - 2
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = 5\\
x = \dfrac{3}{2}
\end{array} \right.\\
c,\\
x = 1\\
c,\\
\left[ \begin{array}{l}
x = 1\\
x =  - 5
\end{array} \right.\\
e,\\
\left[ \begin{array}{l}
x = 0\\
x = 2\\
x = 3
\end{array} \right.\\
f,\\
\left[ \begin{array}{l}
x = 9\\
x = 2\\
x =  - 2
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
5 - \left( {x - 6} \right) = 4.\left( {3 - 2x} \right)\\
 \Leftrightarrow 5 - x + 6 = 12 - 8x\\
 \Leftrightarrow 11 - x = 12 - 8x\\
 \Leftrightarrow 11 - x - 12 + 8x = 0\\
 \Leftrightarrow  - 1 + 7x = 0\\
 \Leftrightarrow 7x = 1\\
 \Leftrightarrow x = \dfrac{1}{7}\\
b,\\
3 - x\left( {1 - 3x} \right) = 5.\left( {1 - 2x} \right)\\
 \Leftrightarrow 3 - \left( {x - 3{x^2}} \right) = 5 - 10x\\
 \Leftrightarrow 3 - x + 3{x^2} = 5 - 10x\\
 \Leftrightarrow 3 - x + 3{x^2} - 5 + 10x = 0\\
 \Leftrightarrow 3{x^2} + 9x - 2 = 0\\
 \Leftrightarrow 3.\left( {{x^2} + 3x - \dfrac{2}{3}} \right) = 0\\
 \Leftrightarrow {x^2} + 3x - \dfrac{2}{3} = 0\\
 \Leftrightarrow \left( {{x^2} + 3x + \dfrac{9}{4}} \right) - \dfrac{{35}}{{12}} = 0\\
 \Leftrightarrow {x^2} + 3x + \dfrac{9}{4} = \dfrac{{35}}{{12}}\\
 \Leftrightarrow {x^2} + 2.x.\dfrac{3}{2} + {\left( {\dfrac{3}{2}} \right)^2} = \dfrac{{35}}{{12}}\\
 \Leftrightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = \dfrac{{35}}{{12}}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{3}{2} = \sqrt {\dfrac{{35}}{{12}}} \\
x + \dfrac{3}{2} =  - \sqrt {\dfrac{{35}}{{12}}} 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{3}{2} = \dfrac{{\sqrt {105} }}{6}\\
x + \dfrac{3}{2} =  - \dfrac{{\sqrt {105} }}{6}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{\sqrt {105} }}{6} - \dfrac{3}{2}\\
x =  - \dfrac{{\sqrt {105} }}{6} - \dfrac{3}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{\sqrt {105}  - 9}}{6}\\
x = \dfrac{{ - \sqrt {105}  - 9}}{6}
\end{array} \right.\\
c,\\
\left( {x - 3} \right)\left( {x + 4} \right) - 2.\left( {3x - 2} \right) = {\left( {x - 4} \right)^2}\\
 \Leftrightarrow \left( {{x^2} + 4x - 3x - 12} \right) - \left( {6x - 4} \right) = {x^2} - 2.x.4 + {4^2}\\
 \Leftrightarrow {x^2} + x - 12 - 6x + 4 = {x^2} - 8x + 16\\
 \Leftrightarrow {x^2} - 5x - 8 = {x^2} - 8x + 16\\
 \Leftrightarrow {x^2} - 5x - 8 - {x^2} + 8x - 16 = 0\\
 \Leftrightarrow 3x - 24 = 0\\
 \Leftrightarrow 3x = 24\\
 \Leftrightarrow x = 8\\
d,\\
\dfrac{{3x + 2}}{2} - \dfrac{{3x + 1}}{6} = \dfrac{5}{3} + 2x\\
 \Leftrightarrow \dfrac{{3.\left( {3x + 2} \right) - \left( {3x + 1} \right)}}{6} = \dfrac{{5 + 2x}}{3}\\
 \Leftrightarrow \dfrac{{9x + 6 - 3x - 1}}{6} = \dfrac{{2.\left( {5 + 2x} \right)}}{6}\\
 \Leftrightarrow \dfrac{{6x + 5}}{6} = \dfrac{{10 + 4x}}{6}\\
 \Leftrightarrow 6x + 5 = 10 + 4x\\
 \Leftrightarrow 6x + 5 - 10 - 4x = 0\\
 \Leftrightarrow 2x - 5 = 0\\
 \Leftrightarrow x = \dfrac{5}{2}\\
e,\\
\dfrac{{2x - 1}}{5} - \dfrac{{x - 2}}{3} = \dfrac{{x + 7}}{{15}}\\
 \Leftrightarrow \dfrac{{3.\left( {2x - 1} \right) - 5.\left( {x - 2} \right)}}{{15}} = \dfrac{{x + 7}}{{15}}\\
 \Leftrightarrow \dfrac{{6x - 3 - 5x + 10}}{{15}} = \dfrac{{x + 7}}{{15}}\\
 \Leftrightarrow \dfrac{{x + 7}}{{15}} = \dfrac{{x + 7}}{{15}}\\
 \Rightarrow Phương\,\,trình\,\,có\,\,vô\,\,số\,\,nghiệm\\
f,\\
\dfrac{{x - 23}}{{24}} + \dfrac{{x - 23}}{{25}} = \dfrac{{x - 23}}{{26}} + \dfrac{{x - 23}}{{27}}\\
 \Leftrightarrow \left( {x - 23} \right).\left( {\dfrac{1}{{24}} + \dfrac{1}{{25}}} \right) = \left( {x - 23} \right).\left( {\dfrac{1}{{26}} + \dfrac{1}{{27}}} \right)\\
 \Leftrightarrow \left( {x - 23} \right).\left( {\dfrac{1}{{24}} + \dfrac{1}{{25}} - \dfrac{1}{{26}} - \dfrac{1}{{27}}} \right) = 0\\
 \Leftrightarrow x - 23 = 0\\
 \Leftrightarrow x = 23\\
2,\\
a,\\
9{x^2} - 1 = \left( {3x + 1} \right)\left( {4x + 1} \right)\\
 \Leftrightarrow {\left( {3x} \right)^2} - {1^2} = \left( {3x + 1} \right)\left( {4x + 1} \right)\\
 \Leftrightarrow \left( {3x - 1} \right)\left( {3x + 1} \right) - \left( {3x + 1} \right)\left( {4x + 1} \right) = 0\\
 \Leftrightarrow \left( {3x + 1} \right).\left[ {\left( {3x - 1} \right) - \left( {4x + 1} \right)} \right] = 0\\
 \Leftrightarrow \left( {3x + 1} \right).\left( {3x - 1 - 4x - 1} \right) = 0\\
 \Leftrightarrow \left( {3x + 1} \right).\left( { - x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
3x + 1 = 0\\
 - x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{1}{3}\\
x =  - 2
\end{array} \right.\\
b,\\
3x - 15 = 2x.\left( {x - 5} \right)\\
 \Leftrightarrow 3\left( {x - 5} \right) = 2x\left( {x - 5} \right)\\
 \Leftrightarrow 3\left( {x - 5} \right) - 2x\left( {x - 5} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {3 - 2x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
3 - 2x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \dfrac{3}{2}
\end{array} \right.\\
c,\\
2x\left( {x - 1} \right) = {x^2} - 1\\
 \Leftrightarrow 2x\left( {x - 1} \right) = {x^2} - {1^2}\\
 \Leftrightarrow 2x\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x + 1} \right)\\
 \Leftrightarrow 2x.\left( {x - 1} \right) - \left( {x - 1} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left[ {2x - \left( {x + 1} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left( {2x - x - 1} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left( {x - 1} \right) = 0\\
 \Leftrightarrow x - 1 = 0\\
 \Leftrightarrow x = 1\\
c,\\
{x^2} + 4x - 5 = 0\\
 \Leftrightarrow \left( {{x^2} + 4x + 4} \right) - 9 = 0\\
 \Leftrightarrow {x^2} + 2.x.2 + {2^2} = 9\\
 \Leftrightarrow {\left( {x + 2} \right)^2} = {3^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 2 = 3\\
x + 2 =  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 5
\end{array} \right.\\
e,\\
{x^3} - 5{x^2} + 6x = 0\\
 \Leftrightarrow x.\left( {{x^2} - 5x + 6} \right) = 0\\
 \Leftrightarrow x.\left[ {\left( {{x^2} - 2x} \right) + \left( { - 3x + 6} \right)} \right] = 0\\
 \Leftrightarrow x.\left[ {x.\left( {x - 2} \right) - 3.\left( {x - 2} \right)} \right] = 0\\
 \Leftrightarrow x.\left( {x - 2} \right)\left( {x - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 2 = 0\\
x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x = 3
\end{array} \right.\\
f,\\
{x^3} + 9{x^2} - 4x - 36 = 0\\
 \Leftrightarrow \left( {{x^3} + 9{x^2}} \right) + \left( { - 4x - 36} \right) = 0\\
 \Leftrightarrow {x^2}.\left( {x + 9} \right) - 4.\left( {x + 9} \right) = 0\\
 \Leftrightarrow \left( {x + 9} \right).\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow \left( {x + 9} \right).\left( {{x^2} - {2^2}} \right) = 0\\
 \Leftrightarrow \left( {x + 9} \right).\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 9 = 0\\
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 9\\
x = 2\\
x =  - 2
\end{array} \right.
\end{array}\)

\(\begin{array}{l}
1,\\
a,\\
5 - \left( {x - 6} \right) = 4.\left( {3 - 2x} \right)\\
 \Leftrightarrow 5 - x + 6 = 12 - 8x\\
 \Leftrightarrow 11 - x = 12 - 8x\\
 \Leftrightarrow 11 - x - 12 + 8x = 0\\
 \Leftrightarrow  - 1 + 7x = 0\\
 \Leftrightarrow 7x = 1\\
 \Leftrightarrow x = \dfrac{1}{7}\\
b,\\
3 - x\left( {1 - 3x} \right) = 5.\left( {1 - 2x} \right)\\
 \Leftrightarrow 3 - \left( {x - 3{x^2}} \right) = 5 - 10x\\
 \Leftrightarrow 3 - x + 3{x^2} = 5 - 10x\\
 \Leftrightarrow 3 - x + 3{x^2} - 5 + 10x = 0\\
 \Leftrightarrow 3{x^2} + 9x - 2 = 0\\
 \Leftrightarrow 3.\left( {{x^2} + 3x - \dfrac{2}{3}} \right) = 0\\
 \Leftrightarrow {x^2} + 3x - \dfrac{2}{3} = 0\\
 \Leftrightarrow \left( {{x^2} + 3x + \dfrac{9}{4}} \right) - \dfrac{{35}}{{12}} = 0\\
 \Leftrightarrow {x^2} + 3x + \dfrac{9}{4} = \dfrac{{35}}{{12}}\\
 \Leftrightarrow {x^2} + 2.x.\dfrac{3}{2} + {\left( {\dfrac{3}{2}} \right)^2} = \dfrac{{35}}{{12}}\\
 \Leftrightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = \dfrac{{35}}{{12}}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{3}{2} = \sqrt {\dfrac{{35}}{{12}}} \\
x + \dfrac{3}{2} =  - \sqrt {\dfrac{{35}}{{12}}} 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{3}{2} = \dfrac{{\sqrt {105} }}{6}\\
x + \dfrac{3}{2} =  - \dfrac{{\sqrt {105} }}{6}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{\sqrt {105} }}{6} - \dfrac{3}{2}\\
x =  - \dfrac{{\sqrt {105} }}{6} - \dfrac{3}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{\sqrt {105}  - 9}}{6}\\
x = \dfrac{{ - \sqrt {105}  - 9}}{6}
\end{array} \right.\\
c,\\
\left( {x - 3} \right)\left( {x + 4} \right) - 2.\left( {3x - 2} \right) = {\left( {x - 4} \right)^2}\\
 \Leftrightarrow \left( {{x^2} + 4x - 3x - 12} \right) - \left( {6x - 4} \right) = {x^2} - 2.x.4 + {4^2}\\
 \Leftrightarrow {x^2} + x - 12 - 6x + 4 = {x^2} - 8x + 16\\
 \Leftrightarrow {x^2} - 5x - 8 = {x^2} - 8x + 16\\
 \Leftrightarrow {x^2} - 5x - 8 - {x^2} + 8x - 16 = 0\\
 \Leftrightarrow 3x - 24 = 0\\
 \Leftrightarrow 3x = 24\\
 \Leftrightarrow x = 8\\
d,\\
\dfrac{{3x + 2}}{2} - \dfrac{{3x + 1}}{6} = \dfrac{5}{3} + 2x\\
 \Leftrightarrow \dfrac{{3.\left( {3x + 2} \right) - \left( {3x + 1} \right)}}{6} = \dfrac{{5 + 2x}}{3}\\
 \Leftrightarrow \dfrac{{9x + 6 - 3x - 1}}{6} = \dfrac{{2.\left( {5 + 2x} \right)}}{6}\\
 \Leftrightarrow \dfrac{{6x + 5}}{6} = \dfrac{{10 + 4x}}{6}\\
 \Leftrightarrow 6x + 5 = 10 + 4x\\
 \Leftrightarrow 6x + 5 - 10 - 4x = 0\\
 \Leftrightarrow 2x - 5 = 0\\
 \Leftrightarrow x = \dfrac{5}{2}\\
e,\\
\dfrac{{2x - 1}}{5} - \dfrac{{x - 2}}{3} = \dfrac{{x + 7}}{{15}}\\
 \Leftrightarrow \dfrac{{3.\left( {2x - 1} \right) - 5.\left( {x - 2} \right)}}{{15}} = \dfrac{{x + 7}}{{15}}\\
 \Leftrightarrow \dfrac{{6x - 3 - 5x + 10}}{{15}} = \dfrac{{x + 7}}{{15}}\\
 \Leftrightarrow \dfrac{{x + 7}}{{15}} = \dfrac{{x + 7}}{{15}}\\
 \Rightarrow Phương\,\,trình\,\,có\,\,vô\,\,số\,\,nghiệm\\
f,\\
\dfrac{{x - 23}}{{24}} + \dfrac{{x - 23}}{{25}} = \dfrac{{x - 23}}{{26}} + \dfrac{{x - 23}}{{27}}\\
 \Leftrightarrow \left( {x - 23} \right).\left( {\dfrac{1}{{24}} + \dfrac{1}{{25}}} \right) = \left( {x - 23} \right).\left( {\dfrac{1}{{26}} + \dfrac{1}{{27}}} \right)\\
 \Leftrightarrow \left( {x - 23} \right).\left( {\dfrac{1}{{24}} + \dfrac{1}{{25}} - \dfrac{1}{{26}} - \dfrac{1}{{27}}} \right) = 0\\
 \Leftrightarrow x - 23 = 0\\
 \Leftrightarrow x = 23\\
2,\\
a,\\
9{x^2} - 1 = \left( {3x + 1} \right)\left( {4x + 1} \right)\\
 \Leftrightarrow {\left( {3x} \right)^2} - {1^2} = \left( {3x + 1} \right)\left( {4x + 1} \right)\\
 \Leftrightarrow \left( {3x - 1} \right)\left( {3x + 1} \right) - \left( {3x + 1} \right)\left( {4x + 1} \right) = 0\\
 \Leftrightarrow \left( {3x + 1} \right).\left[ {\left( {3x - 1} \right) - \left( {4x + 1} \right)} \right] = 0\\
 \Leftrightarrow \left( {3x + 1} \right).\left( {3x - 1 - 4x - 1} \right) = 0\\
 \Leftrightarrow \left( {3x + 1} \right).\left( { - x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
3x + 1 = 0\\
 - x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{1}{3}\\
x =  - 2
\end{array} \right.\\
b,\\
3x - 15 = 2x.\left( {x - 5} \right)\\
 \Leftrightarrow 3\left( {x - 5} \right) = 2x\left( {x - 5} \right)\\
 \Leftrightarrow 3\left( {x - 5} \right) - 2x\left( {x - 5} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {3 - 2x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
3 - 2x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \dfrac{3}{2}
\end{array} \right.\\
c,\\
2x\left( {x - 1} \right) = {x^2} - 1\\
 \Leftrightarrow 2x\left( {x - 1} \right) = {x^2} - {1^2}\\
 \Leftrightarrow 2x\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x + 1} \right)\\
 \Leftrightarrow 2x.\left( {x - 1} \right) - \left( {x - 1} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left[ {2x - \left( {x + 1} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left( {2x - x - 1} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left( {x - 1} \right) = 0\\
 \Leftrightarrow x - 1 = 0\\
 \Leftrightarrow x = 1\\
c,\\
{x^2} + 4x - 5 = 0\\
 \Leftrightarrow \left( {{x^2} + 4x + 4} \right) - 9 = 0\\
 \Leftrightarrow {x^2} + 2.x.2 + {2^2} = 9\\
 \Leftrightarrow {\left( {x + 2} \right)^2} = {3^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 2 = 3\\
x + 2 =  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 5
\end{array} \right.\\
e,\\
{x^3} - 5{x^2} + 6x = 0\\
 \Leftrightarrow x.\left( {{x^2} - 5x + 6} \right) = 0\\
 \Leftrightarrow x.\left[ {\left( {{x^2} - 2x} \right) + \left( { - 3x + 6} \right)} \right] = 0\\
 \Leftrightarrow x.\left[ {x.\left( {x - 2} \right) - 3.\left( {x - 2} \right)} \right] = 0\\
 \Leftrightarrow x.\left( {x - 2} \right)\left( {x - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 2 = 0\\
x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x = 3
\end{array} \right.\\
f,\\
{x^3} + 9{x^2} - 4x - 36 = 0\\
 \Leftrightarrow \left( {{x^3} + 9{x^2}} \right) + \left( { - 4x - 36} \right) = 0\\
 \Leftrightarrow {x^2}.\left( {x + 9} \right) - 4.\left( {x + 9} \right) = 0\\
 \Leftrightarrow \left( {x + 9} \right).\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow \left( {x + 9} \right).\left( {{x^2} - {2^2}} \right) = 0\\
 \Leftrightarrow \left( {x + 9} \right).\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 9 = 0\\
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 9\\
x = 2\\
x =  - 2
\end{array} \right.
\end{array}\)