#andy

$\begin{array}{l}
1)\\
a)\left( {x + 1} \right)\left( {2x - 3} \right)\\
 = 2{x^2} - 3x + 2x - 3\\
 = 2{x^2} - x - 3\\
b)\left( {{x^3} - 3{x^2} + x - 3} \right):\left( {x - 3} \right)\\
 = \left[ {{x^2}\left( {x - 3} \right) + \left( {x - 3} \right)} \right]:\left( {x - 3} \right)\\
 = \left( {{x^2} + 1} \right)\left( {x - 3} \right):\left( {x - 3} \right)\\
 = {x^2} + 1\\
2)\\
a)6x{y^2} - 3{x^2}y\\
 = 3xy\left( {2y - x} \right)\\
b){x^2} + 5x - {y^2} - 5y\\
 = \left( {{x^2} - {y^2}} \right) - \left( {5x + 5y} \right)\\
 = \left( {x - y} \right)\left( {x + y} \right) + 5\left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {x - y + 5} \right)
\end{array}$

$\begin{array}{l}
2)\\
a)\\
B = \dfrac{3}{{x + 5}} + \dfrac{{20 - 2x}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}\\
B = \dfrac{{3\left( {x - 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} + \dfrac{{20 - 2x}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}\\
B = \dfrac{{3x - 15 + 20 - 2x}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}\\
B = \dfrac{{x + 5}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}\\
B = \dfrac{1}{{x - 5}}\\
b)\dfrac{{x + 2}}{{x - 5}} + \dfrac{1}{{x - 5}}\\
 = \dfrac{{x + 2 + 1}}{{x - 5}}\\
 = \dfrac{{x + 3}}{{x + 5}}\\
3)\\
a)3x + 2\left( {5 - x} \right) = 0\\
 \Leftrightarrow 3x + 10 - 2x = 0\\
 \Leftrightarrow x + 10 = 0\\
 \Leftrightarrow x =  - 10\\
b)x\left( {x - 2} \right) + 5\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x + 5} \right)\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 5 = 0\\
x - 2 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 5\\
x = 2
\end{array} \right.\\
 \Rightarrow S \in \{  - 5;2\} 
\end{array}$

Đáp án:

Giải thích các bước giải:

 Câu 1 :

$\ a) (x + 1)(2x - 3) $

$\ = 2x² - 3x + 2x - 3 $
$\ = 2x² - x - 3 $

$\ b) (x³ - 3x² + x - 3) : (x - 3) $

$\ = [(x³ - 3x²) + (x - 3)] : (x - 3) $

$\ = [x²(x - 3) + (x - 3)] : (x - 3) $
$\ = [(x - 3)(x² + 1)] : (x - 3) $
$\ = x² + 1 $

Câu 2 :

$\ a) 6xy² - 3x²y $

$\ = 3xy(2y - x) $

$\ b) x² + 5x - y² - 5y $
$\ = (x² - y²) + (5x - 5y) $

$\ = (x - y)(x + y) + 5(x - y) $

$\ = (x - y)(x + y + 5) $

Bài 2 : 

 a) B = $\frac{3}{x + 5}$ + $\frac{20 - 2x}{(x - 5)(x + 5)}$ 

 B = $\frac{3(x - 5)}{(x + 5)(x - 5)}$ + $\frac{20 - 2x}{(x - 5)(x + 5)}$ 

 B = $\frac{3x - 15}{(x + 5)(x - 5)}$ + $\frac{20 - 2x}{(x - 5)(x + 5)}$ 

 B = $\frac{3x - 15 + 20 - 2x}{(x + 5)(x - 5)}$

B = $\frac{x + 5}{(x + 5)(x - 5)}$

B = $\frac{1}{x - 5}$

b) Ta có : $\ P = A + B $

⇒ P = $\frac{x + 2}{x - 5}$ + $\frac{1}{x - 5}$ 

P = $\frac{x + 2 + 1}{x - 5}$ 

P = $\frac{x + 3}{x - 5}$ 

Bài 3 :

$\ a) 3x + 2(5 - x) = 0 $
$\ 3x + 10 - 2x = 0 $

$\ x + 10 = 0 $
$\ ⇒ x = - 10 $

$\ b) x(x - 2) + 5(x - 2) = 0 $

$\ (x - 2)(x + 5) = 0 $

⇒ \(\left[ \begin{array}{l}x-2=0\\x+5=0\end{array} \right.\) 

⇒ \(\left[ \begin{array}{l}x=2\\x=-5\end{array} \right.\) 

Vậy $\ x = 2 $ hoặc $\ x = - 5$