Đáp án:

B22:

a) 0

Giải thích các bước giải:

\(\begin{array}{l}
B22:\\
a)\left( { - \dfrac{3}{4} + \dfrac{2}{3} - \dfrac{1}{4} + \dfrac{1}{3}} \right):\dfrac{5}{{11}}\\
 = \left( { - 1 + 1} \right):\dfrac{5}{{11}} = 0\\
b)5.\left( {\dfrac{2}{5} - \dfrac{3}{2}} \right):\left( { - 8.\dfrac{{11}}{8}} \right)\\
 = 5.\left( { - \dfrac{{11}}{{10}}} \right):\left( { - 11} \right)\\
 = \dfrac{{11}}{2}.\dfrac{1}{{11}} = \dfrac{1}{2}\\
C23:\\
a)\left| {3 - 2x} \right| = \dfrac{4}{5} - \dfrac{3}{5}\\
 \to \left| {3 - 2x} \right| = \dfrac{1}{5}\\
 \to \left[ \begin{array}{l}
3 - 2x = \dfrac{1}{5}\\
3 - 2x =  - \dfrac{1}{5}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{7}{5}\\
x = \dfrac{8}{5}
\end{array} \right.\\
b){\left( {2x - 1} \right)^2} = 100\\
 \to \left| {2x - 1} \right| = 10\\
 \to \left[ \begin{array}{l}
2x - 1 = 10\\
2x - 1 =  - 10
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{11}}{2}\\
x =  - \dfrac{9}{2}
\end{array} \right.\\
c)\dfrac{2}{5}x - \dfrac{3}{2} =  - \dfrac{{16}}{{13}}.\dfrac{{13}}{5}\\
 \to \dfrac{2}{5}x - \dfrac{3}{2} =  - \dfrac{{16}}{5}\\
 \to \dfrac{2}{5}x =  - \dfrac{{17}}{{10}}\\
 \to x =  - \dfrac{{17}}{4}
\end{array}\)

Câu `22` :

`a)` `(-3/4 + 2/3 ) : 5/11 + ( -1/4 + 1/3 ) : 5/11`

`=` `(-1)/12 : 5/11 + 1/12 : 5/11`

`=` `5/11 . ((-1)/12 + 1/12)`

`=` `5/11 . 0`

`=` `0`

`b)` `\sqrt{25}` . `(0,4 - 1 1/2 ) : [(-2)^3 . 11/8]`

`=` `5 . (-11)/10 : (-11)`

`=` `(-11)/2 : (-11)`

`=` `1/2`

`a)` `|3-2x|` `+` `\sqrt{9/25}` `=` `\sqrt{16/25}`

`=` `|3-2x|` `=` `7/25`

`⇒` \(\left[ \begin{array}{l}3-2x=7/25\\3-2x=-7/25\end{array} \right.\)

`⇒` \(\left[ \begin{array}{l}x=34/25\\x=41/25\end{array} \right.\)

Vậy `x∈{34/25;41/25}`

`b)` `(2x-1)/2 = (-50)/(2x-1)`

`⇒` `(2x-1)^2 = 100`

`⇒` `2x-1=±10`

`⇒` \(\left[ \begin{array}{l}2x-1=10\\2x-1=-10\end{array} \right.\)

`⇒` \(\left[ \begin{array}{l}x=11/2\\x=-9/2\end{array} \right.\)

Vậy `x∈{11/2;(-9)/2}`

`c)` `(2/5 x - 1,5) : 2,6 = -1 3/13`

`2/5 x - 1,5  = (-16)/5`

`2/5 x = (-17)/10`

`x=(-17)/4`