Đáp án:

`35 . B` 

`36 . B`

Giải thích các bước giải:

Câu `35.`

`(3x+12)/(4x-16) . (8-2x)/(x+4)`

`= [3(x+4)]/[4(x-4)] . [2(4-x)]/(x+4)`

`=[3(x+4).2(4-x)]/[4(x-4)(x+4)]`

`=[-6(x+4)(x-4)]/[4(x+4)(x-4)]`

`=-6/4`

`=-3/2`

`-> B`

Câu `36.`

`(6x-3)/(2x^3y^2) : (12x-6)/(4x^2y^3)`

`=[3(2x-1)]/(2x^3y^2) : [6(2x-1)]/(4x^2y^3)`

`=[3(2x-1)]/(2x^3y^2) . (4x^2y^3)/[6(2x-1)]`

`=[3(2x-1) . 4x^2y^3]/[6(2x-1) . 2x^3y^2]`

`=(2y)/(2x)`

`=y/x`

`-> B`

#andy

\[\begin{array}{l}
35)\\
\dfrac{{3x + 12}}{{4x - 16}}.\dfrac{{8 - 2x}}{{x + 4}}\\
 = \dfrac{{3\left( {x + 4} \right)}}{{4\left( {x + 4} \right)}}.\dfrac{{ - 2\left( {x + 4} \right)}}{{x + 4}}\\
 = \dfrac{3}{4}.\left( { - 2} \right)\\
 =  - \dfrac{3}{2}\\
 \Rightarrow B\\
36)\\
\dfrac{{6x - 3}}{{2{x^3}{y^3}}}:\dfrac{{12x - 6}}{{4{x^2}{y^3}}}\\
 = \dfrac{{3\left( {2x - 1} \right)}}{{2{x^3}{y^3}}}.\dfrac{{4{x^2}{y^3}}}{{6\left( {2x - 1} \right)}}\\
 = \dfrac{{3\left( {2x - 1} \right).4{x^2}{y^3}}}{{2{x^3}{y^2}.6\left( {2x - 1} \right)}}\\
 = \dfrac{{2y}}{{2x}}\\
 = \dfrac{y}{x}\\
 \Rightarrow B
\end{array}\]