Đáp án:
B1:
\(A = \dfrac{{14}}{{11}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
A = \dfrac{{4{x^2} - 4}}{{x + 3}}:\dfrac{{2\left( {x - 1} \right)}}{1}\\
= \dfrac{{4\left( {x - 1} \right)\left( {x + 1} \right)}}{{x + 3}}.\dfrac{1}{{2\left( {x - 1} \right)}}\\
= \dfrac{{2\left( {x + 1} \right)}}{{x + 3}}\\
Thay:x = \dfrac{5}{2}\\
\to A = \dfrac{{2\left( {\dfrac{5}{2} + 1} \right)}}{{\dfrac{5}{2} + 3}} = \dfrac{{14}}{{11}}\\
B2:\\
\left[ {\dfrac{2}{{3x}} - \dfrac{2}{{x + 1}}.\left( {\dfrac{{x + 1}}{{3x}} - x - 1} \right)} \right]:\dfrac{{x - 1}}{x}\\
= \left[ {\dfrac{2}{{3x}} - \dfrac{2}{{x + 1}}.\left( {\dfrac{{x + 1 - 3{x^2} - 3x}}{{3x}}} \right)} \right].\dfrac{x}{{x - 1}}\\
= \left( {\dfrac{2}{{3x}} - \dfrac{2}{{x + 1}}.\dfrac{{ - 3{x^2} - 2x + 1}}{{3x}}} \right).\dfrac{x}{{x - 1}}\\
= \left( {\dfrac{2}{{3x}} - \dfrac{2}{{x + 1}}.\dfrac{{\left( {1 - 3x} \right)\left( {x + 1} \right)}}{{3x}}} \right).\dfrac{x}{{x - 1}}\\
= \left( {\dfrac{2}{{3x}} - \dfrac{{2\left( {1 - 3x} \right)}}{{3x}}} \right).\dfrac{x}{{x - 1}}\\
= \dfrac{{2 - 2 + 6x}}{{3x}}.\dfrac{x}{{x - 1}}\\
= 2.\dfrac{x}{{x - 1}} = \dfrac{{2x}}{{x - 1}}\\
\to dpcm
\end{array}\)
