Cho biểu thức: A=(x+5/x-2+3x/+2-4x^2/x^2-4)×x^2/x+10 a)rút gọn A b)tính giá trị a biết x^2-x-6=0

Đáp án:

 b) \(A = \dfrac{9}{5}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \left\{ { – 10; – 2;2} \right\}\\
A = \left( {\dfrac{{x + 5}}{{x – 2}} + \dfrac{{3x}}{{x + 2}} – \dfrac{{4{x^2}}}{{{x^2} – 4}}} \right).\dfrac{{{x^2}}}{{x + 10}}\\
 = \left[ {\dfrac{{\left( {x + 5} \right)\left( {x + 2} \right) + 3x\left( {x – 2} \right) – 4{x^2}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}} \right].\dfrac{{{x^2}}}{{x + 10}}\\
 = \dfrac{{{x^2} + 7x + 10 + 3{x^2} – 6x – 4{x^2}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}.\dfrac{{{x^2}}}{{x + 10}}\\
 = \dfrac{{x + 10}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}.\dfrac{{{x^2}}}{{x + 10}} = \dfrac{{{x^2}}}{{{x^2} – 4}}\\
b){x^2} – x – 6 = 0\\
 \to \left( {x – 3} \right)\left( {x + 2} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 3\\
x =  – 2\left( l \right)
\end{array} \right.\\
Thay:x = 3\\
 \to A = \dfrac{{{3^2}}}{{{3^2} – 4}} = \dfrac{9}{5}
\end{array}\)