Toán Lớp 8: Cho A= 2+x/2-x – 4x^2/x^2-4 – 2-x/2+x và B= x^2-3x/2x^2-x^3
a. Rút gọn P= A.1/B
b. Tìm giá trị nguyên của x để P chia hết cho 4
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Toán Lớp 8: Cho A= 2+x/2-x – 4x^2/x^2-4 – 2-x/2+x và B= x^2-3x/2x^2-x^3
a. Rút gọn P= A.1/B
b. Tìm giá trị nguyên của x để P chia hết cho 4
Comments ( 1 )
a)Dkxd:\left\{ \begin{array}{l}
x\# 2\\
x\# – 2\\
x\# 0\\
x\# 3
\end{array} \right.\\
A = \dfrac{{2 + x}}{{2 – x}} – \dfrac{{4{x^2}}}{{{x^2} – 4}} – \dfrac{{2 – x}}{{2 + x}}\\
= \dfrac{{\left( {2 + x} \right)\left( {2 + x} \right) + 4{x^2} – \left( {2 – x} \right)\left( {2 – x} \right)}}{{\left( {2 + x} \right)\left( {2 – x} \right)}}\\
= \dfrac{{{x^2} + 4x + 4 + 4{x^2} – 4 + 4x – {x^2}}}{{\left( {2 + x} \right)\left( {2 – x} \right)}}\\
= \dfrac{{4{x^2} + 8x}}{{\left( {2 + x} \right)\left( {2 – x} \right)}}\\
= \dfrac{{4x\left( {x + 2} \right)}}{{\left( {2 + x} \right)\left( {2 – x} \right)}}\\
= \dfrac{{4x}}{{2 – x}}\\
B = \dfrac{{{x^2} – 3x}}{{2{x^2} – {x^3}}} = \dfrac{{x\left( {x – 3} \right)}}{{{x^2}\left( {2 – x} \right)}}\\
= \dfrac{{x – 3}}{{x\left( {2 – x} \right)}}\\
P = A.\dfrac{1}{B}\\
= \dfrac{{4x}}{{2 – x}}.\dfrac{{x\left( {2 – x} \right)}}{{x – 3}} = \dfrac{{4{x^2}}}{{x – 3}}\\
b)P \vdots 4\\
\Leftrightarrow \dfrac{{4{x^2}}}{{x – 3}} \vdots 4\\
\Leftrightarrow 4.\dfrac{{{x^2}}}{{x – 3}} \vdots 4\\
\Leftrightarrow \dfrac{{{x^2}}}{{x – 3}} \in Z\\
\Leftrightarrow \dfrac{{{x^2} – 9 + 9}}{{x – 3}} \in Z\\
\Leftrightarrow x + 3 + \dfrac{9}{{x – 3}} \in Z\\
\Leftrightarrow \dfrac{9}{{x – 3}} \in Z\\
\Leftrightarrow \left( {x – 3} \right) \in \left\{ { – 9; – 3; – 1;1;3;9} \right\}\\
\Leftrightarrow x \in \left\{ { – 6;0;2;4;6;12} \right\}\\
Do:x\# 0;x\# 2\\
\Leftrightarrow x \in \left\{ { – 6;4;6;12} \right\}\\
Vay\,x \in \left\{ { – 6;4;6;12} \right\}
\end{array}$