Toán học 19 Tháng Tư, 2022 No Comments By Uyên Trâm Toán Lớp 8: x/x^2-2x-x^2+4x/x^3-4x-2/x^2+2x phép trừ nha mn
Giải đáp: \(\dfrac{{ – 4x + 4}}{{x\left( {x – 2} \right)\left( {x + 2} \right)}}\) Lời giải và giải thích chi tiết: \(\begin{array}{l}DK:x \ne \left\{ { – 2;0;2} \right\}\\\dfrac{x}{{{x^2} – 2x}} – \dfrac{{{x^2} + 4x}}{{{x^3} – 4x}} – \dfrac{2}{{{x^2} + 2x}}\\ = \dfrac{x}{{x\left( {x – 2} \right)}} – \dfrac{{{x^2} + 4x}}{{x\left( {x – 2} \right)\left( {x + 2} \right)}} – \dfrac{2}{{x\left( {x + 2} \right)}}\\ = \dfrac{{x\left( {x + 2} \right) – {x^2} – 4x – 2\left( {x – 2} \right)}}{{x\left( {x – 2} \right)\left( {x + 2} \right)}}\\ = \dfrac{{{x^2} + 2x – {x^2} – 4x – 2x + 4}}{{x\left( {x – 2} \right)\left( {x + 2} \right)}}\\ = \dfrac{{ – 4x + 4}}{{x\left( {x – 2} \right)\left( {x + 2} \right)}}\end{array}\) Trả lời
DK:x \ne \left\{ { – 2;0;2} \right\}\\
\dfrac{x}{{{x^2} – 2x}} – \dfrac{{{x^2} + 4x}}{{{x^3} – 4x}} – \dfrac{2}{{{x^2} + 2x}}\\
= \dfrac{x}{{x\left( {x – 2} \right)}} – \dfrac{{{x^2} + 4x}}{{x\left( {x – 2} \right)\left( {x + 2} \right)}} – \dfrac{2}{{x\left( {x + 2} \right)}}\\
= \dfrac{{x\left( {x + 2} \right) – {x^2} – 4x – 2\left( {x – 2} \right)}}{{x\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} + 2x – {x^2} – 4x – 2x + 4}}{{x\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{ – 4x + 4}}{{x\left( {x – 2} \right)\left( {x + 2} \right)}}
\end{array}\)