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\(\frac{a-2}{a+2}\) - \(\frac{a+2}{a-2}\) +
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. a: \(A=\left(x^2+x+1-x\right):\dfrac{1-x^2}{\left(1-x\right)-x^2\left(1-x\right)}\) \(=\left(x^2+1\right)\cdot\left(1-x\right)\) b: Để A<0 thì 1-x<0 =>x>1 c: |x-4|=5 =>x-4=5 hoặc x-4=-5 =>x=9(nhận) hoặc x=-1(loại) Thay x=9 vào A, ta được: \(A=\left(9^2+1\right)\left(1-9\right)=82\cdot\left(-8\right)=-656\) a: ĐKXĐ: \(a\notin\left\{0;1;-1\right\}\) \(A=\dfrac{a^2}{\left(a-1\right)\left(a+1\right)}-\dfrac{a^2}{a^2+1}\cdot\dfrac{a^2+1}{a\left(a+1\right)}\) \(=\dfrac{a^2}{\left(a-1\right)\left(a+1\right)}-\dfrac{a}{a+1}\) \(=\dfrac{a^2-a^2+a}{\left(a-1\right)\left(a+1\right)}=\dfrac{a}{\left(a-1\right)\left(a+1\right)}=\dfrac{a}{a^2-1}\) b: Để A=3 thì \(3a^2-3=a\) \(\Leftrightarrow2a^2=3\) hay \(a\in\left\{\dfrac{\sqrt{6}}{2};-\dfrac{\sqrt{6}}{2}\right\}\) a)ĐKXĐ:x>=0;x khác 9 A=[\(\frac{\sqrt{x}}{\sqrt{x}-3}\) - \(\frac{3\sqrt{x}+9}{x-9}\)+ \(\frac{2\sqrt{x}}{\sqrt{x}+3}\)] \(\div\) [\(\frac{2\sqrt{x}-2}{\sqrt{x}-3}\)-1] A=[\(\frac{\sqrt{x}\left(\sqrt{x}-3\right)-3\sqrt{x}-9+2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}\)] \(\div\) [\(\frac{\left(2\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-x+9}{x-9}\)] A=[\(\frac{3x-12\sqrt{x}-9}{x-9}\)].[\(\frac{x-9}{x-4\sqrt{x}+3}\)] A=\(\frac{3x-12\sqrt{x}-9}{x-4\sqrt{x}+3}\) a) ĐKXĐ: \(x\ge0;x\ne9\) \(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\) \(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{x-9}\) \(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{x-9}\) \(B=\frac{2x-6+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\) \(B=\frac{3x-6+15\sqrt{x}}{x-9}\) a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\) b: \(M=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\dfrac{x^2-4+10-x^2}{x+2}\) \(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\) \(=\dfrac{-1}{x-2}\) d: Để M nguyên thì \(x-2\in\left\{1;-1\right\}\) hay \(x\in\left\{3;1\right\}\)
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