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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 \(x\in\left\{3;1\right\}\)
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: \(A=\left(x^2+x+1-x\right):\dfrac{1-x^2}{\left(1-x\right)-x^2\left(1-x\right)}\)
\(=\left(x^2+1\right)\cdot\left(1-x\right)\)
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:
\(A=\left(9^2+1\right)\left(1-9\right)=82\cdot\left(-8\right)=-656\)
b: Để N là số nguyên dương thì \(\sqrt{x}-3>0\)
\(\Leftrightarrow x>9\)
mà x là số nguyên
nên \(\left\{{}\begin{matrix}x\in Z\\x>9\end{matrix}\right.\)
\(M=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\right)\left(x^4-\frac{\left(x^2-1\right)\left(x^2+1\right)}{x^2+1}\right)\)
\(=\left(\frac{x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\right)\left(x^4-x^2+1\right)\)
\(=\frac{x^2-2}{x^2+1}\)
b/ \(M=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
Do \(x^2+1\ge1\Rightarrow\frac{3}{x^2+1}\le3\Rightarrow1-\frac{3}{x^2+1}\ge1-3=-2\)
\(\Rightarrow M_{min}=-2\) khi \(x=0\)