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PT
1
H9
HT.Phong (9A5)
CTVHS
31 tháng 10 2023
\(A=\left(\dfrac{1}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1-\sqrt{x}}{x+4\sqrt{x}+4}\left(x>0;x\ne1\right)\)
\(A=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right]:\dfrac{1-\sqrt{x}}{\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot2+2^2}\)
\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{1-\sqrt{x}}{\left(\sqrt{x}+2\right)^2}\)
\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}\)
\(A=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
Vậy: ...
19 tháng 1 2022
Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
11 tháng 8 2021
a: Ta có: \(M=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1\)
\(=a-\sqrt{a}\)
ta có
\(A=\left(\frac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)=-\frac{6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}=-\frac{1}{x-2}\)
Với \(x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}=-\frac{6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}=-\frac{1}{x-2}=\frac{1}{2-x}\)