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ĐKXĐ: x>=1 và x<>2

\(A=\dfrac{\sqrt{x-1}+\left|\sqrt{x-1}-1\right|+1}{\left|x-2\right|}\)

Trường hợp 1: \(\sqrt{x-1}>1\Leftrightarrow x>2\)

=>\(A=\dfrac{2\sqrt{x-1}}{\left|x-2\right|}\)

Trường hợp 2: 1<x<2

\(A=\dfrac{2}{\left|x-2\right|}\)

20 tháng 8 2016

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=\(\frac{3x-12\sqrt{x}-9}{x-4\sqrt{x}+3}\)

 

 

1 tháng 9 2016

a) ĐKXĐ: \(x\ge0;x\ne9\)

\(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

\(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{x-9}\)

\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{x-9}\)

\(B=\frac{2x-6+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)

\(B=\frac{3x-6+15\sqrt{x}}{x-9}\)

18 tháng 6 2016

ĐK:\(x>0\)

\(C=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}=\frac{\sqrt{x}.\left[\left(\sqrt{x}\right)^3+1\right]}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)

=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-2\sqrt{x}-1\)

\(=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)

a: ĐKXĐ: x>=0; x<>1

\(A=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

b: \(A-\dfrac{2}{3}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2}{3}\)

\(=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}< =0\)

Do đó: A<=2/3

23 tháng 9 2017

toán 8 ạ mình lộn mất TvT