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Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
\(A=4\sqrt{x}-\frac{\left(\sqrt{x}+3\right)^2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=4\sqrt{x}-\left(\sqrt{x}+3\right)\)
\(=3\sqrt{x}-3\)
\(B=\frac{\sqrt{\left(3x+2\right)^2}}{3x+2}=\frac{|3x+2|}{3x+2}\)
\(TH1:3x+2>0\Rightarrow B=1\)
\(TH2:3x+2< 0\Rightarrow B=-1\)
A <=> 4√x - [ ( (√x )^2 + 2√x3+ 3^2)*( √x -3)]/ (x-9)
<=> 4√x - [(√x+3)^2×(√x-3)]/( x-9)
<=> 4√x - [(√x+3)*(x-9)]/(x-9)
<=> 4√x - √x -3
<=> 3√x -3
b, <=> √[(3*x) ^2+2*3x*2+2^2]/(3x+2)
<=> √[( 3x+2)^2] /(3x+2)
<=> (3x+2)/(3x+2) = 1
Bài làm:
Ta có:
\(P=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{\sqrt{x}-9}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(P=\frac{x-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\left[\frac{\left(9-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(P=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{-x+6\sqrt{x}+27+x-4\sqrt{x}+2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\div\frac{x+2\sqrt{x}+20}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x+2\sqrt{x}+20}\)
\(P=\frac{3\left(\sqrt{x}-2\right)}{x+2\sqrt{x}+20}=\frac{3\sqrt{x}-6}{x+2\sqrt{x}+20}\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
Lời giải:
a) \(A=4\sqrt{x}-\frac{(\sqrt{x}+3)^2(\sqrt{x}-3)}{x-9}=4\sqrt{x}-\frac{(\sqrt{x}+3)(x-9)}{x-9}=4\sqrt{x}-(\sqrt{x}+3)\)
\(=3\sqrt{x}-3\)
b)
\(B=\frac{\sqrt{9x^2+12x+4}}{3x+2}=\frac{\sqrt{(3x)^2+2.3x.2+2^2}}{3x+2}=\frac{\sqrt{(3x+2)^2}}{3x+2}=\frac{|3x+2|}{3x+2}\)
\(B=1\) nếu $x>\frac{-2}{3}$
$B=-1$ nếu $x< \frac{-2}{3}$