\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\right)\) (ĐK: \(x>0;x\ne1;x\ne9\))

\(=\left[\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\right]\)

\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1+x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{2x-10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\)

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

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

\(=\dfrac{\sqrt{x}-3}{2x\sqrt{x}-10\sqrt{x}}\)

\(A>0\) khi

\(\dfrac{\sqrt{x}-3}{2x\sqrt{x}-10\sqrt{x}}>0\)

TH1: 

\(\sqrt{x}-3>0\) và \(2x\sqrt{x}-10\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}>3\) và \(2\sqrt{x}\left(x-5\right)>0\)

\(\Leftrightarrow x>9\) và \(x>5\)

\(\Leftrightarrow x>9\)

TH2: 

\(\sqrt{x}-3< 0\) và \(2x\sqrt{x}-10\sqrt{x}< 0\)

\(\Leftrightarrow\sqrt{x}< 3\) và \(2\sqrt{x}\left(x-5\right)< 0\)

\(\Leftrightarrow x< 9\) và \(x< 5\)

\(\Leftrightarrow x< 5\)

Vậy A > 0 khi \(\left[{}\begin{matrix}x>9\\x< 5\end{matrix}\right.\) 

Ta có: 

\(x=3-2\sqrt{2}=\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2=\left(\sqrt{2}-1\right)^2\)

\(A=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-3}{2\cdot\left(\sqrt{2}-1\right)^2\cdot\sqrt{\left(\sqrt{2}-1\right)^2}-10\cdot\sqrt{\left(\sqrt{2}-1\right)^2}}\)

\(A=\dfrac{\left|\sqrt{2}-1\right|-3}{2\cdot\left(3-2\sqrt{2}\right)\cdot\left|\sqrt{2}-1\right|-10\cdot\left|\sqrt{2}-1\right|}\)

\(A=\dfrac{\sqrt{2}-1-3}{\left(6-4\sqrt{2}\right)\left(\sqrt{2}-1\right)-10\left(\sqrt{2}-1\right)}\)

\(A=\dfrac{\sqrt{2}-4}{6\sqrt{2}-6-8+4\sqrt{2}-10\sqrt{2}+10}\)

\(A=\dfrac{\sqrt{2}-4}{-4}\)

\(A=\dfrac{4-\sqrt{2}}{4}\)

a) Điều kiện: \(x\ne\pm1\)

 \(B=\frac{x-1}{x+1}-\frac{x+1}{x-1}-\frac{4}{1-x^2}\)

\(B=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}-\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{-4}{\left(x-1\right).\left(x+1\right)}\)

\(B=\frac{x^2-x-x+1-x^2-x-x-1+4}{\left(x-1\right).\left(x+1\right)}\)

\(B=\frac{-4x+4}{\left(x-1\right).\left(x+1\right)}=\frac{-4.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}=\frac{-4}{x+1}\)

b) \(x^2-x=0\Leftrightarrow x.\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Khi  \(x=0\Leftrightarrow\frac{-4}{0-1}=\frac{-4}{-1}=4\)

Khi \(x=1\Leftrightarrow\frac{-4}{1-1}=0\)

c) \(\frac{-4}{x+1}=-3\Leftrightarrow-3.\left(x+1\right)=-4\Leftrightarrow x+1=\frac{4}{3}\Leftrightarrow x=\frac{1}{3}\)

1: \(B=\dfrac{6x+x^2-3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x^2+3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x}{x-3}\)

Bài 1:

a: Sửa đề \(x^3y-2x^2y+xy\)

\(=y\left(x^3-2x^2+x\right)\)

\(=x\cdot y\cdot\left(x^2-2x+1\right)\)

\(=xy\left(x-1\right)^2\)

b: Sửa đề: \(x^2-9-2xy+y^2\)

\(=\left(x^2-2xy+y^2\right)-9\)

\(=\left(x-y\right)^2-9\)

\(=\left(x-y-3\right)\left(x-y+3\right)\)

Bài 2:

a: ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)

b: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)

\(=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{2x+6-x-5}{x+3}\)

\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)

\(=\dfrac{x^2-3x-2x-6-x^2+1}{x-3}\cdot\dfrac{1}{x+1}\)

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

c: \(x^2-x-2=0\)

=>\(\left(x-2\right)\left(x+1\right)=0\)

=>\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)

Thay x=2 vào A, ta được:

\(A=\dfrac{-5}{2-3}=\dfrac{-5}{-1}=5\)

mình không biết làm:)

a: \(B=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-2x}{x-9}=\dfrac{x-3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

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

Để |P|>P thì P<0

=>căn x-2<0

=>0<x<4

=>x=1

\(a,ĐK:x\ne3;x\ge1\\ b,A=\dfrac{\left(\sqrt{x-1}+\sqrt{2}\right)\left(\sqrt{x-1}-\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x-1}+\sqrt{2}\\ b,A=4\left(2-\sqrt{3}\right)\\ \Leftrightarrow\sqrt{x-1}+\sqrt{2}=8-4\sqrt{3}\\ \Leftrightarrow\sqrt{x-1}=8-4\sqrt{3}-\sqrt{2}\\ \Leftrightarrow x-1=\left(8-4\sqrt{3}-\sqrt{2}\right)^2\\ \Leftrightarrow x=\left(8-4\sqrt{3}-\sqrt{2}\right)^2+1=...\\ d,A=\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\\ A_{min}=\sqrt{2}\Leftrightarrow x-1=0\Leftrightarrow x=1\)

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