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1: ĐKXĐ: -2/2x-2>=0
=>2x-2<0
=>x<1
2: ĐKXĐ: 2/3x-1>=0
=>3x-1>0
=>x>1/3
3: ĐKXĐ: 2x-2/(-2)>=0
=>2x-2<=0
=>x<=1
4: ĐKXĐ: (3x-2)/5>=0
=>3x-2>=0
=>x>=2/3
5: ĐKXĐ: (x-2)/(x+3)>=0
=>x>=2 hoặc x<-3
\(A=\dfrac{4x\sqrt{x}+3x+9+x-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}:\dfrac{x+2\sqrt{x}-4\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4x\sqrt{x}+4x}{x-2\sqrt{x}-3}=\dfrac{4x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{4x}{\sqrt{x}-3}\)
sửa đề:
\(M=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)
ĐKXĐ: \(x\ge0\);\(x\ne\pm1\)
\(M=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\\M=\left[\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right].\dfrac{\sqrt{x}}{x+\sqrt{x}+2} \) \(M=\dfrac{x+\sqrt{x}+2}{x-1}.\dfrac{\sqrt{x}}{x+\sqrt{x}+2}=\dfrac{\sqrt{x}}{x-1}\)
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
ĐKXĐ: \(x>0,x\ne4\)
B= \(\left(x-\sqrt{x}-2\right).\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{4-\sqrt{x}}{x-2\sqrt{x}}\right)\)
= \(\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right).\dfrac{3\sqrt{x}-4+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{4\left(x-1\right)}{\sqrt{x}}\)
ĐKXĐ: \(x\ge0\)
\(K=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
\(K=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right]:\left[\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right]-1\)
\(K=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}:\dfrac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-1\)
\(K=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-2\sqrt{x}+1}-1\\K=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}-1\\ K=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-1 \)
\(K=\dfrac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{x+2}{\sqrt{x}-1}\)
a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b:Sửa đề: 2A
2A=2căn x+5
=>(2căn x+2)/căn x=2căn x+5
=>2x+5căn x-2căn x-2=0
=>2x+3căn x-2=0
=>(căn x+2)(2căn x-1)=0
=>x=1/4
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x+1\ge0\\x-2>0\\x+2>0\\x\ge0\end{matrix}\right.\) và \(4-x\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x>2\\x>-2\\x\ge0\end{matrix}\right.\) và \(x\ne4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x\ne4\end{matrix}\right.\)