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a) A= (\(\left(\frac{1+\sqrt{x}}{1+\sqrt{x}}-\frac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x-2}\right)}+\frac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right)\)
A=\(\left(\frac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}\right)\)
A= \(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
a: \(A=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
b: Để A<0 thì \(\sqrt{x}-2< 0\)
hay 0<x<4
\(ĐKXĐ:x\ge0;x\ne1\)
\(B=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}\)
\(B=\frac{1}{2\left(\sqrt{x}-1\right)}-\frac{1}{2\left(\sqrt{x}+1\right)}+\frac{4\sqrt{x}}{2\left(\sqrt{x}+1\right).2\left(\sqrt{x}-1\right)}\)
\(B=\frac{2\sqrt{x}+2-2\sqrt{x}+2+4\sqrt{x}}{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{4\sqrt{x}+4}{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{4\left(\sqrt{x}+1\right)}{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)
là \(\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}nha toi bi nham\)
ĐKXĐ : \(x\ne\pm1\)
a/ \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{2}{x^2-1}-\frac{x}{x-1}+\frac{1}{x+1}\right)\)
\(=\frac{x^2+2x+1-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}:\frac{2-x\left(x+1\right)+\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{1-x^2}=\frac{4x}{1-x^2}\)
b/ Ta có \(3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\Rightarrow\sqrt{3+\sqrt{8}}=\sqrt{2}+1\)
Suy ra : Nếu x = \(\sqrt{2}+1\) thì \(A=\frac{4\left(\sqrt{2}+1\right)}{1-\left(\sqrt{2}+1\right)^2}=\frac{4\left(\sqrt{2}+1\right)}{-\sqrt{2}.\sqrt{2}\left(\sqrt{2}+1\right)}=-\frac{4}{2}=-2\)
c/ \(A=\sqrt{5}\Rightarrow4x=\sqrt{5}\left(1-x^2\right)\Leftrightarrow\sqrt{5}x^2+4x-\sqrt{5}=0\)
Nhân cả hai vế của pt trên với \(\sqrt{5}\ne0\)
Được \(5x^2+4\sqrt{5}x-5=0\) . Đặt \(t=x\sqrt{5}\) pt trở thành \(t^2+4t-5=0\Leftrightarrow\left(t+5\right)\left(t-1\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}t=1\\t=-5\end{array}\right.\)
Với t = 1 thì \(x=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}\)
Với t = -5 thì \(x=-\frac{5}{\sqrt{5}}=-\sqrt{5}\)
\(A=\left[\frac{x^2+2x+1-x^2+2x-1}{x^2-1}\right]:\left[\frac{2-x^2-x+x-1}{x^2-1}\right]=\left[\frac{4x}{x^2-1}\right].\left[\frac{x^2-1}{1-x^2}\right]=\frac{4x}{1-x^2}\)
ĐKXĐ:...
\(P=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\sqrt{x}-1}\right)\frac{\left(\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x+\sqrt{x}+1}-\sqrt{x}\right)\frac{1}{\left(2\sqrt{x}-1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\frac{2x\sqrt{x}+x-\sqrt{x}-x\sqrt{x}-x-\sqrt{x}}{x+\sqrt{x}+1}\right)\frac{1}{\left(2\sqrt{x}-1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\frac{x\sqrt{x}-2\sqrt{x}}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\frac{x\sqrt{x}-2\sqrt{x}+x\sqrt{x}+x+\sqrt{x}}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}=\frac{2x\sqrt{x}+x-\sqrt{x}}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(=\frac{\left(x+\sqrt{x}\right)\left(2\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}\)