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a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
a/ ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x\ge0\end{cases}}\)
\(A=\left[\frac{1}{\sqrt{x}-1}+\frac{1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\left[\frac{2\left(\sqrt{x}-1\right)-\sqrt{x}+4}{\sqrt{x}-1}\right]\)
\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}+1}\)
b/
Ta có: \(A=\frac{1}{\sqrt{x}+1}\ge1\)
Vậy Min A = 1 .Dấu "=" xảy ra khi x = 0
a , rút gọn : A= \(\left(\frac{1}{\sqrt{x}+1}+\frac{1}{x-1}\right):\left(2-\frac{\sqrt{x}-4}{\sqrt{x}-1}\right)\)
A= \(\left(\frac{1\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\frac{\sqrt{x}-4}{\sqrt{x}-1}\right)\)
A= \(\left(\frac{\sqrt{x}+1+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+4}{\sqrt{x}-1}\right)\)
A= \(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
A=\(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
A = \(\frac{1}{\sqrt{x}+1}\)
các pro giúp em TvT
\(A=\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}:\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(ĐKXĐ:x\ge0;x\ne1\right)\)
\(< =>A=\frac{1}{x-\sqrt{x}}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\frac{1}{x-\sqrt{x}}+\sqrt{x}\)
\(< =>A=\frac{1+\sqrt{x}\left(x-\sqrt{x}\right)}{x-\sqrt{x}}=\frac{1+x\sqrt{x}-x}{x-\sqrt{x}}\)
Với \(x=\frac{18}{4+\sqrt{7}}\)thì \(A=\frac{1+\frac{18}{4+\sqrt{7}}.\sqrt{\frac{18}{4+\sqrt{7}}}-\frac{18}{4+\sqrt{7}}}{\frac{18}{4+\sqrt{7}}-\sqrt{\frac{18}{4+\sqrt{7}}}}\)
\(=\frac{1}{18+\frac{4}{7}-\sqrt{18+\frac{4}{7}}}+\sqrt{18+4\sqrt{7}}\)
Em mới lớp 7 nên chỉ làm được thế thôi ạ :3