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a: ĐKXĐ: x>=0; x<>1
\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)
\(=\frac{3+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\left(\sqrt{x}+1\right)\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
b: Thay \(x=3+2\sqrt2=\left(\sqrt2+1\right)^2\) vào P, ta được:
\(P=\frac{\sqrt{\left(\sqrt2+1\right)^2}+2}{\sqrt{\left(\sqrt2+1\right)^2}-1}=\frac{\sqrt2+1+2}{\sqrt2+1-1}=\frac{3+\sqrt2}{\sqrt2}=\frac{3\sqrt2+2}{2}\)
c: P<0
=>\(\frac{\sqrt{x}+2}{\sqrt{x}-1}<0\)
=>\(\sqrt{x}-1<0\)
=>\(\sqrt{x}<1\)
=>0<=x<1
a, \(A=\left(\frac{1}{1-\sqrt{x}}+\frac{1}{1+\sqrt{x}}\right):\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right)+\frac{1}{1-\sqrt{x}}\)ĐK : \(x>0;x\ne1\)
\(=\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{1-x}\right):\left(\frac{1+\sqrt{x}-1+\sqrt{x}}{1-x}\right)+\frac{1}{1-\sqrt{x}}\)
\(=\frac{2}{1-x}.\frac{1-x}{2\sqrt{x}}+\frac{1}{1-\sqrt{x}}=\frac{1}{\sqrt{x}}+\frac{1}{1-\sqrt{x}}=\frac{1-\sqrt{x}+\sqrt{x}}{-x+\sqrt{x}}=\frac{1}{\sqrt{x}-x}\)
b, Ta có : \(x=7+4\sqrt{3}=7+2.2\sqrt{3}=\left(\sqrt{4}+\sqrt{3}\right)^2\)
\(A=\frac{1}{\sqrt{4}+\sqrt{3}-7+4\sqrt{3}}\)
a:
ĐKXĐ: x>=0; x<>1
\(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b: P=-1
=>\(\sqrt{x}-1=-\sqrt{x}-1\)
=>\(2\sqrt{x}=0\)
=>\(\sqrt{x}=0\)
=>x=0(nhận)
c: Để P là số nguyên thì \(\sqrt{x}-1\) ⋮\(\sqrt{x}+1\)
=>\(\sqrt{x}+1-2\) ⋮\(\sqrt{x}+1\)
=>-2⋮\(\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\lbrace1;2\right\rbrace\)
=>\(\sqrt{x}\in\left\lbrace0;1\right\rbrace\)
=>x∈{0;1}
Kết hợp ĐKXĐ, ta được: x=0