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a. ĐK \(x\ge0\)và \(x\ne1\)
A =\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\cdot\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+2\sqrt{x}+1+\sqrt{x}-x-1+\sqrt{x}}\)
\(=\frac{x+1}{4\sqrt{x}}\)
b. Thay \(x=\frac{2-\sqrt{3}}{2}\Rightarrow A=\frac{\frac{2-\sqrt{3}}{2}+1}{4\sqrt{\frac{2-\sqrt{3}}{2}}}=\frac{4-\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\frac{4-\sqrt{3}}{4-4\sqrt{3}}=-\frac{1+3\sqrt{3}}{8}\)
c . Ta có \(A-\frac{1}{2}=\frac{x+1}{4\sqrt{x}}-\frac{1}{2}=\frac{x-2\sqrt{x}+1}{4\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}>0\)với \(\forall x>0\)và \(x\ne1\)
Vậy A >1/2
ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le1\\x\ne0\end{matrix}\right.\)
\(A=\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{\sqrt{1-x}^2}{\sqrt{1-x}\left(\sqrt{1+x}-\sqrt{1-x}\right)}-\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)
\(=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}-\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)
\(=\frac{1+x+1-x+2\sqrt{1-x^2}-\left(1+x+1-x-2\sqrt{1-x^2}\right)}{2x}\)
\(=\frac{2\sqrt{1-x^2}}{x}\)
\(\sqrt{1-x^2}=\sqrt{1-\frac{4+2\sqrt{3}}{8}}=\sqrt{\frac{4-2\sqrt{3}}{8}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\(\Rightarrow A=\frac{\sqrt{3}-1}{\sqrt{2}}.\frac{2\sqrt{2}}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}-1\right)^2}{2}=4-2\sqrt{3}\)
a, x = \(\frac{4\left(\sqrt{3}+1\right)}{3-1}-\frac{4\left(\sqrt{3}-1\right)}{3-1}\)
x = \(\left(2\sqrt{3}+2\right)-\left(2\sqrt{3}-2\right)\)
x = \(2\sqrt{3}+2-2\sqrt{3}+2\)
x = 4 (TMĐK)
=> A = \(\frac{2\sqrt{4}+1}{3\sqrt{4}+1}\)
=> A = \(\frac{5}{7}\)
Vậy x = \(\frac{4}{\sqrt{3}-1}-\frac{4}{\sqrt{3}+1}\) thì A = \(\frac{5}{7}\)
b, B = \(\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)
B = \(\frac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{1}{\sqrt{x}-1}\)
B = \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
c, \(\frac{B}{A}>2\) <=> \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}:\frac{2\sqrt{x}+1}{3\sqrt{x}+1}\) > 2
<=> \(\frac{3\sqrt{x}+1}{\sqrt{x}+1}>2\)
<=> \(\frac{3\sqrt{x}+1}{\sqrt{x}+1}-2>0\)
<=> \(\frac{3\sqrt{x}+1-2\sqrt{x}-2}{\sqrt{x}+1}>0\)
<=> \(\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)
mà \(\sqrt{x}+1>0\) \(\forall\) \(x\in\) ĐKXĐ
=> \(\sqrt{x}-1>0\)
<=> \(\sqrt{x}>1\)
<=> \(x>1\)
Kết hợp ĐKXĐ : x \(\ge0\) ; x \(\ne\) 1
=> x > 1 thì \(\frac{B}{A}>2\)
1) Sửa đề: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Ta có: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
Ta có: \(x=3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\), ta được:
\(A=\frac{-5\cdot\sqrt{\left(\sqrt{2}-1\right)^2}+2}{\sqrt{\left(\sqrt{2}-1\right)^2}+3}\)
\(=\frac{-5\cdot\left(\sqrt{2}-1\right)+2}{\sqrt{2}-1+3}\)
\(=\frac{-5\sqrt{2}+5+2}{\sqrt{2}+2}\)
\(=\frac{-5\sqrt{2}+7}{\sqrt{2}+2}\)
Vậy: Khi \(x=3-2\sqrt{2}\) thì \(A=\frac{-5\sqrt{2}+7}{\sqrt{2}+2}\)
2) Ta có: \(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{\left(x+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}+x+2\sqrt{x}+2+x+x\sqrt{x}-\sqrt{x}-1-\left(2x+2\sqrt{x}+x\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2x+2x\sqrt{x}+\sqrt{x}+1-2x-2\sqrt{x}-x\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(x-1\right)}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
Ta có: \(x=7-2\sqrt{6}\)
\(=6-2\sqrt{6}\cdot1+1\)
\(=\left(\sqrt{6}-1\right)^2\)
Thay \(x=\left(\sqrt{6}-1\right)^2\) vào biểu thức \(B=\frac{\sqrt{x}}{x+\sqrt{x}+1}\), ta được:
\(B=\frac{\sqrt{\left(\sqrt{6}-1\right)^2}}{\left(\sqrt{6}-1\right)^2+\sqrt{\left(\sqrt{6}-1\right)^2}+1}\)
\(=\frac{\sqrt{6}-1}{7-2\sqrt{6}+\sqrt{6}-1+1}\)
\(=\frac{\sqrt{6}-1}{7-\sqrt{6}}\)
Vậy: Khi \(x=7-2\sqrt{6}\) thì \(B=\frac{\sqrt{6}-1}{7-\sqrt{6}}\)
3) Ta có: \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\frac{\sqrt{x}\left(x-3\sqrt{x}-x-9\right)}{\left(\sqrt{x}+3\right)\left(2\sqrt{x}+4\right)}\)
\(=\frac{\sqrt{x}\left(-3\sqrt{x}-9\right)}{\left(\sqrt{x}+3\right)\cdot2\cdot\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(2\sqrt{x}+4\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Ta có: \(x=7-4\sqrt{3}\)
\(=4-2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2-\sqrt{3}\right)^2\)
Thay \(x=\left(2-\sqrt{3}\right)^2\) vào biểu thức \(C=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\), ta được:
\(C=\frac{-3\cdot\sqrt{\left(2-\sqrt{3}\right)^2}}{2\cdot\sqrt{\left(2-\sqrt{3}\right)^2}+4}\)
\(=\frac{-3\cdot\left(2-\sqrt{3}\right)}{2\cdot\left(2-\sqrt{3}\right)+4}\)
\(=\frac{-6+3\sqrt{3}}{4-2\sqrt{3}+4}\)
\(=\frac{-6+3\sqrt{3}}{8-2\sqrt{3}}\)
Vậy: Khi \(x=7-4\sqrt{3}\) thì \(C=\frac{-6+3\sqrt{3}}{8-2\sqrt{3}}\)