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\(B=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right).\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(B=\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(B=\frac{-\sqrt{x}-1}{\sqrt{x}}\). Vậy ....
a, \(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}\) (ĐKXĐ: \(x>0\))
\(=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
b, \(\frac{A}{B}=\frac{2+\sqrt{x}}{\sqrt{x}}:\frac{\sqrt{x}+2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(\frac{A}{B}>\frac{3}{2}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{3}{2}>0\)
\(\Leftrightarrow\frac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)
\(\Leftrightarrow2-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với điều kiện \(x>0\)ta có: \(0< x< 4\)
Vậy với \(0< x< 4\)thì \(\frac{A}{B}>\frac{3}{2}\)
=\(\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\):\(\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=\(\left(\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right)\):\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)=\(\frac{-3\left(\sqrt{x}+1\right)}{x-9}\).\(\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=\(\frac{-3}{\sqrt{x}+3}\)
\(\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)^2\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(DKXD:x>0;x\ne1\right)\)
\(\Leftrightarrow\left(\frac{\sqrt{x}.\sqrt{x}-1}{2\sqrt{x}}\right)^2\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\left(2\sqrt{x}\right)^2}\left(\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right)\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{4x}.\frac{\left(\sqrt{x}-1-\sqrt{x}-1\right)\left(\sqrt{x}-1+\sqrt{x}-1\right)}{x-1}\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{4x}.\frac{-2.2\sqrt{x}}{x-1}\)
\(\Leftrightarrow\frac{\left(x-1\right)^2.-4\sqrt{x}}{4x.\left(x-1\right)}\)
\(\Leftrightarrow\frac{x-1}{-\sqrt{x}}\Leftrightarrow\frac{1+x}{\sqrt{x}}\Leftrightarrow\frac{\left(1+x\right).\sqrt{x}}{\sqrt{x}.\sqrt{x}}\Leftrightarrow\frac{\sqrt{x}+x\sqrt{x}}{x}\)
\(VT=\left(\frac{x\sqrt{x}+3\sqrt{3}}{x-\sqrt{3x}+3}-2\sqrt{x}\right)\left(\frac{\sqrt{x}+\sqrt{3}}{3-x}\right)\)
\(=\left(\frac{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3x}+3\right)}{x-\sqrt{3x}+3}-2\sqrt{x}\right)\left(\frac{\sqrt{x}+\sqrt{3}}{\left(\sqrt{3}-\sqrt{x}\right)\left(\sqrt{3}+\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}+\sqrt{3}-2\sqrt{x}\right)\left(\frac{1}{\sqrt{3}-\sqrt{x}}\right)\)
\(=\frac{\sqrt{3}-\sqrt{x}}{\sqrt{3}-\sqrt{x}}=1=VP\left(dpcm\right)\)
a) Đk \(x>0\)và \(x\ne4\)
=\(\left(\frac{\sqrt{x}-2+\sqrt{x}+2}{x-4}\right)\).\(\frac{\sqrt{x}-2}{\sqrt{x}}\)
=\(\frac{2\sqrt{x}}{x-4}\).\(\frac{\sqrt{x}-2}{\sqrt{x}}\)
=\(\frac{2}{\sqrt{x}+2}\)
b) Để \(\frac{2}{\sqrt{x}+2}>\frac{1}{2}\)
\(\Leftrightarrow\frac{4-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)\(>0\)
\(\Leftrightarrow\frac{-\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)\(>0\)
Vì \(2\left(\sqrt{x}+2\right)>0\)
mà\(\frac{-\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)\(>0\)
nên \(-\sqrt{x}+2>0\)\(\Leftrightarrow x< 4\)
Vậy vs \(0< x< 4\)thì \(A>\frac{1}{2}\)
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