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ĐK: \(-1\le x\le1\)\(;\)\(x\ne0\)
\(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\sqrt{2}\)
\(\Leftrightarrow\)\(\frac{\left(\sqrt{1+x}+\sqrt{1-x}\right)^2}{\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(\sqrt{1+x}+\sqrt{1-x}\right)}=\sqrt{2}\)
\(\Leftrightarrow\)\(\frac{1+x+1-x+2\sqrt{\left(1+x\right)\left(1-x\right)}}{1+x-1+x}=\sqrt{2}\)
\(\Leftrightarrow\)\(\sqrt{1-x^2}=\sqrt{2}x-1\)
\(\Leftrightarrow\)\(1-x^2=2x^2-2\sqrt{2}x+1\)
\(\Leftrightarrow\)\(x^2-\frac{2\sqrt{2}}{3}x=0\)
\(\Leftrightarrow\)\(x\left(x-\frac{2\sqrt{2}}{3}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\left(l\right)\\x=\frac{2\sqrt{2}}{3}\left(tm\right)\end{cases}}\)
\(\frac{x-1}{x+1}=\frac{\frac{2\sqrt{2}}{3}-1}{\frac{2\sqrt{2}}{3}+1}=\frac{\frac{2\sqrt{2}-3}{3}}{\frac{2\sqrt{2}+3}{3}}=\frac{2\sqrt{2}-3}{2\sqrt{2}+3}=12\sqrt{2}-17\) ( giống như tìm x ở trên )
\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-x}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(A=\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)\(\div\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(A=\left(\frac{x+2\sqrt{x}+1+x-\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{2x+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}}\)
\(A=\frac{2x+1}{4\sqrt{x}}\)
c, \(A=\frac{2x+1}{4\sqrt{x}}=\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\)
ap dụng cô si ta có \(\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\ge2\sqrt{\frac{\sqrt{x}}{2}\cdot\frac{1}{4\sqrt{x}}}=\frac{\sqrt{2}}{2}\)
dấu = xảy ra khi \(\frac{\sqrt{x}}{2}=\frac{1}{4\sqrt{x}}\Leftrightarrow x=\frac{1}{2}\) (tm)
a, ĐK \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(P=\frac{x-1}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
Ta thấy \(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}>0\forall x>0,x\ne1\)
b, P=\(\frac{x+2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\frac{2}{2+\sqrt{3}}+2\sqrt{\frac{2}{2+\sqrt{3}}}+1}{\sqrt{\frac{2}{2+\sqrt{3}}}-1}\)
=\(\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\sqrt{\left(\frac{2}{\left(\sqrt{3}+1\right)^2}\right)}+1}{\sqrt{\left(\frac{2}{2+\sqrt{3}}\right)^2}-1}=\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\frac{2}{\sqrt{3}+1}+1}{\frac{2}{\sqrt{3}+1}-1}\)
\(=\frac{12+6\sqrt{3}}{1-3}=-6-3\sqrt{3}\)
\(E=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-x}{x-\sqrt{x}}\right)\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\) \(\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\sqrt{x}}+\frac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\)\(\left[\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(E=\frac{x}{\sqrt{x}-1}\)
b) \(E>1\Leftrightarrow\frac{x}{\sqrt{x}-1}>1\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-1>0\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{x-2\sqrt{x}+1+\sqrt{x}}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-1\right)^2+\sqrt{x}}{\sqrt{x}-1}>0\)
\(\Rightarrow\sqrt{x}-1>0\) vì tử của phân số luôn \(\ge0\forall x\ge0\)
\(\Rightarrow x>1\)
kết hợp với ĐKXĐ \(x\ge0\Rightarrow x>1\)
vậy \(x>1\) thì \(E>1\)