Rút gon
\(J=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right):\left(1-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(K=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{1+\sqrt{x}}+\frac{2}{x-1}\right)\)
\(I=\left(\frac{\sqrt{a}-2}{a-1}-\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}\right).\left(1+\frac{1}{\sqrt{a}}\right)\)
Giúp mk với các cao nhân ơi,mk xin cảm on nhiều. Mk sẽ tick cho.
a/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(J=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right):\left(1-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}+1-\sqrt{x}+1\right)\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{2}{\sqrt{x}+1}\)
\(=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{2}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}-1}\)
Vậy...
b/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(K=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{1+\sqrt{x}}+\frac{2}{x-1}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x-1}}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{x-1}{\sqrt{x}}\)
Vậy...
c/ Tương tự
