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ĐKXĐ: \(x\ge0\)
\(\frac{1}{\sqrt{x}+1}-\frac{3}{x\sqrt{x}+1}+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
a)\(P=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}-1}-1\right)ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}.\)
\(=\left(\frac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\frac{\sqrt{x}-1}{\sqrt{x}-\sqrt{x}+1}\right)\)
=\(\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b)P=3/2 <=>\(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{3}{2}\Leftrightarrow2\sqrt{x}+1=\frac{3}{2}\sqrt{x}+\frac{3}{2}.\)
\(\Leftrightarrow\frac{1}{2}\sqrt{x}=\frac{1}{2}\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
Với x=1 thoả nãm yêu cầu
M= \(\sqrt{2}+1-\) \(\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}+1-\sqrt{2}+1=2\)
N=\(\sqrt{1+2\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{1+2\left(\sqrt{2}+1\right)}=\) \(\sqrt{1+2\sqrt{2}+2}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
P= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}+\frac{2\sqrt{x}.\sqrt{x}}{\sqrt{x}}\) (dk \(x>0\))
=\(\sqrt{x}+1+2\sqrt{x}=3\sqrt{x}+1\)
Q= \(\sqrt{\left(\sqrt{x}+1\right)^2}+\sqrt{\left(\sqrt{x}-1\right)^2}\) (dk \(x\ge0\) )
=\(\left|\sqrt{x}+1\right|+\left|\sqrt{x}-1\right|\)
th1 \(\sqrt{x}\ge1\Leftrightarrow x\ge1\) Q=\(\sqrt{x}+1+\sqrt{x}-1=2\sqrt{x}\)
th2 \(0\le x< 1\) Q=\(\sqrt{x}+1+1-\sqrt{x}=2\)
a) \(M=\sqrt{2}+1-\sqrt{1,5.2-2.\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2.\left(1,5-\sqrt{2}\right)}\)\(=\sqrt{2}+1-\sqrt{2}.\sqrt{1,5-\sqrt{2}}\)
\(=\sqrt{2}.\left(1+1,5-\sqrt{2}\right)+1=\sqrt{2}.\left(2,5-\sqrt{2}\right)+1\)
\(=\sqrt{2}.2,5-2+1=\sqrt{2}.2,5-1\)
P/s: Theo em thì em nghĩ là đúng '-' Khoảng 90% :)
Sửa lại đề nha , đề đúng nè :
\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{x-1}\right):\)\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{x+2\sqrt{x}+1}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\)\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{x}{\left(\sqrt{x}+1\right)^2}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{x-\sqrt{x}-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{x+\sqrt{x}-x}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\sqrt{x}}=-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(\)
a) \(\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\left(\sqrt{x}-\sqrt{y}\right)}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}-x+2\sqrt{xy}-y\)
\(=3\sqrt{xy}\)
b) \(\frac{x-y}{\sqrt{y}-1}.\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-y}{\sqrt{y}-1}.\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\left(x-y\right)\left(\sqrt{y}-1\right)}{\left(x-1\right)^2}\)
a) \(=\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=x+\sqrt{xy}+y-x+2\sqrt{xy}-y=3\sqrt{xy}\)
Giúp với mình gấp quá
\(=\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x+1}{x-1}\)
\(=\frac{2\sqrt{x}\left(x-1\right)-\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)
\(=\frac{2x\sqrt{x}-2\sqrt{x}-x\sqrt{x}+x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}+1}{x\sqrt{x}-\sqrt{x}-x+1}=1-\frac{x}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)