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#)Giải :
a) \(A=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\frac{x-1}{2\sqrt{x}}\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x-1}{2\sqrt{x}}.\frac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\frac{-4}{2\sqrt{x}}=-2\sqrt{x}\)
1,
\(D=\frac{1}{\sqrt{h+2\sqrt{h-1}}}+\frac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\frac{1}{\sqrt{h-1+2\sqrt{h-1}+1}}+\frac{1}{\sqrt{h-1-2\sqrt{h-1}+1}}\)
\(=\frac{1}{\sqrt{h-1}+1}+\frac{1}{\sqrt{h-1}-1}\)
\(=\frac{\sqrt{h-1}-1+\sqrt{h-1}+1}{h-1-1}\)
\(=\frac{2\sqrt{h-1}}{h-2}\)
Thay \(h=3\)vào D ta có:
\(D=\frac{2\sqrt{3-1}}{3-2}=2\sqrt{2}\)
Vậy với \(h=3\)thì \(D=2\sqrt{2}\)
2,
a, \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)(ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\left(TM\right)\)
Vậy PT có nghiệm là \(x=2\)
b, \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)(ĐK: \(-\sqrt{2}\le x\le\sqrt{2}\))
\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}=-3\)
\(\Leftrightarrow0=-3\)(vô lí)
Vậy PT đã cho vô nghiệm.
\(1,\\ a,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{2x+1}=\dfrac{2}{3}\Leftrightarrow2x+1=\dfrac{4}{9}\Leftrightarrow x=-\dfrac{5}{18}\left(tm\right)\\ b,PT\Leftrightarrow\left|x-3\right|=2\Leftrightarrow\left[{}\begin{matrix}x-3=2\\3-x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\\ 2,\\ a,=\left|5-x\right|=x-5\\ b,=\sqrt{4a\cdot44a}=\sqrt{176a^2}=4\left|a\right|\sqrt{11}=4a\sqrt{11}\\ c,=\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)