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1. b) \(\left(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\left(x\sqrt{\dfrac{6x}{x^2}}+\sqrt{\dfrac{6x}{9}}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\left(\sqrt{6x}+\dfrac{1}{3}\sqrt{6x}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\dfrac{7}{3}\sqrt{6x}:\sqrt{6x}=\dfrac{7}{3}\)
2.
P=\(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)(bn có ghi sai đề ko)
a) ĐKXĐ : \(x\ge1,x\ge2,x\ge0\)
b) P=\(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{x-3\sqrt{x}-\sqrt{x}+3-2x+\sqrt{x}+4\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-2}\)
c) thay x= \(4-2\sqrt{3}\)vào P ta có :
\(\dfrac{1}{\sqrt{4-2\sqrt{3}}-2}=\dfrac{1}{\sqrt{3}-1-2}=\dfrac{1}{\sqrt{3}-3}\)
Bài 3:
a: \(=\left(4\sqrt{2}-6\sqrt{2}\right)\cdot\dfrac{\sqrt{2}}{2}=-2\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=-2\)
b: \(=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-2\left(\sqrt{6}-1\right)\)
\(=\sqrt{6}-2\sqrt{6}+2=2-\sqrt{6}\)
\(\dfrac{\sqrt{x-1}}{x^2}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ge0\\x^2\ne0\end{matrix}\right.\Leftrightarrow x\ge1\)
\(\sqrt{\dfrac{x}{\left(x-1\right)^2}}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x-1\ne0\end{matrix}\right.\) \(\Leftrightarrow x\ge0\)
\(\sqrt{x+5}-\sqrt{2x+1}\)
ĐKXĐ:\(\left\{{}\begin{matrix}x+5\ge0\\2x+1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x\ge\dfrac{-1}{2}\end{matrix}\right.\)\(\Leftrightarrow x\ge\dfrac{-1}{2}\)
\(\sqrt{3-x^2}\)
ĐKXĐ: \(3-x^2\ge0\Leftrightarrow x\le\pm\sqrt{3}\)
\(G=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=-\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b: \(P=\left(\sqrt{x}+\sqrt{x-1}-\left(\sqrt{x-1}+\sqrt{2}\right)\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(=-\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}}\)
c: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=-\dfrac{\sqrt{2}-1-\sqrt{2}}{\sqrt{2}-1}=\dfrac{1}{\sqrt{2}-1}=\sqrt{2}+1\)
* \(\sqrt{\dfrac{x^2}{2}}\) . ĐKXĐ: mọi x
* \(\sqrt{\dfrac{x-1}{-2}}\) . ĐKXĐ: \(\dfrac{x-1}{-2}\ge0\Leftrightarrow x-1\le0\) (vì -2<0) <=> x \(\le\) 1
* \(\sqrt{x^2-4}\) . ĐKXĐ: \(x^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
* \(\sqrt{\dfrac{x-1}{2x^2}}\) .ĐKXĐ: \(\dfrac{x-1}{2x^2}\ge0\Leftrightarrow x-1\ge0\)(vì 2x^2 > 0 với mọi x) <=> x \(\ge\) 1