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\(\left(\frac{4\sqrt{a}}{\sqrt{a}+2}+\frac{8a}{4-a}\right):\left(\frac{\sqrt{a}-1}{a-2\sqrt{a}}-\frac{2}{\sqrt{a}}\right)\) (ĐKXĐ : \(a>0;a\ne4;a\ne9\))
\(=\left[\frac{4\sqrt{a}\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\frac{8a}{a-4}\right]:\left[\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-2\right)}-\frac{2\left(\sqrt{a}-2\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}\right]\)
\(=\frac{4a-8\sqrt{a}-8a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}:\frac{\sqrt{a}-1-2\sqrt{a}+4}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{-4\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}:\frac{-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{-4\sqrt{a}}{\sqrt{a}-2}.\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{3-\sqrt{a}}=-\frac{4a}{3-\sqrt{a}}\)
\(A=\left(\dfrac{6x+4}{3\sqrt{3x^3}-8}-\dfrac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right).\left(\dfrac{1+3\sqrt{3x^3}}{1+\sqrt{3x}}-\sqrt{3x}\right)\)
Điều kiện tự làm nha:
Đặt \(\sqrt{3x}=a\) thì ta có:
\(A=\left(\dfrac{2a^2+4}{a^3-8}-\dfrac{a}{a^2+2a+4}\right).\left(\dfrac{1+a^3}{1+a}-a\right)\)
\(=\left(\dfrac{2a^2+4}{\left(a-2\right)\left(a^2+2a+4\right)}-\dfrac{a}{a^2+2a+4}\right).\left(\dfrac{\left(1+a\right)\left(1-a+a^2\right)}{1+a}-a\right)\)
\(=\dfrac{a^2+2a+4}{\left(a-2\right)\left(a^2+2a+4\right)}.\left(1-2a+a^2\right)\)
\(=\dfrac{\left(a-1\right)^2}{a-2}=\dfrac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}\)
Sửa đề; \(D=\left(\dfrac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}-2\sqrt{y}}-\dfrac{2\sqrt{xy}}{x-y}\right)\cdot\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(D=\dfrac{x+2\sqrt{xy}+y-4\sqrt{xy}}{2\left(x-y\right)}\cdot\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}\cdot\dfrac{\sqrt{x}}{x-y}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(A=\left(\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right)\left(ĐK:x\ge0;\ne1\right)\)
\(=\left[\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right]\)
\(=\frac{3\left(\sqrt{x}+1\right)-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)
\(=\frac{2\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}=\frac{2\left(\sqrt{x}+3\right)}{\sqrt{x}+1}\)
ĐKXĐ: x>=1 và x<>2
\(A=\dfrac{\sqrt{x-1}+\left|\sqrt{x-1}-1\right|+1}{\left|x-2\right|}\)
Trường hợp 1: \(\sqrt{x-1}>1\Leftrightarrow x>2\)
=>\(A=\dfrac{2\sqrt{x-1}}{\left|x-2\right|}\)
Trường hợp 2: 1<x<2
\(A=\dfrac{2}{\left|x-2\right|}\)
\(=\dfrac{a\sqrt{a}-3-2\left(a-6\sqrt{a}+9\right)-a-4\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{a-1}{a+8}\)
\(=\dfrac{a\sqrt{a}-a-4\sqrt{a}-6-2a+12\sqrt{a}-18}{\left(\sqrt{a}-3\right)}\cdot\dfrac{\sqrt{a}-1}{a+8}\)
\(=\dfrac{a\sqrt{a}-3a+8\sqrt{a}-24}{\left(\sqrt{a}-3\right)}\cdot\dfrac{\sqrt{a}-1}{a+8}=\sqrt{a}-1\)
Ta có \(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\left(\sqrt{10}-\sqrt{2}\right)\)
= \(2\sqrt{4+\sqrt{\sqrt{5}^2-2\sqrt{5}.1+1}}\sqrt{2}\left(\sqrt{5}-1\right)\)
= \(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\sqrt{2}\left(\sqrt{5}-1\right)\)
= \(\sqrt{2}\sqrt{4+\sqrt{5}-1}.\left(\sqrt{5}-1\right)2\)
= \(\sqrt{2\left(3+\sqrt{5}\right)}\left(\sqrt{5}-1\right)2\)
= \(\sqrt{6+2\sqrt{5}}\left(\sqrt{5}-1\right)2\)
= \(\sqrt{\left(\sqrt{5}+1\right)^2}\left(\sqrt{5}-1\right)2\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)2\)
= \(\left(\sqrt{5}^2-1\right)2\)
= 4.2
= 8
Chúc bạn làm bài tốt :)