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a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
\(A=\sqrt{\dfrac{18-3\sqrt{3}}{11}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{6}+\sqrt{2}}{2}\)
\(=\dfrac{2\sqrt{11\left(18-3\sqrt{3}\right)}-11\sqrt{6}-11\sqrt{2}}{22}\)
b: \(=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-4x-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}\)
1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)
\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)
\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)
Ta có: \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\dfrac{8\sqrt{x}-8x+8x}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(=\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
ta có : \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
=\(\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{4-x}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-x\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
=\(\dfrac{8\sqrt{x}-4x+8x}{4-x}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\dfrac{8\sqrt{x}+4x}{4-x}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x-2}\right)}\) =\(\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\dfrac{4\sqrt{x}}{2-\sqrt{x}}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\) =\(\dfrac{4x\left(\sqrt{x}-2\right)}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)
=\(-\dfrac{4x\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\) =\(-\dfrac{4x}{3-\sqrt{x}}\) =\(\dfrac{4x}{\sqrt{x}-3}\)
này mới đúng !!
a) \(M=\dfrac{x+\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}:\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(x+1\right)}.\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) \(x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2+\sqrt{3}+2-\sqrt{3}=4\)
\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{4}+1}{\sqrt{4}-1}=\dfrac{2+1}{2-1}=3\)
Câu 1: \(\sqrt{8}\) − \(\sqrt{18}\) + \(2\sqrt{32}\) = \(\sqrt{4\text{×}2}\) − \(\sqrt{\text{9×2}}\) + 2\(\sqrt{\text{16×2}}\)
=2\(\sqrt{2}\) − 3\(\sqrt{2}\) + 2×4\(\sqrt{2}\)
=(2− 3+ 8)\(\sqrt{2}\)
=7\(\sqrt{2}\)
Câu 2: Mik ko chắc làm đúng hay ko
nên ko làm
hảo hán