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8 tháng 10 2017

\(=\left(\frac{x-2\sqrt{x}-1}{x-4}-\frac{x-4}{x-4}\right):\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right]\)

\(=\frac{x-2\sqrt{x}-1-x+4}{x-4}:\left[\frac{\sqrt{x}-2}{\sqrt{x}+3}+\frac{\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right]\)

\(=\frac{3-2\sqrt{x}}{x-4}:\frac{\left(x-4\right)\left(\sqrt{x}-3\right)+\left(x-4\right)\left(\sqrt{x}+3\right)-\left(x-9\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)\left(\sqrt{x}+2\right)}\)

bạn làm tiếp nha! làm bằng máy tính phức tạp lắm

3 tháng 10 2020

a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)

\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)

\(=-2+2\sqrt{5}-\sqrt{5}\)

\(=-2+\sqrt{5}\)

3 tháng 10 2020

b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)

\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)

\(=\frac{27\sqrt{2}}{4}\cdot8\)

\(=54\sqrt{2}\)

18 tháng 9 2019

a)\(\frac{3\sqrt{6}-\sqrt{2}}{1-3\sqrt{3}}=\frac{3\sqrt{3}.\sqrt{2}-\sqrt{2}}{1-3\sqrt{3}}=\frac{\sqrt{2}.\left(3\sqrt{3}-1\right)}{-\left(3\sqrt{3}-1\right)}=-\sqrt{2}\)

b)\(\frac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\frac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}=\frac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2.\left(\sqrt{2}-\sqrt{3}\right)}=\frac{\sqrt{5}}{2}\)

c)\(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}\)

d)\(\frac{5\sqrt{6}-6\sqrt{5}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{5^2.6}-\sqrt{6^2.5}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{30}.\sqrt{5}-\sqrt{30}.\sqrt{6}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{30}.\left(\sqrt{5}-\sqrt{6}\right)}{\sqrt{5}-\sqrt{6}}=\sqrt{30}\)

e)\(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{6}}=\frac{\sqrt{2^2.3}-\sqrt{3^2.2}}{\sqrt{6}}=\frac{\sqrt{6}.\sqrt{2}-\sqrt{6}.\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{6}.\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}}=\sqrt{2}-\sqrt{3}\)

f)\(\frac{6\sqrt{2}-4}{\sqrt{2}}=\frac{6\sqrt{2}-\sqrt{16}}{\sqrt{2}}=\frac{6\sqrt{2}-\sqrt{2}.2\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}.\left(6-2\sqrt{2}\right)}{\sqrt{2}}=6-2\sqrt{2}\)

g)\(\frac{6-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{36}-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}.2\sqrt{3}-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}.\left(2\sqrt{3}-5\right)}{\sqrt{3}}=2\sqrt{3}-5\)

19 tháng 9 2019

Cảm ơn bạn nha

13 tháng 9 2015

P2\(=\left(\frac{1-A\sqrt{A}}{1-\sqrt{A}}+\sqrt{A}\right).\left(\frac{1-\sqrt{A}}{1-A}\right)^2\)\(=\left(\frac{1-A\sqrt{A}+\sqrt{A}-A}{1-\sqrt{A}}\right).\frac{\left(1-\sqrt{A}\right)^2}{\left(1-A\right)^2}\)\(=\frac{\left(\sqrt{A}+1\right)\left(1-A\right)}{1-\sqrt{A}}.\frac{\left(1-\sqrt{A}\right)^2}{\left(1-\sqrt{A}\right)^2\left(1+\sqrt{A}\right)^2}\)

\(=\left(\sqrt{A}+1\right)^2.\frac{1}{\left(1+\sqrt{A}\right)^2}=1\)

18 tháng 8 2015

C=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{\left(\sqrt{x}+2\right).\left(x-1\right)-\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{x\sqrt{x}-\sqrt{x}+2x-2-\left(x-1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{x-1+x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{\left(x-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

C=\(\frac{1}{\sqrt{x}}=\frac{\sqrt{x}}{x}\)