ai giúp với .......please
\(B=\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)
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27 tháng 7 2016
a) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
b)\(\frac{x-4}{2\left(\sqrt{x}+2\right)}\) (ĐK:x\(\ge0\))
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2}{2}\)
c)\(\frac{x-5\sqrt{x}+6}{3\sqrt{x}-6}\) (ĐK:x\(\ge0;x\ne4\))
\(=\frac{x-3\sqrt{x}-2\sqrt{x}+6}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)-2\left(\sqrt{x}-3\right)}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-3}{3}\)
27 tháng 7 2016
b) Tử \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\) (hằng đăngt thức số 3 )
NC
29 tháng 9 2020
Ta có: \(\frac{1}{3}\left(\sqrt{6}+\sqrt{5}\right)^2-\frac{1}{4}\sqrt{120}-2\sqrt{\frac{15}{2}}\)
\(=\frac{1}{3}\left(11+2\sqrt{30}\right)-\frac{\sqrt{30}}{2}-\sqrt{30}\)
\(=\frac{11}{3}+\frac{2}{3}\sqrt{30}-\frac{\sqrt{30}}{2}-\sqrt{30}\)
\(=\frac{11}{3}-\frac{5}{6}\sqrt{30}\)
\(=\frac{22-5\sqrt{30}}{6}\)
NC
29 tháng 9 2020
Ta có: \(\left(\frac{1}{2}\sqrt{\frac{2}{3}}-\frac{3}{4}\sqrt{54}+\frac{1}{3}\sqrt{\frac{8}{3}}\right)\div\sqrt{\frac{81}{6}}\)
\(=\left(\frac{\sqrt{6}}{6}-\frac{9\sqrt{6}}{4}+\frac{2\sqrt{6}}{9}\right)\div\frac{3\sqrt{6}}{2}\)
\(=-\frac{67\sqrt{6}}{36}\cdot\frac{2}{3\sqrt{6}}\)
\(=-\frac{67}{54}\)
26 tháng 7 2019
\( 2)2\sqrt {\dfrac{{16}}{3}} - 3\sqrt {\dfrac{1}{{27}}} - 6\sqrt {\dfrac{4}{{75}}} \\ = 2.\dfrac{4}{{\sqrt 3 }} - 3.\dfrac{1}{{\sqrt {27} }} - 6\dfrac{2}{{\sqrt {75} }}\\ = \dfrac{8}{{\sqrt 3 }} - \dfrac{3}{{3\sqrt 3 }} - \dfrac{{12}}{{5\sqrt 2 }}\\ = \dfrac{{8\sqrt 3 }}{3} - \dfrac{{\sqrt 3 }}{3} - \dfrac{{4\sqrt 3 }}{5}\\ = \dfrac{{23\sqrt 3 }}{{15}}\\ 3)2\sqrt {27} - 6\sqrt {\dfrac{4}{3}} + \dfrac{3}{5}\sqrt {75} \\ = 6\sqrt 3 - \dfrac{{12}}{{\sqrt 3 }} + 3\sqrt 3 \\ = 9\sqrt 3 - 4\sqrt 3 \\ = 5\sqrt 3 \)
7 tháng 8 2016
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(2+\sqrt{6}+\sqrt{8}\right)}..\)
= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}.\)
= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(1+\sqrt{2}\right).\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}=\frac{1}{1+\sqrt{2}}=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right).\left(\sqrt{2}-1\right)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1.\)
\(B=\frac{2+\sqrt{2}-\sqrt{3}-\sqrt{2.3}+\sqrt{2.4}+2}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{2+\sqrt{2}-\sqrt{3}+\sqrt{2}\left(2+\sqrt{2}-\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}}\)
=\(\frac{\left(2+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}\right)}{2+\sqrt{2}-\sqrt{3}}\)
= \(1+\sqrt{2}\)