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a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
1) Ta có: \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)
\(=1+\sqrt{2}\)
2) Ta có: \(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)
\(=\sqrt{108}-\sqrt{36\cdot\frac{4}{3}}+\sqrt{75\cdot\frac{9}{25}}\)
\(=\sqrt{108}-\sqrt{48}+\sqrt{27}\)
\(=\sqrt{3}\left(6-4+3\right)\)
\(=5\sqrt{3}\)
3) Sửa đề: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)
Ta có: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)
\(=\sqrt{2}\cdot\sqrt{4}\cdot\sqrt{3}-10\sqrt{4}\cdot\sqrt{3}+16\cdot\sqrt{4}\cdot\sqrt{3}\)
\(=\sqrt{2}\cdot\sqrt{12}-10\sqrt{12}+16\sqrt{12}\)
\(=\sqrt{12}\left(\sqrt{2}-10+16\right)\)
\(=2\sqrt{3}\left(\sqrt{2}-6\right)\)
\(=2\sqrt{6}-12\sqrt{3}\)
4) Ta có: \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{\sqrt{12}}{6}-\frac{2\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6\left(2-\sqrt{3}\right)+2\sqrt{3}-6+2\sqrt{3}}{6}\)
\(=\frac{12-6\sqrt{3}+2\sqrt{3}-6+2\sqrt{3}}{6}\)
\(=\frac{6-2\sqrt{3}}{6}\)
\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{2\sqrt{3}\cdot\sqrt{3}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{3}}\)
5) Ta có: \(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\frac{\sqrt{3}\left(2+5+3\right)}{\sqrt{15}}=\frac{10}{\sqrt{5}}=2\sqrt{5}\)
6) Ta có: \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\sqrt{48\cdot\frac{1}{4}}-\sqrt{75\cdot4}-\sqrt{3}+5\sqrt{\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{25\cdot\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{\frac{100}{3}}\)
\(=\sqrt{3}\left(2-10-1+\frac{10}{3}\right)\)
\(=-\frac{17\sqrt{3}}{3}=-\frac{17}{\sqrt{3}}\)
a/ \(\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)
b/ \(\frac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\frac{-2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=\frac{-3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}\)
c/ \(\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}+\frac{\sqrt{\left(2+\sqrt{3}\right)^2}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}=2-\sqrt{3}+2+\sqrt{3}=4\)
d/ \(\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\frac{\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)}{8}=\frac{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}=1\)
e/ \(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\frac{\sqrt{2}}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{\sqrt{2}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\frac{\sqrt{2}}{3+\sqrt{3}}+\frac{\sqrt{2}}{3-\sqrt{3}}=\frac{\sqrt{2}\left(3-\sqrt{3}+3+\sqrt{3}\right)}{6}=\sqrt{2}\)
f/ \(\frac{9+4\sqrt{5}-8\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{9-4\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-2}{2}\)
Nhân cả tử và mẫu với biểu thức liên hợp của mẫu (câu a mẫu cuối kì kì)
\(A=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\frac{1}{4}=\sqrt{3}-\frac{3}{4}\)
\(B=-\left(\sqrt{1}+\sqrt{2}-\sqrt{2}-\sqrt{3}+\sqrt{3}+\sqrt{4}-...+\sqrt{7}+\sqrt{8}-\sqrt{8}-\sqrt{9}\right)\)
\(B=-\left(\sqrt{1}-\sqrt{9}\right)=2\)
Trả lời:
Đặt \(B=\sqrt[3]{\frac{1}{4}+\frac{\sqrt{5}}{8}}-\sqrt[3]{\frac{\sqrt{5}}{8}-\frac{1}{4}}\)
\(4B=4.\sqrt[3]{\frac{1}{4}+\frac{\sqrt{5}}{8}}-4.\sqrt[3]{\frac{\sqrt{5}}{8}-\frac{1}{4}}\)
\(4B=\sqrt[3]{64.\left(\frac{1}{4}+\frac{\sqrt{5}}{8}\right)}-\sqrt[3]{64.\left(\frac{\sqrt{5}}{8}-\frac{1}{4}\right)}\)
\(4B=\sqrt[3]{16+8\sqrt{5}}-\sqrt[3]{5\sqrt{8}-16}\)
\(4B=\sqrt[3]{1+3\sqrt{5}+15+5\sqrt{5}}-\sqrt[3]{-\left(16-5\sqrt{8}\right)}\)
\(4B=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{-\left(1-3\sqrt{5}+15-5\sqrt{5}\right)}\)
\(4B=1+\sqrt{5}-\sqrt[3]{-\left(1-\sqrt{5}\right)^3}\)
\(4B=1+\sqrt{5}-\left[-\left(1-\sqrt{5}\right)\right]\)
\(4B=1+\sqrt{5}+1-\sqrt{5}\)
\(4B=2\)
\(B=\frac{1}{2}\)