Tính: a
\(\sqrt{60}-\sqrt{135}+\frac{1}{3}\sqrt{15}\)
b.\(\sqrt{28}-\frac{1}{2}\sqrt{343}+2\sqrt{63}\)
c.\(\sqrt{12}-\frac{2}{3}\sqrt{27}+\sqrt{243}\)
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21 tháng 7 2020
Bài 1:
a) Ta có: \(\sqrt{243}-\frac{1}{2}\sqrt{12}-2\sqrt{75}+\sqrt{27}\)
\(=\sqrt{3}\cdot9-\frac{1}{2}\cdot\sqrt{3}\cdot2-2\cdot\sqrt{3}\cdot5+\sqrt{3}\cdot3\)
\(=\sqrt{3}\left(9-1-10+3\right)\)
\(=\sqrt{3}\cdot1=\sqrt{3}\)
b) Ta có: \(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{5}{1+\sqrt{6}}-6\sqrt{\frac{1}{6}}\)
\(=\frac{\left(2\sqrt{3}-3\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(\sqrt{3}+\sqrt{2}\right)}+\frac{5\cdot\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\sqrt{36\cdot\frac{1}{6}}\)
\(=-\sqrt{6}+\frac{5\left(\sqrt{6}-1\right)}{5}-\sqrt{6}\)
\(=-2\sqrt{6}+\sqrt{6}-1\)
\(=-\sqrt{6}-1\)
Bài 2: Rút gọn
Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
10 tháng 8 2020
a) Ta có: \(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
\(=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}-\frac{\sqrt{5}}{2}\)
\(=\frac{2\sqrt{5}-\sqrt{5}}{2}=\frac{\sqrt{5}}{2}\)
b) Ta có: \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{3}-1}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}\)
\(=\frac{\left(2\sqrt{8}-\sqrt{12}\right)\left(\sqrt{3}+1\right)}{3-1}-\frac{\left(\sqrt{5}+\sqrt{27}\right)\left(\sqrt{30}+\sqrt{2}\right)}{30-2}\)
\(=\frac{4\sqrt{6}+4\sqrt{2}-6-2\sqrt{3}}{2}-\frac{5\sqrt{6}+\sqrt{10}+9\sqrt{10}+3\sqrt{6}}{28}\)
\(=\frac{4\sqrt{2}\left(\sqrt{3}+1\right)-2\sqrt{3}\left(\sqrt{3}+1\right)}{2}-\frac{10\sqrt{10}+8\sqrt{6}}{28}\)
\(=\frac{2\cdot\left(\sqrt{3}+1\right)\left(2\sqrt{2}-\sqrt{3}\right)}{2}-\frac{\sqrt{1000}+\sqrt{384}}{28}\)
\(=2\sqrt{6}-3+2\sqrt{2}-\sqrt{3}-\frac{\sqrt{2}\cdot\left(5\sqrt{5}+4\sqrt{3}\right)}{14}\)
\(=\frac{28\sqrt{6}-42+28\sqrt{2}-14\sqrt{3}-10\sqrt{10}-8\sqrt{6}}{14}\)
\(=\frac{20\sqrt{6}-42+28\sqrt{2}-14\sqrt{3}-10\sqrt{10}}{14}\)
c) Ta có: \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}\)
\(=\frac{2\sqrt{3}+2-2\sqrt{3}+2}{2}\)
\(=\frac{4}{2}=2\)
a) \(\sqrt{60}-\sqrt{135}+\frac{1}{3}\sqrt{15}\)
\(=2\sqrt{15}-3\sqrt{15}+\frac{1}{3}\sqrt{15}\)
\(=-\frac{2}{3}\sqrt{15}\)
b) \(\sqrt{28}-\frac{1}{2}\sqrt{343}+2\sqrt{63}\)
\(=2\sqrt{7}-\frac{7}{2}\sqrt{7}+6\sqrt{7}\)
\(=\frac{9}{2}\sqrt{7}\)
c) \(\sqrt{12}-\frac{2}{3}\sqrt{27}+\sqrt{243}\)
\(=2\sqrt{3}-2\sqrt{3}+9\sqrt{3}\)
\(=9\sqrt{3}\)