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25 tháng 8 2020
a) Ta có: \(\sqrt{45}:\sqrt{80}\)
\(=\sqrt{\frac{45}{80}}=\sqrt{\frac{9}{20}}\)
\(=\frac{3}{2\sqrt{5}}\)
b) Ta có: \(\sqrt{\frac{3}{15}}:\sqrt{\frac{36}{45}}\)
\(=\sqrt{\frac{1}{5}:\frac{4}{5}}\)
\(=\sqrt{\frac{1}{5}\cdot\frac{5}{4}}\)
\(=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
c) Ta có: \(\sqrt{\frac{72}{9}}:\sqrt{8}\)
\(=\frac{\sqrt{8}}{\sqrt{8}}=1\)
d) Ta có: \(\sqrt{\frac{288}{169}}:\sqrt{\frac{8}{225}}\)
\(=\sqrt{\frac{288}{169}:\frac{8}{225}}\)
\(=\sqrt{\frac{288}{169}\cdot\frac{225}{8}}\)
\(=\sqrt{\frac{8100}{169}}=\frac{90}{13}\)
NC
3 tháng 10 2020
a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
NC
3 tháng 10 2020
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
NT
25 tháng 6 2019
\(1.\)
\(x+6\sqrt{x}+8\\ =\sqrt{x}^2+2\sqrt{x}.3+9-1\\ =\left(\sqrt{x}+3\right)^2-1\\ =\left(\sqrt{x}+2\right)\left(\sqrt{x}+4\right)\)
\(2.\)
\(x-2\sqrt{x}-3\\ =\sqrt{x}^2-2\sqrt{x}+1-4\\ =\left(\sqrt{x}-1\right)^2-2^2\\ =\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)
\(4.\)
\(x^2-2\sqrt{2}x+2\\ =\left(x-\sqrt{2}\right)^2\)
\(5.\)
\(x^2+2\sqrt{13}x+13=\left(x+\sqrt{13}\right)^2\)
21 tháng 8 2020
a) Ta có: \(A=\frac{8+2\sqrt{15}+\sqrt{21}+\sqrt{35}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{5}\right)^2+\sqrt{7}\cdot\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)}{\sqrt{3}+\sqrt{5}+\sqrt{7}}\)
\(=\sqrt{3}+\sqrt{5}\)
b) Ta có: \(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}+\sqrt{3}\right)\left(\sqrt{4}-\sqrt{3}\right)}+\frac{\sqrt{5}-\sqrt{4}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\frac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}+\sqrt{5}-2+\sqrt{6}-\sqrt{5}\)
\(=-1+\sqrt{6}\)
VH
10 tháng 10 2019
\(a,\text{ ĐKXĐ : }x>1\)
\(P=\left(\frac{-8\sqrt{x}-8}{x+2\sqrt{x}-3}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right):\left(2-\frac{\sqrt{x}+4}{\sqrt{x}+3}\right)\)
\(=\left[\frac{-8\sqrt{x}-8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\right]:\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
\(=\frac{-8\sqrt{x}-8+x+6\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(=\frac{\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(\text{b,}\)
\(P< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}< \frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}-\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}-2-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-4}{2\left(\sqrt{x}+2\right)}< 0\)
\(\Leftrightarrow\sqrt{x}-4< 0\)
\(\Leftrightarrow1< x< 16\)
\(c,\)
\(\frac{1}{P}\in Z\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}\in Z\)
\(\Leftrightarrow\frac{\sqrt{x}+2-3}{\sqrt{x}+2}\in Z\)
\(\Leftrightarrow1-\frac{3}{\sqrt{x}+2}\in Z\)
\(\Leftrightarrow\sqrt{x}+2\inƯ_3=\left\{1;3\right\}\)
\(\Rightarrow x=1\)