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Bài 2:
a, Ta có
\(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}\)
= \(3\left|-2\right|+\left|-5\right|\)
=\(6+5\)
= 11
Vậy \(3\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}=11\)
b, Ta có
\(\sqrt{6+2\sqrt{5}}-\sqrt{5}\)
= \(\sqrt{5+2\sqrt{5}+1}-\sqrt{5}\)
= \(\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}\)
= \(\left|\sqrt{5}+1\right|-\sqrt{5}\)
= \(\sqrt{5}+1-\sqrt{5}=1\)
Vậy \(\sqrt{6+2\sqrt{5}}-\sqrt{5}=1\)
Bài 1 :
\(6xy\cdot\sqrt{\frac{9x^2}{16y^2}}=6xy\cdot\frac{3x}{4y}=\frac{18x^2y}{4y}=\frac{9}{2}x^2\)
\(\sqrt{\frac{4+20a+25a^2}{b^4}}=\sqrt{\frac{\left(2+5a\right)^2}{\left(b^2\right)^2}}=\frac{2+5a}{b^2}\)
\(\left(m-n\right).\sqrt{\frac{m-n}{\left(m-n\right)^2}}=\sqrt{\left(m-n\right)^2}\cdot\sqrt{\frac{1}{m-n}}=\sqrt{\frac{\left(m-n\right)^2}{m-n}}=\sqrt{m-n}\)
Bài 2 :
1. \(\left(2\sqrt{3}-\sqrt{12}\right):5\sqrt{3}=\left(2\sqrt{3}-2\sqrt{3}\right):5\sqrt{3}=0:5\sqrt{3}=0\)
2. \(\sqrt{\frac{317^2-302^2}{1013^2-1012^2}}=\frac{\sqrt{\left(317+302\right)\left(317-302\right)}}{\sqrt{\left(1013+1012\right)\left(1013-1012\right)}}=\frac{\sqrt{619}\cdot\sqrt{15}}{\sqrt{2025}}=\sqrt{\frac{619}{135}}\)(check lại)
3. \(\sqrt{27\left(1-\sqrt{3}\right)^2}:3\sqrt{75}\)
\(=\sqrt{27}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=3\sqrt{3}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=\frac{1-\sqrt{3}}{5}\)
4.\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\frac{5}{\sqrt{5}}+\frac{\sqrt{20}}{2}-\frac{\frac{5}{4}\cdot2}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\frac{2\sqrt{5}}{2}-\frac{\frac{5}{2}}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\sqrt{5}+\frac{\sqrt{5}}{2}+\sqrt{5}\right):2\sqrt{5}\)
\(=\frac{7}{2}\sqrt{5}:2\sqrt{5}\)
\(=\frac{7}{4}\)
a) \(A=\frac{1}{2}\sqrt{32}+\sqrt{98}-\frac{1}{6}\sqrt{18}=\frac{1}{2}\sqrt{4^2.2}+\sqrt{7^2.2}-\frac{1}{6}.\sqrt{3^2.2}\)
\(=\frac{1}{2}\sqrt{4^2}.\sqrt{2}+\sqrt{7^2}.\sqrt{2}-\frac{1}{6}.\sqrt{3^2}.\sqrt{2}\)\(=\frac{1}{2}.4\sqrt{2}+7\sqrt{2}-\frac{1}{6}.3.\sqrt{2}\)\(=2.\sqrt{2}+7\sqrt{2}-\frac{1}{2}\sqrt{2}=\left(2+7-\frac{1}{2}\right)\sqrt{2}=\frac{17}{2}\sqrt{2}\)
P2\(=\left(\frac{1-A\sqrt{A}}{1-\sqrt{A}}+\sqrt{A}\right).\left(\frac{1-\sqrt{A}}{1-A}\right)^2\)\(=\left(\frac{1-A\sqrt{A}+\sqrt{A}-A}{1-\sqrt{A}}\right).\frac{\left(1-\sqrt{A}\right)^2}{\left(1-A\right)^2}\)\(=\frac{\left(\sqrt{A}+1\right)\left(1-A\right)}{1-\sqrt{A}}.\frac{\left(1-\sqrt{A}\right)^2}{\left(1-\sqrt{A}\right)^2\left(1+\sqrt{A}\right)^2}\)
\(=\left(\sqrt{A}+1\right)^2.\frac{1}{\left(1+\sqrt{A}\right)^2}=1\)