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Cách 1: 3√1728 : 3√64 = 12:4 = 3
Cách 2: 3√1728:3√64 = 3√(1728/64) = 3√27 = 3
Cách 1: 3√1728 : 3√64 = 12:4 = 3
Cách 2: 3√1728:3√64 = 3√(1728/64) = 3√27 = 3
~Học tốt~
\(pt\Leftrightarrow\sqrt{x^2+x-6}=\sqrt{x^2+2}\)
Ta thấy 2 vế luôn dương bình phương lên ta có:
\(\sqrt{\left(x^2+x-6\right)^2}=\sqrt{\left(x^2+2\right)^2}\)
\(\Rightarrow x^2+x-6=x^2+2\)
\(\Rightarrow x^2-x^2+x=6+2\)
\(\Rightarrow x=8\)
\(=\frac{2\sqrt{2}\left(1-\sqrt{3}\right)}{3\sqrt{2-\sqrt{3}}}\)
\(=\frac{2.\left(1-\sqrt{3}\right).\sqrt{2}.\sqrt{2+\sqrt{3}}}{3.\sqrt{2-\sqrt{3}}.\sqrt{2+\sqrt{3}}}\)
\(=\frac{2.\left(1-\sqrt{3}\right).\sqrt{2\left(2+\sqrt{3}\right)}}{3.\sqrt{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}}\)
\(=\frac{2.\left(1-\sqrt{3}\right).\sqrt{4+2\sqrt{3}}}{3.\sqrt{2^2-\left(\sqrt{3}\right)^2}}\)
\(=\frac{2\left(1-\sqrt{3}\right)\sqrt{\left(1+\sqrt{3}\right)^2}}{3.\sqrt{4-3}}\)
\(=\frac{2\left(1-\sqrt{3}\right)|1+\sqrt{3}|}{3\sqrt{1}}\)
\(=\frac{2\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}{3}\)
\(=\frac{2\left(1^2-\left(\sqrt{3}\right)^2\right)}{3}\)
\(=\frac{2.\left(-2\right)}{3}=\frac{-4}{3}\)
a: Sửa đề: căn 6+2căn 5-căn 5
\(a=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}=\dfrac{2}{1}=2\)
b: \(a^3=2-\sqrt{3}+2+\sqrt{3}+3a\)
=>a^3-3a-4=0
=>a^3-3a=4
\(\dfrac{64}{\left(a^2-3\right)^3}-3a=\left(\dfrac{4}{a^2-3}\right)^3-3a\)
\(=\left(\dfrac{a^3-3a}{a^2-3}\right)^3-3a=a^3-3a\)
=4
b: Ta có: \(\sqrt[3]{-0.008}-\dfrac{1}{5}\cdot\sqrt[3]{64}+5\cdot\sqrt[3]{\left(-5\right)^3}\)
\(=-\dfrac{1}{5}-\dfrac{1}{5}\cdot4+5\cdot\left(-5\right)\)
\(=-\dfrac{1}{5}-\dfrac{4}{5}-25\)
=-26
\(a=\dfrac{4\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}}{2}\)
\(=2\sqrt{\sqrt{5}-\sqrt{5}+1}=2\)
\(P=\left(2^5-7\cdot2^2-3\right)^{81}+19=1+19=20\)