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Bài 2:
a: \(\sqrt{\dfrac{130}{200}}:\dfrac{\sqrt{5}}{\sqrt{8}}=\dfrac{\sqrt{130}}{10\sqrt{2}}\cdot\dfrac{2\sqrt{2}}{\sqrt{5}}=\dfrac{\sqrt{26}}{5}\)
b: \(=\sqrt{\dfrac{1}{5}:\dfrac{4}{5}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)
c: \(=\left(3\sqrt{5}+3\sqrt{5}-5\sqrt{5}\right):3\sqrt{5}\)
\(=\dfrac{\sqrt{5}}{3\sqrt{5}}=\dfrac{1}{3}\)
\(a)\sqrt{9\times^2}-2\times\)
\(=\sqrt{3^2\times^2}-2\times\)
\(=\sqrt{(3\times)^2}-2\times\)
\(=3\times-2\times\)
\(=\times\)
\(\sqrt{x^2\left(x-1\right)^2}=\left|x\left(x-1\right)\right|\)
\(x< 0\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x< 0\end{matrix}\right.\Leftrightarrow x\left(x-1\right)>0\Rightarrow\left|x\left(x-1\right)\right|=x\left(x-1\right)=x^2-x\)
\(b,\sqrt{13x}.\sqrt{\frac{52}{x}}=\sqrt{\frac{13.52.x}{x}}=\sqrt{13.52}=\sqrt{13^2.2^2}=\sqrt{26^2}=26\)
Lời giải :
a) \(\sqrt{x^2\left(x-1\right)^2}=\left|x\right|\cdot\left|x-1\right|=-x\left(1-x\right)=x^2-x\)
b) \(\sqrt{13x}\cdot\sqrt{\frac{52}{x}}=\sqrt{\frac{13x\cdot52}{x}}=\sqrt{676}=26\)
c) \(5xy\cdot\sqrt{\frac{25x^2}{y^6}}=5xy\cdot\sqrt{\left(\frac{5x}{y^3}\right)^2}=5xy\cdot\frac{-5x}{y^3}=\frac{-25x^2}{y^2}\)
d) \(\sqrt{\frac{9+12x+4x^2}{y^2}}=\sqrt{\frac{\left(2x+3\right)^2}{y^2}}=\frac{2x+3}{-y}=\frac{-2x-3}{y}\)
Lời giải:
Áp dụng BĐT AM-GM:
$A=a^2b^2(a^2+b^2)$
$4A=2ab.2ab(a^2+b^2)\leq \left(\frac{2ab+2ab+a^2+b^2}{3}\right)^3$
$=[\frac{(a+b)^2+2ab}{3}]^3=(\frac{16+2ab}{3})^3$
Mà:
$2ab\leq 2(\frac{a+b}{2})^2=2(\frac{4}{2})^2=8$
$\Rightarrow 4A\leq (\frac{16+8}{3})^3=512$
$\Rightarrow A\leq 128$
Dấu "=" xảy ra khi $a=b=2$
a) \(2\sqrt{5a^2}=2\sqrt{5}\left|a\right|=-2a\sqrt{5}\)
b)\(2\sqrt{18a^2}=2.3\sqrt{2}.\left|a\right|=6a\sqrt{2}\)
c)\(\sqrt{-9b^3}=\sqrt{9.\left(-b\right)^3}=3\sqrt{-b}.\left|b\right|=-3b\sqrt{-b}\)
d)\(\sqrt{24a^4b^8}=\sqrt{6.\left(4a^2b^4\right)^2}=2a^2b^4\sqrt{6}\)