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Ta có: \(P=\frac{x^2+y^2}{x-y}=\frac{\left(x-y\right)^2+2xy}{x-y}=\left(x-y\right)+\frac{2xy}{x-y}\)
\(=x-y+\frac{16}{x-y}\ge2.4=8\)
Đặt \(t=x^2+y^2\) thì ta có :
\(P^2=\frac{\left(x^2+y^2\right)^2}{\left(x-y\right)^2}=\frac{t^2}{t-16}=\frac{1}{\frac{t-16}{t^2}}=\frac{1}{-\frac{16}{t^2}+\frac{1}{t}}=\frac{1}{-16\left(\frac{1}{t}-\frac{1}{32}\right)^2+\frac{1}{64}}\ge\frac{1}{\frac{1}{64}}=64\)
\(\Rightarrow P\ge8\). Đẳng thức xảy ra khi \(\hept{\begin{cases}x^2+y^2=32\\xy=8\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=2+2\sqrt{2}\\y=-2+2\sqrt{3}\end{cases}}\)
a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
b: \(x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
c: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)
a) \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)
b) \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)
d) \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
\(\sqrt{x^2\left(x-1\right)^2}=\left|x\left(x-1\right)\right|\)
\(x< 0\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x< 0\end{matrix}\right.\Leftrightarrow x\left(x-1\right)>0\Rightarrow\left|x\left(x-1\right)\right|=x\left(x-1\right)=x^2-x\)
\(b,\sqrt{13x}.\sqrt{\frac{52}{x}}=\sqrt{\frac{13.52.x}{x}}=\sqrt{13.52}=\sqrt{13^2.2^2}=\sqrt{26^2}=26\)
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) \(\sqrt{\frac{9+12x+4x^2}{y^2}}=\sqrt{\frac{\left(2x+3\right)^2}{y^2}}=\frac{2x+3}{-y}=\frac{-2x-3}{y}\)