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a. Ta có:\(\frac{x}{y}\sqrt{\frac{y^2}{x^4}=}\) \(\frac{x}{y}.\frac{\left|y\right|}{x^2}=\frac{x.y}{x^2y}\)\(=\frac{1}{x}\)(Vì \(x\ne0;y>0\))
b \(3x^2\sqrt{\frac{8}{x^2}}=3x^2\frac{2\sqrt{2}}{\left|x\right|}=\frac{6x^2\sqrt{2}}{-x}=-6x\sqrt{2}\)( Vì \(x< 0\))
2.
a/ Áp dụgn hệ quả bđt cô si,ta có :
\(A=xy+yz+zx\le\dfrac{\left(x+y+z\right)}{3}=\dfrac{a^2}{3}\)
Vậy GTLN A =a^2/3 khi x= y =z =a/3
b/Áp dụng BĐT Cô-Si dạng Engel,ta có :
\(B=\dfrac{x^2}{1}+\dfrac{y^2}{1}+\dfrac{z^2}{z}\ge\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{a^2}{3}\)
Vậy GTNN của B = a^2/2 khi x=y=z =a/3
\(B=\dfrac{3x}{1-x}+\dfrac{4\left(1-x\right)}{x}+7\ge2\sqrt{\dfrac{3x}{1-x}.\dfrac{4\left(1-x\right)}{x}}+7=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
Vậy min B = \(\left(2+\sqrt{3}\right)^2\) khi \(\dfrac{3x}{1-x}=\dfrac{4\left(1-x\right)}{x}\Leftrightarrow x=\left(\sqrt{3}-1\right)^2\)
1: \(=\dfrac{cotx+1+tanx+1}{\left(tanx+1\right)\left(cotx+1\right)}\)
\(=\dfrac{\dfrac{1}{cotx}+cotx+2}{2+tanx+cotx}\)
\(=1\)
2: \(VT=\dfrac{cos^2x+cosxsinx+sin^2x-sinx\cdot cosx}{sin^2x-cos^2x}\)
\(=\dfrac{1}{sin^2x-cos^2x}\)
\(VP=\dfrac{1+cot^2x}{1-cot^2x}=\left(1+\dfrac{cos^2x}{sin^2x}\right):\left(1-\dfrac{cos^2x}{sin^2x}\right)\)
\(=\dfrac{1}{sin^2x}:\dfrac{sin^2x-cos^2x}{sin^2x}=\dfrac{1}{sin^2x-cos^2x}\)
=>VT=VP
2b
\(\left\{{}\begin{matrix}\sqrt{3}x-2\sqrt{2}y=7\\\sqrt{2}x+3\sqrt{3}y=-2\sqrt{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{6}x-4y=7\sqrt{2}\\\sqrt{6}x+9y=-6\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-13y=13\sqrt{2}\\\sqrt{3}x-2\sqrt{2}y=7\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=-\sqrt{2}\\x=\sqrt{3}\end{matrix}\right.\)
2 a)
\(\left\{{}\begin{matrix}2x-y=3\\3x+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=10\\2x-7=3\end{matrix}\right.\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)