A/ \(2\left(x+4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=3\end{matrix}\right.\)

KL:...........

B/ \(\left(x-1\right)^2\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)

KL:..................

C/ \(\left(\frac{2x}{3}+4\right)\left(2x-3\right)\left(\frac{x}{2}-1\right)=0\Leftrightarrow\left[{}\begin{matrix}\frac{2x}{3}+4=0\\2x-3=0\\\frac{x}{2}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\frac{3}{2}\\x=2\end{matrix}\right.\)

KL:.....................

\(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-4\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5\)

\(\Leftrightarrow30x=60\)

\(\Leftrightarrow x=2\)

\(\left|x-1\right|=2x+3\)

Với \(x-1\ge0\Leftrightarrow x\ge1\)

\(\Leftrightarrow x-1=2x+3\)

\(\Leftrightarrow-x=4\)

\(\Leftrightarrow x=-4\left(L\right)\)

Với \(x-1< 0\Leftrightarrow x< 1\)

\(\Leftrightarrow-x+1=2x+3\)

\(\Leftrightarrow-3x=2\)

\(\Leftrightarrow x=-\dfrac{2}{3}\left(N\right)\)

Vậy \(S=\left\{-\dfrac{2}{3}\right\}\)

Ta thấy:

\(\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge\frac{2\sqrt{xy}2}{1\sqrt{xy}}=4\) 

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\left(dpcm\right)\)

đây là 1 BĐT

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}<=>\frac{x+y}{xy}\ge\frac{4}{x+y}<=>\left(x+y\right)^2\ge4xy\)

\(<=>x^2+2xy+y^2-4xy\ge0<=>x^2-2xy+y^2\ge0<=>\left(x-y\right)^2\ge0\) (luôn đúng)

Vậy ta có đpcm

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