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

Vậy tập nghiệm \(S=\left\{-2;3\right\}\)

2) \(\left(2x+3\right)\left(-x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=7\end{matrix}\right.\)

Vậy...

3) \(\left(x-1\right)\left(x+5\right)\left(-3x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\\-3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\\x=\dfrac{8}{3}\end{matrix}\right.\)

Vậy...

4) \(x\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy...

đk: x\(\ne1,x\ne2\)

\(\dfrac{1}{x-1}-\dfrac{3}{x-2}=\dfrac{-1}{\left(x-1\right)\left(x-2\right)}\)

\(\Leftrightarrow\dfrac{x-2}{\left(x-1\right)\left(x-2\right)}-\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=\dfrac{-1}{\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow x-2-3\left(x-1\right)=-1\)

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

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

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

\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{6x+24}{x^2-9}\)       \(\left(x\ne\pm3\right)\)

\(\Leftrightarrow\dfrac{\left(x+3\right)^2}{x^2-9}-\dfrac{\left(x-3\right)^2}{x^2-9}=\dfrac{6x+24}{x^2-9}\)

\(\Rightarrow\left(x+3\right)^2-\left(x-3\right)^2=6x+24\)

\(\Leftrightarrow x^2+6x+9-\left(x^2-6x+9\right)=6x+24\)

\(\Leftrightarrow x^2+6x+9-x^2+6x-9=6x+24\)

\(\Leftrightarrow x^2+6x-x^2+6x-6x=24-9+9\)

\(\Leftrightarrow6x=24\Leftrightarrow x=4\)

Vậy ..

Ta có: 

\(x^2+9x+2x=\left(x+4\right)\left(x+5\right)\)

\(x^2+11x+30=\left(x+6\right)\left(x+5\right)\)

\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)

ĐK: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)

pt \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)

\(\Rightarrow18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)\)

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)  (tm)

Ta có: \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt{xy}}\)

\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}\right)+\sqrt{y}\left(x-\sqrt{xy}\right)}{x^2-xy}\)

\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}+x-\sqrt{xy}\right)}{x\left(x-y\right)}=\dfrac{2x\sqrt{y}}{x\left(x-y\right)}\)

\(=\dfrac{2\sqrt{y}}{x-y}=\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(\Rightarrow A=\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}.\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\dfrac{\sqrt{x}+\sqrt{y}-1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\dfrac{\sqrt{x}+\sqrt{y}-1+1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}}{x}\)