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Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-2}\)

\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-2}\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=100\)

\(\Leftrightarrow x^2-5x+6-100=0\)

\(\Leftrightarrow x^2-5x-94=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5-\sqrt{401}}{2}\\x=\dfrac{5+\sqrt{401}}{2}\end{matrix}\right.\)

a: Ta có: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2-9-x^2-3x+10=6\)

\(\Leftrightarrow-3x=5\)

hay \(x=-\dfrac{5}{3}\)

c: \(4x^2-9=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\)

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

\(a,\Leftrightarrow x^2-9-x^2-3x+10=6\\ \Leftrightarrow-3x=5\Leftrightarrow x=-\dfrac{5}{3}\\ b,\Leftrightarrow2x^2+3x^2-3=5x^2+5x\\ \Leftrightarrow5x=-3\Leftrightarrow x=-\dfrac{3}{5}\\ c,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(5-2x\right)^2-4=0\\ \Leftrightarrow\left(5-2x-2\right)\left(5-2x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\\ e,\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

\(f,\Leftrightarrow\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{9}{2}\end{matrix}\right.\\ g,\Leftrightarrow\left(x^2-4\right)\left(3x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=\dfrac{4}{3}\end{matrix}\right.\\ h,\Leftrightarrow\left(x+1\right)\left(x^4+x^2+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^4+2x^2+1-x^2\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\)

\(x^2-5\)

\(=x^2-\left(\sqrt{5}\right)^2\)

\(=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

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