`1)`Đề có sai không bạn ơi, bạn này là pt bậc 3 lận ý
`2)`\(9x^2-4-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2-3x+2\right)=0\)
\(\Leftrightarrow4\left(3x-2\right)=0\)
\(\Leftrightarrow3x-2=0\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(S=\left\{\dfrac{2}{3}\right\}\)
`3)`\(\left(5x-4\right)^2-49x^2=0\)
\(\Leftrightarrow\left(5x-4+7x\right)\left(5x-7-7x\right)=0\)
\(\Leftrightarrow\left(12x-4\right)\left(-2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)
`4)`\(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy \(S=\left\{-\dfrac{3}{2}\right\}\)
`5)`\(9x^2-16\left(x-1\right)^2=0\)
\(\Leftrightarrow9x^2-\left[4\left(x-1\right)\right]^2=0\)
\(\Leftrightarrow\left(3x-4x+4\right)\left(3x+4x-4\right)=0\)
\(\Leftrightarrow\left(4-x\right)\left(7x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{7}\end{matrix}\right.\)
Vậy \(S=\left\{4;\dfrac{4}{7}\right\}\)
`6)`\(x^2-5x-14=0\)
\(\Leftrightarrow x^2+2x-7x-14=0\)
\(\Leftrightarrow x\left(x+2\right)-7\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)
Vậy \(S=\left\{-2;7\right\}\)
