\(\left(3x+5\right)\left(5-3x\right)+3\left(x+2\right)^2=20\)
\(\Leftrightarrow\left(5+3x\right)\left(5-3x\right)+3\left(x^2+4x+4\right)=20\)
\(\Leftrightarrow25-9x^2+3x^2+12x+12=20\)
\(\Leftrightarrow37-6x^2+12x=20\)
\(\Leftrightarrow37-6x^2+12x-20=0\)
\(\Leftrightarrow17-6x^2+12x=0\)
\(\Leftrightarrow-6x^2+12x+17=0\)
\(\Leftrightarrow6x^2-12x-17=0\)
\(\Leftrightarrow x=\dfrac{-\left(-12\right)\pm\sqrt{\left(-12\right)^2-4\cdot6\cdot\left(-17\right)}}{2\cdot6}\)
\(\Leftrightarrow x=\dfrac{12\pm\sqrt{144+408}}{12}\)
\(\Leftrightarrow x=\dfrac{12\pm\sqrt{552}}{12}\)
\(\Leftrightarrow x=\dfrac{12\pm2\sqrt{138}}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6+\sqrt{138}}{6}\\x=\dfrac{6-\sqrt{138}}{6}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{6+\sqrt{138}}{6};x_2=\dfrac{6-\sqrt{138}}{6}\)
mk đg cần nhưng còn nhìu bài để làm cho nên giúp mk vs nha Thanks trước