Ta có: \(\left(5x-2\right)\left(3x+2\right)=10x\left(x+7\right)\)
\(\Leftrightarrow15x^2+10x-6x-4=10x^2+70x\)
\(\Leftrightarrow15x^2+4x-4-10x^2-70x=0\)
\(\Leftrightarrow5x^2+66x-4=0\)
\(\Delta=66^2-4\cdot5\cdot\left(-4\right)=4436\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-b-\sqrt{\Delta}}{2a}\\x_2=\dfrac{-b+\sqrt{\Delta}}{2a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-66-2\sqrt{1109}}{10}=\dfrac{-33-\sqrt{1109}}{5}\\x_2=\dfrac{-66+2\sqrt{1109}}{10}=\dfrac{-33+\sqrt{1109}}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{-33-\sqrt{1109}}{5};\dfrac{-33+\sqrt{1109}}{5}\right\}\)
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