Giải theo cách lớp 9 thì cũng đơn giản
\(2x^2+6x=3\)
\(\Rightarrow x^2+3x=\dfrac{3}{2}\)
\(\Rightarrow x^2+3x+\dfrac{9}{4}=\dfrac{15}{4}\)
\(\Rightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{15}{4}\)
\(\Rightarrow\left(x+\dfrac{3}{2}-\sqrt{\dfrac{15}{4}}\right)\left(x+\dfrac{3}{2}+\sqrt{\dfrac{15}{4}}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{15}{4}}-\dfrac{3}{2}\\x=-\sqrt{\dfrac{15}{4}}-\dfrac{3}{2}\end{matrix}\right.\)
Đúng 0
Bình luận (0)