\(4\cdot2-12x+9=x+7\)
\(\Leftrightarrow8-12x+9=x+7\)
\(\Leftrightarrow-12x-x=7-8-9\)
\(\Leftrightarrow-13x=-10\)
\(\Leftrightarrow x=\dfrac{10}{13}\)
Vậy \(x=\dfrac{10}{13}\)
Ta có: \(4x^2-12x+9=x+7\)
\(\Leftrightarrow4x^2-13x+2=0\)
\(\text{Δ}=\left(-13\right)^2-4\cdot4\cdot2=137\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{13-\sqrt{137}}{8}\\x_2=\dfrac{13+\sqrt{137}}{8}\end{matrix}\right.\)
Ta có: 4x2-12x+9=x+7
⇔ 4x2-13x+2=0
Ta có:Δ=(-13)2-4.4.2=137
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{13-\sqrt{137}}{8}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{13+\sqrt{137}}{8}\end{matrix}\right.\)
Ta có: \(\sqrt{4x^2-12x+9}=x+7\)
\(\Leftrightarrow\sqrt{\left(2x-3\right)^2}=x+7\)
\(\Leftrightarrow2x-3=x+7\)
\(\Leftrightarrow x=10\)
