Theo Wolfram ta có: (tự viết đề lại nhé)
\(3x^2+22x+40=0\)
\(\Leftrightarrow4x^2+24x+49=x^2+6x+9\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-\frac{10}{3}\end{cases}}\)
Ps: chả biết đúng hay sai!
\(\left(2x+7\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\left(2x+7\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(2x+7-x-3\right)\left(2x+7+x+3\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\3x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=-\frac{10}{3}\end{cases}}\)
Vậy pt có 2 nghiệm x=-4,x=-10/3
\(\left(2x+7\right)^2=\left(x+3\right)^2\)
\(\left(2x+7\right)^2-\left(x+3\right)^2=0\)
\(\left(2x+7-x-3\right)\left(2x+7+x+3\right)=0\)
\(\left(x+4\right)\left(3x+10\right)=0\)
\(\orbr{\begin{cases}x+4=0\\3x+10=0\end{cases}=>\orbr{\begin{cases}x=-4\\x=\frac{-10}{3}\end{cases}}}\)
vậy \(x=-4\) hoặc \(x=\frac{-10}{3}\)
Ta có: ( 2x + 7 )2 = ( x + 3 )2
\(\Leftrightarrow\orbr{\begin{cases}2x+7=x+3\\2x+7=-\left(x+3\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\3x+10=0\end{cases}}\)
\(\orbr{\begin{cases}x=-4\\x=\frac{-10}{3}\end{cases}}\)
Vậy x=-4;\(\frac{-10}{3}\)