b) \(x^2-11x+28=0\)
\(\Leftrightarrow\left(x^2-4x\right)-\left(7x-28\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-7\right)=0\)
a) \(\left(4x+3\right)^2=\left(x^2-2x+1\right)\)
\(\Leftrightarrow\left(4x+3\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(4x+3+x-1\right)\left(4x-3-x+1\right)=0\)
\(\Leftrightarrow\left(5x+2\right)\left(3x-2\right)=0\)
Ta có :
\(\left(4x+3\right)^2=x^2-2x+1\)
\(\Leftrightarrow\)\(\left(4x+3\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow\)\(4x+3=x-1\)
\(\Leftrightarrow\)\(4x-x=-1-3\)
\(\Leftrightarrow\)\(3x=-4\)
\(\Leftrightarrow\)\(x=\frac{-4}{3}\)
Vậy \(x=\frac{-4}{3}\)
Mk giải theo kiểu bình thường thui vì mk mới lớp 7
a) Ta có :
\(\left(4x+3\right)^2=x^2-2x+1\)
\(\Leftrightarrow\)\(\left(4x+3\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow\)\(\left(4x+3\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\)\(\left(4x+3+x-1\right)\left(4x+3-x+1\right)=0\)
\(\Leftrightarrow\)\(\left(5x+4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}5x+4=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=-4\\3x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{-2}{3}\end{cases}}}\)
Vậy \(x=\frac{-4}{5}\) hoặc \(x=\frac{-2}{3}\)
