Question

Giai phương trình sau :

  \(\left(x+3\right)\left(4-3x\right)+\left(x^2+6x+9\right)=0\)

Answer 1

\(\left(x+3\right)\left(4-3x\right)+\left(x^2+6x+9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(4-3x\right)+\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(x+3\right)\left[\left(4-3x\right)+\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(4-3x+x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(7-2x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\\7-2x=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=\frac{7}{2}\end{array}\right.\)

Vậy phương trình có tập nghiệm là \(\left\{-3;\frac{7}{2}\right\}\)

Answer 2

(x+3)(4-3x)+(x²+6x+9)=0

(x+3)(4-3x)+(x+3)²=0

(x+3)(4-3x)+(x+3)(x+3)=0

(x+3)(4-3x+x+3)=0

(x+3)(7-2x)=0

    \(\Rightarrow\left[\begin{array}{nghiempt}x+3=0\\7-2x=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=-3\\2x=7\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=-3\\x=\frac{7}{2}\end{array}\right.\)

              Vậy x=3;\(\frac{7}{2}\)