Câu hỏi của Bich Hoang

Question

(3x+2)(x2+1)=(9x2-4)(x+1)

Hoàng Yến · Accepted Answer

$\left(3x+2\right)\left(x^2+1\right)=\left(9x^2-4\right)\left(x+1\right)\
\Leftrightarrow3x^3+2x^2+3x+2=9x^3+9x^2-4x-4\\Leftrightarrow
-6x^3-7x^2+7x+6=0\\Leftrightarrow
-6\left(x^3+\frac{7}{6}x^2-\frac{7}{6}x-1\right)=0\\Leftrightarrow
x^3+\frac{7}{6}x^2-\frac{7}{6}x-1=0\\Leftrightarrow
x^3-x^2+\frac{2}{3}x^2+\frac{3}{2}x^2-\frac{2}{3}x-\frac{3}{2}x+x-1=0\\Leftrightarrow
x^3-x^2+\frac{2}{3}x^2-\frac{2}{3}x+\frac{3}{2}x^2-\frac{3}{2}x+x-1=0\
\Leftrightarrow\left(x^2+\frac{2}{3}x+\frac{3}{2}x+1\right)\left(x-1\right)=0\\Leftrightarrow
\left(x+\frac{2}{3}\right)\left(x+\frac{3}{2}\right)\left(x-1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x+\frac{2}{3}=0\x+\frac{3}{2}=0\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\x=-\frac{3}{2}\x=1\end{matrix}\right.$

Vậy nghiệm của phương trình trên là $S=\left\{-\frac{2}{3};-\frac{3}{2};1\right\}$Câu trả lời: $\left(3x+2\right)\left(x^2+1\right)=\left(9x^2-4\right)\left(x+1\right)\
\Leftrightarrow3x^3+2x^2+3x+2=9x^3+9x^2-4x-4\\Leftrightarrow
-6x^3-7x^2+7x+6=0\\Leftrightarrow
-6\left(x^3+\frac{7}{6}x^2-\frac{7}{6}x-1\right)=0\\Leftrightarrow
x^3+\frac{7}{6}x^2-\frac{7}{6}x-1=0\\Leftrightarrow
x^3-x^2+\frac{2}{3}x^2+\frac{3}{2}x^2-\frac{2}{3}x-\frac{3}{2}x+x-1=0\\Leftrightarrow
x^3-x^2+\frac{2}{3}x^2-\frac{2}{3}x+\frac{3}{2}x^2-\frac{3}{2}x+x-1=0\
\Leftrightarrow\left(x^2+\frac{2}{3}x+\frac{3}{2}x+1\right)\left(x-1\right)=0\\Leftrightarrow
\left(x+\frac{2}{3}\right)\left(x+\frac{3}{2}\right)\left(x-1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x+\frac{2}{3}=0\x+\frac{3}{2}=0\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\x=-\frac{3}{2}\x=1\end{matrix}\right.$

Vậy nghiệm của phương trình trên là $S=\left\{-\frac{2}{3};-\frac{3}{2};1\right\}$

Vũ Minh Tuấn · Answer

Sửa lại đề là:

$\left(3x+2\right).\left(x^2+1\right)=\left(9x^2-4\right).\left(x+1\right)$

$\Leftrightarrow\left(3x+2\right).\left(x^2+1\right)-\left(9x^2-4\right).\left(x+1\right)=0$

$\Leftrightarrow\left(3x+2\right).\left(x^2+1\right)-\left[\left(3x\right)^2-2^2\right].\left(x+1\right)=0$

$\Leftrightarrow\left(3x+2\right).\left(x^2+1\right)-\left(3x-2\right).\left(3x+2\right).\left(x+1\right)=0$

$\Leftrightarrow\left(3x+2\right).\left[x^2+1-\left(3x-2\right).\left(x+1\right)\right]=0$

$\Leftrightarrow\left(3x+2\right).\left[x^2+1-\left(3x^2+3x-2x-2\right)\right]=0$

$\Leftrightarrow\left(3x+2\right).\left(x^2+1-3x^2-3x+2x+2\right)=0$

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

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

$\Leftrightarrow\left(3x+2\right).\left[\left(-2x^2-3x\right)+\left(2x+3\right)\right]=0$

$\Leftrightarrow\left(3x+2\right).\left[-x.\left(2x+3\right)+\left(2x+3\right)\right]=0$

$\Leftrightarrow\left(3x+2\right).\left(2x+3\right).\left(1-x\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}3x+2=0\2x+3=0\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-2\2x=-3\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\x=-\frac{3}{2}\x=1\end{matrix}\right.$

Vậy phương trình có tập hợp nghiệm là: $S=\left\{-\frac{2}{3};-\frac{3}{2};1\right\}.$

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