Question

giải phương trình tích sau:

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

Answer 1

Answer 2

Lời giải:

Ta có:

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

\(\Leftrightarrow x(x^2-9)-(x+2)(x^2-2x)-4(x+2)=0\)

\(\Leftrightarrow x(x^2-9)-x(x^2-4)-4(x+2)=0\)

\(\Leftrightarrow -9x-(-4x)-4(x+2)=0\)

\(\Leftrightarrow -9x-8=0\Leftrightarrow x=\frac{-8}{9}\)

Answer 3

x( x + 3 ) ( x - 3 ) - (x + 2 ) ( x2 - x + 4 ) = 0

\(\Leftrightarrow\) x ( x2 -9 ) - (x3 + 8 ) = 0

\(\Leftrightarrow\) x3 - 9x -x3 - 8 = 0

\(\Leftrightarrow\) -9x - 8 = 0

\(\Leftrightarrow\) -9x = 8

\(\Leftrightarrow\) x = \(-\dfrac{8}{9}\)

Vậy x = \(-\dfrac{8}{9}\)

Chúc bạn học tốt !

Answer 4

Thank các bn!