Question

giải pt

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

Answer 1

mk chỉ biết kq là -14/3 thôi

Answer 2

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

\(\Leftrightarrow\left[\left(x^2+4x\right)-\left(2x+8\right)\right]-\left(x^2+3x+2x+6\right)=0\)

\(\Leftrightarrow x^2+4x-2x-8-x^2-3x-2x-6=0\)

\(\Leftrightarrow\left(x^2-x^2\right)+\left[4x-2x-2x-3x\right]+\left(-8-6\right)=0\)

\(\Leftrightarrow0+0+\left(-3x\right)+\left(-14\right)=0\)

\(\Leftrightarrow-3x-14=0\)

\(\Leftrightarrow-3x=0+14\)

\(\Leftrightarrow-3x=14\)

\(\Leftrightarrow x=14\div\left(-3\right)\)

\(\Leftrightarrow x=-\frac{14}{3}\)

Vậy S = { \(-\frac{14}{3}\) }