\(\left(x-5\right)\left(x+4\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=0+5\\x=0-4\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
\(\Rightarrow x=5\) hoặc \(x=-4\)
\(\left|-15\right|-\left|x\right|=\left|-12\right|\)
\(15-\left|x\right|=12\)
\(\left|x\right|=15-12\)
\(\left|x\right|=3\)
\(\Rightarrow x=\pm3\)
5x+9=2x+15<=>5x-2x=15-9<=>3x=6<=>x=3
(x-5).(x+4)=0
Th1x-5=0<=>x=5
Th2x+4=0<=>x=-4
Vậy x=0;-4
2.(x-3)-4(x+4)=3.(-7)+5<=>2x-6-4x-16=-21+5<=>-2x-22=-16<=>-2x=-16+22
<=>-2x=6<=>2x=-6<=>x=-3
/-15/-/x/=/-12/
Th1:x<0=>/-15/-/x/=/-12/<=>15-(-x)=12<=>15+x=12<=>x=27
Th2:x> hoặc= 0=>/-15/-/x/=/-12/<=>15-x=12<=>15-12=x<=>x=3
