a) (x+1)(x+9)=(x+3)(x+5)
➜\(x^2+9x+x+9=x^2+5x+3x+15\)
➜\(x^2+9x+x-x^2-5x-3x=15-9\)
➜2x=6
➜\(x=3\)
b) (x+2)^2 + 2(x-4)=(x-4)(x-2)
➜\(x^2+4x+4+2x-8=x^2-2x-4x+8\)
➞Bạn giải ra là xong
a) Ta có: (x+1)(x+9)=(x+3)(x+5)
\(\Leftrightarrow x^2+10x+9=x^2+8x+15\)
\(\Leftrightarrow x^2+10x+9-x^2-8x-15=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow2\left(x-3\right)=0\)
Vì 2>0
nên x-3=0
hay x=3
Vậy: x=3
b) Ta có: \(\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+4x+4+2x-8=x^2-6x+8\)
\(\Leftrightarrow x^2+6x-4-x^2+6x-8=0\)
\(\Leftrightarrow12x-12=0\)
\(\Leftrightarrow12\left(x-1\right)=0\)
Vì 12>0
nên x-1=0
hay x=1
Vậy: x=1
