Bài 3:
a:=>x^2-3x+36=x^2
=>-3x=-36
=>x=12
b: \(\Leftrightarrow\left(x+7\right)\cdot\left(4x-5\right)=0\)
=>x=5/4 hoặc x=-7
c: \(\Leftrightarrow x^2=\left(10x-5\right)^2\)
=>(10x-5-x)(10x-5+x)=0
=>(9x-5)(11x-5)=0
=>x=5/9 hoặc x=5/11
d: \(\Leftrightarrow8x^2+12x-10x-15=0\)
=>(2x+3)(4x-5)=0
=>x=5/4 hoặc x=-3/2