c)(2x+5)^2 = (x+2)^2
=>(2x+5)^2 - (x+2)^2=0
<=>(2x+5+x+2).(2x+5-x-2)=0
<=>(3x+7).(x+3)=0
=> 3x+7=0 =>x=-7/3
hoặc x+3=0=>x=-3
d) x^2-5x+6=0
<=>x^2-4x+4-x+2=0
<=>(x-2)^2-(x-2)=0
<=>(x-2)(x-3)=0
=> x-2=0 =>x=2
hoặc x-3=0 =>x=3
e)x^3+27+(x+3)(x-9)=0
<=>x^3+3^3+(x+3)(x-9)=0
<=>(x+3)(x^2-3x+3^2)+(x+3)(x-9)=0
<=>(x+3)(x^2-3x+3^2+x-9)=0
<=>(x+3)(x^2-2x)=0
<=>x(x+3)(x-2)=0
=>x=0
hoặc x+3=0 =>x=-3
hoặc x-2=0 =>x=2
f)4(x-2)^2=25(1-2x)^2
<=>[2(x-2)]^2=[5(1-2x)]^2
<=>(2x-4)^2=(5-10x)^2
=>(2x-4)^2-(5-10x)^2=0
<=>(2x-4+5-10x)(2x-4-5+10x)=0
<=>(1-8x)(12x-9)=0
=> 1-8x=0 => x=1/8
hoặc 12x-9=0 =>x=3/4
Chúc em học tốt nha😁😊😉
