Question

Tìm x

(x+1)²=x+1                                             (x+3)²=9

x²+8x+16=0                                           4x²-12x=-9

4(x-1)²-9(x+2)²=0                                   (x-2)²=(3x+1)²                                        64=(x-3)²

Answer 1

(x+1)²=x+1

x²+2x+1-x-1=0

x²+x=0

x(x+1)=0

*x=0

*x+1=0 =>x=-1

Answer 2

(x+1)²=x+1

<=>(x+1)²-(x+1)=0

<=>(x+1)(x+1-1)=0

<=>(x+1).x=0

<=>x+1=0 hoặc x=0

<=>x=-1 hoặc x=0

x²+8x+16=0

<=>(x+4)²=0

<=>x+4=0

<=>x=-4

4.(x-1)²-9.(x+2)²=0

<=>[2.(x-1)]²-[3.(x+2)]²=0

<=>[(2x-2)-(3x+6)][(2x-2)+(3x+6)]=0

<=>(2x-2-3x-6)(2x-2+3x+6)=0

<=>(-x-8)(5x+4)=0

<=>-x-8=0 hoặc 5x+4=0

<=>x=-8 hoặc x=-4/5

(x+3)²=9

<=>(x+3)²-9=0

<=>(x+3-3)(x+3+3)=0

<=>x.(x+6)=0

<=>x=0 hoặc x+6=0

<=>x=0 hoặc x=-6

4x²-12x=-9

<=>4x²-12x+9=0

<=>(2x-3)²=0

<=>2x-3=0

<=>x=3/2

(x-2)²=(3x+1)²

<=>(x-2)²-(3x+1)²=0

<=>[(x-2)-(3x+1)][(x-2)+(3x+1)]=0

<=>(x-2-3x-1)(x-2+3x+1)=0

<=>(-2x-3)(4x-1)=0

<=>-2x-3=0 hoặc 4x-1=0

<=>x=-3/2 hoặc x=1/4

64=(x-3)²

<=>64-(x-3)²=0

<=>[8-(x-3)][8+(x-3)]=0

<=>(8-x+3)(8+x-3)=0

<=>(11-x)(x+5)=0

<=>11-x=0 hoặc x+5=0

<=>x=11 hoặc x=-5