Đáp án:
`x=2` hoặc `x=-4.`
Giải thích các bước giải:
`i)(x^2+2x+1)(x^2+2x+3)=99`
`<=>(x^2+2x+1)(x^2+2x+1+2)=99`
`<=>(x^2+2x+1)(x^2+2x+1)+2(x^2+2x+1)=99`
`<=>(x^2+2x+1)^2+2(x^2+2x+1)-99=0(1)`
Ta có:`x^2+2x+1`
`\qquad =x^2+x+x+1`
`\qquad =x(x+1)+x+1`
`\qquad =(x+1)(x+1)=(x+1)^2>=0`
`(1)<=>(x^2+2x+1)^2-9(x^2+2x+1)+11(x^2+2x+1)-99=0`
`<=>(x^2+2x+1)(x^2+2x+1-9)+11(x^2+2x+1-9)=0`
`<=>(x^2+2x+1-9)(x^2+2x+1+11)=0`
Vì `x^2+2x+1=(x+1)^2>=0=>x^2+2x+1+11>=11>0`
`=>x^2+2x+1-9=0`
`<=>x^2+2x+1=9`
`<=>(x+1)^2=9=3^2=(-3)^2`
`<=>[(x+1=3),(x+1=-3):}`
`<=>[(x=2),(x=-4):}`
Vậy `x=2` hoặc `x=-4.`