Toán 8(x+1/x)^2+4(x^2+1/x^2)^2=(x+4)^2+4(x^2+1/x^2)(x+1/x)^2 17/09/2021 By Samantha 8(x+1/x)^2+4(x^2+1/x^2)^2=(x+4)^2+4(x^2+1/x^2)(x+1/x)^2
(`x≠0`) Đặt `t=x+1/x` `<=>t^2-2=x^2+1/(x^2)` Pt `<=>8t^2+4(t^2-2)^2=(x+4)^2+4(t^2-2)t^2` `<=>8t^2+4t^4-16t^2+16-4t^4+8t^2=(x+4)^2` `<=>(x+4)^2=16` `<=>x=0` hoặc `x=-8` Vậy `S={-8}` Trả lời
(`x≠0`)
Đặt `t=x+1/x`
`<=>t^2-2=x^2+1/(x^2)`
Pt `<=>8t^2+4(t^2-2)^2=(x+4)^2+4(t^2-2)t^2`
`<=>8t^2+4t^4-16t^2+16-4t^4+8t^2=(x+4)^2`
`<=>(x+4)^2=16`
`<=>x=0` hoặc `x=-8`
Vậy `S={-8}`