Toán [18×(x-y)^3+6×(y-x)^4-15×(x-y)^3+(y-x)^2]÷(x-y)^2 17/09/2021 By Piper [18×(x-y)^3+6×(y-x)^4-15×(x-y)^3+(y-x)^2]÷(x-y)^2
Đáp án: [18.$(x-y)^{3}$ +6.$(y-x)^{4}$ -15. $(x-y)^{3}$ +$(y-x)^{2}$ ]:$(x-y)^{2}$ =[18.$(x-y)^{3}$ +6.$(x-y)^{4}$ -15. $(x-y)^{3}$ +$(x-y)^{2}$ ]:$(x-y)^{2}$ =$(x-y)^{2}$ [18(x-y)+6$(x-y)^{2}$ -15.(x-y)+1]:$(x-y)^{2}$ =18(x-y)-15(x-y)+6$(x-y)^{2}$ +1 =6$(x-y)^{2}$ +3(x-y)+1 Trả lời
`[18(x-y)^3+6(y-x)^4-15(x-y)^3+(y-x)^2]/(x-y)^2`
`<=>(3(x-y)^3+6(y-x)^4+(y-x)^2)/((x-y)^2)`
Đáp án:
[18.$(x-y)^{3}$ +6.$(y-x)^{4}$ -15. $(x-y)^{3}$ +$(y-x)^{2}$ ]:$(x-y)^{2}$
=[18.$(x-y)^{3}$ +6.$(x-y)^{4}$ -15. $(x-y)^{3}$ +$(x-y)^{2}$ ]:$(x-y)^{2}$
=$(x-y)^{2}$ [18(x-y)+6$(x-y)^{2}$ -15.(x-y)+1]:$(x-y)^{2}$
=18(x-y)-15(x-y)+6$(x-y)^{2}$ +1
=6$(x-y)^{2}$ +3(x-y)+1