cho x+y-2=0 tính M=x^3+x^2y-2x^2-xy-y^2+3y+x+2020 08/09/2021 Bởi Piper cho x+y-2=0 tính M=x^3+x^2y-2x^2-xy-y^2+3y+x+2020
$M=x^3+x^2y-2x^2-xy-y^2+3y+x+2020$ $M=(x^3+x^2y-2x^2)-xy-y^2+2y+y+x-2+2022$ $M=(x^3+x^2y-2x^2)-(xy+y^2-2y)+(y+x-2)+2022$ $M=x^2(x+y-2)-y(x+y-2)+(x+y-2)+2022$ $\text{mà:x+y-2=0}$ $=>M=x^2.0-y.0+0+2022$ $M=0-0+0+2022$ $M=2022$ Bình luận
từ x+y-2=0=>x+y=2 ta có: M=x³+x²y-2x²-xy-y²+3y+x+2020 =x²(x+y)-2x(x+y)+y(x+y)-2y²+2y+y+x+2020 =(x+y)(x²-2x+y+1)-2y²+2y+2020 =2(x²-3x+x+y+1)-2y²+2y+2020 =2(x²-3x+3)-2y²+2y+2020 =2x²-6x+6-2y²+2y+2020 =2(x+y)(x-y)-6x+2y+2026 =4x-4y-6x+2y+2026 =-2x-2y+2026 =-2(x+y)+2026 =2022 Bình luận
$M=x^3+x^2y-2x^2-xy-y^2+3y+x+2020$
$M=(x^3+x^2y-2x^2)-xy-y^2+2y+y+x-2+2022$
$M=(x^3+x^2y-2x^2)-(xy+y^2-2y)+(y+x-2)+2022$
$M=x^2(x+y-2)-y(x+y-2)+(x+y-2)+2022$
$\text{mà:x+y-2=0}$
$=>M=x^2.0-y.0+0+2022$
$M=0-0+0+2022$
$M=2022$
từ x+y-2=0=>x+y=2
ta có:
M=x³+x²y-2x²-xy-y²+3y+x+2020
=x²(x+y)-2x(x+y)+y(x+y)-2y²+2y+y+x+2020
=(x+y)(x²-2x+y+1)-2y²+2y+2020
=2(x²-3x+x+y+1)-2y²+2y+2020
=2(x²-3x+3)-2y²+2y+2020
=2x²-6x+6-2y²+2y+2020
=2(x+y)(x-y)-6x+2y+2026
=4x-4y-6x+2y+2026
=-2x-2y+2026
=-2(x+y)+2026
=2022