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\(M=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)
\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-xy-y^2+2y+y+x-2+2019\)
\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(y+x-2\right)+2019\)
\(\Rightarrow M=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)+2019\)
\(\Rightarrow M=\left(x^2-y+1\right)\left(x+y-2\right)+2019\)
\(\Rightarrow M=\left(x^2-y+1\right).0+2019\)
\(\Rightarrow M=0+2019\)
\(\Rightarrow M=2019\)
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M = x3 + x2y - 2x2 - xy - y2 + 3y + x + 2017
M = (x3 + x2y - 2x2) - (xy + y2 - 2y) + (x + y - 2) + 2019
M = x2. (x + y - 2) - y(x + y - 2) + (x + y - 2) + 2019 = 2019
\(M = x^3 + x^2y - 2x^2 - xy - y^2 + 3y + x + 2017.\)
\(M=(x^3+x^2y-2x^2)-(xy-y^2+2y)+(x+y-2)+2019\)
\(M=x^2.(x+y-2)-y.(x-y+2)+(x+y-2)+2019\)
\(M=x^2.0-y.0+0+2019\)
\(M=0-0+0+2019\)
\(M=2019\)
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\(A=x^3+x^2y-2x^2-xy-y^2+3y+x+2019\)
\(=x^3+x^2\left(2-x\right)-2x^2-y\left(x+y\right)+3y+x+2019\)
\(=x^3+2x^2-x^3-2x^2-2y+3y+x+2019\)
\(=x+y+2019=2021\)
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p = x3 + x2y - 2x2- y ( x + y ) + 3y+ x + 2018
= x2( x + y - 2 ) - y ( x + y ) + 3y +x + 2018
= 0- y x2 + 3y + x + 2018
= y ( -2 + 3 ) + x + 2018
= ( y + x ) + 2018
= 2 + 2018
= 2020
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Bài 1:
Ta thấy: $(x+\frac{1}{2})^2\geq 0$ với mọi $x\in\mathbb{R}$
$\Rightarrow (x+\frac{1}{2})^2+\frac{5}{4}\geq \frac{5}{4}$
Vậy gtnn của biểu thức là $\frac{5}{4}$
Giá trị này đạt tại $x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}$
Bài 2:
$x+y-3=0\Rightarrow x+y=3$
\(M=x^2(x+y)-(x+y)x^2-y(x+y)+4y+x+2019\)
\(=-3y+4y+x+2019=x+y+2019=3+2019=2022\)
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a: \(=3x^4+3x^2y^2+2x^2y^2+2y^4+y^2\)
\(=\left(x^2+y^2\right)\left(3x^2+2y^2\right)+y^2\)
\(=3x^2+3y^2=3\)
b: \(=7\left(x-y\right)+4a\left(x-y\right)-5=-5\)
c: \(=\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(y-x\right)+3=3\)
d: \(=\left(x+y\right)^2-4\left(x+y\right)+1\)
=9-12+1
=-2
ta có:P= x3 + x2y - 2x2 - y (x + y) + 3y +x +2018
Suy ra P= x2(x+y-2)-y(x+y)+3y+x+2018
Thay x+y=2 Vào biểu thức P= x2(x+y-2)-y(x+y)+3y+x+2018
Suy ra :P=x2(2-2)-y2+3y+x+2018
P=0-y(-2+3)+x+2018
ta có:P= x3 + x2y - 2x2 - y (x + y) + 3y +x +2018
Suy ra P= x2(x+y-2)-y(x+y)+3y+x+2018
Thay x+y=2 Vào biểu thức P= x2(x+y-2)-y(x+y)+3y+x+2018
Suy ra :P=x2(2-2)-y2+3y+x+2018
P=0-y(-2+3)+x+2018
P=0-(-y)+x+2018
P= y+x+2018
ta có:P= x3 + x2y - 2x2 - y (x + y) + 3y +x +2018
Suy ra P= x2(x+y-2)-y(x+y)+3y+x+2018
Thay x+y=2 Vào biểu thức P= x2(x+y-2)-y(x+y)+3y+x+2018
Suy ra :P=x2(2-2)-y2+3y+x+2018
P=0-y(-2+3)+x+2018
P=0-(-y)+x+2018
ta có:P= x3 + x2y - 2x2 - y (x + y) + 3y +x +2018
Suy ra P= x2(x+y-2)-y(x+y)+3y+x+2018
Thay x+y=2 Vào biểu thức P= x2(x+y-2)-y(x+y)+3y+x+2018
Suy ra :P=x2(2-2)-y2+3y+x+2018
P=0-y(-2+3)+x+2018
P=0-(-y)+x+2018
P=y+x+2018
P=2+2018
P=2020
Vậy P=2020 do x+y=2
Ta có x + y = 2 => x = 2 - y
Thay x = 2 - y vào biểu thức P, ta có:
\(\left(2-y\right)^3+\left(2-y\right)^2y-2\left(2-y\right)-y\left(2-y+y\right)+3y+2-y+2018\)
= \(\left(2-y\right)^2\left(2-y+y\right)-4+2y-2y+3y+2-y+2018\)
= \(2\left(2-y\right)^2-4+2y+2+2018\)
= \(2\left(2-y\right)^2+2016+2y\)
Vậy giá trị của biểu thức P là 2 (2 - y)2 + 2016 + 2y khi x + y = 2.