\(M=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)

\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-xy-y^2+2y+y+x-2+2019\)

\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(y+x-2\right)+2019\)

\(\Rightarrow M=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)+2019\)

\(\Rightarrow M=\left(x^2-y+1\right)\left(x+y-2\right)+2019\)

\(\Rightarrow M=\left(x^2-y+1\right).0+2019\)

\(\Rightarrow M=0+2019\)

\(\Rightarrow M=2019\)

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

\(M = x^3 + x^2y - 2x^2 - xy - y^2 + 3y + x + 2017.\)

\(M=(x^3+x^2y-2x^2)-(xy-y^2+2y)+(x+y-2)+2019\)

\(M=x^2.(x+y-2)-y.(x-y+2)+(x+y-2)+2019\)

\(M=x^2.0-y.0+0+2019\)

\(M=0-0+0+2019\)

\(M=2019\)

$M={}^{}+{}^{}y-2{}^{}-xy-{}^{}+3y+x+2017.$

$M=({}^{}+{}^{}y-2{}^{})-(xy-{}^{}+2y)+(x+y-2)+2019$

$M={}^{}.(x+y-2)-y.(x-y+2)+(x+y-2)+2019$

$M={}^{}.0-y.0+0+2019$

\(P=\left[\left(\dfrac{-1}{3}\right)^2x^3+\left(2x^2\right)^2+\dfrac{1}{2}\right]-\left[x\left(\dfrac{1}{3}x\right)^2+\dfrac{3}{2^3}+x^4\right]+\left(y-2013\right)^2=\left(\dfrac{1}{9}x^3+4x^4+\dfrac{1}{2}\right)-\left(\dfrac{1}{9}x^3+x^4+\dfrac{3}{8}\right)+\left(y-2013\right)^2=3x^4+\dfrac{1}{8}+\left(y-2013\right)^2\ge\dfrac{1}{8}\).

Dấu "=" xảy ra khi x = 0; y = 2013.

ko trả lời thì thôi đừng nhắn bậy

đúng ko trả lời cứ nhắn bậy

M=(x3+x2y-2x2)+(2y-y2-xy)+(x+y-2)+2020

M=x2(x+y-2)+y(2-y-x)+(x+y-2)+2020

M=x2.0+y.0+0+2020

M=2020

Vậy M=2020

không hiểu chỗ nào hỏi mình nha!

khó quá

Đề sai r bn, nếu x,y thay đổi thì tổng biểu thức cũng thay đổi

-Đề sai.

\(25P=\dfrac{x\left(2+3\right)^2}{2x+x+y+z}+\dfrac{y\left(2+3\right)^2}{2y+x+y+z}+\dfrac{z\left(2+3\right)^2}{2z+x+y+z}\)

\(25P\le x\left(\dfrac{2^2}{2x}+\dfrac{3^2}{x+y+z}\right)+y\left(\dfrac{2^2}{2y}+\dfrac{3^2}{x+y+z}\right)+z\left(\dfrac{2^2}{2z}+\dfrac{3^2}{x+y+z}\right)\)

\(25P\le6+\dfrac{9\left(x+y+z\right)}{x+y+z}=15\)

\(\Rightarrow P\le\dfrac{3}{5}\)

Dấu "=" xảy ra khi \(x=y=z\)

\(A=x^3+x^2y-2x^2-xy-y^2+3y+x+2019\)

\(=x^3+x^2\left(2-x\right)-2x^2-y\left(x+y\right)+3y+x+2019\)

\(=x^3+2x^2-x^3-2x^2-2y+3y+x+2019\)

\(=x+y+2019=2021\)

1q

Lời giải:

a. $=(x-y)(x+y)=[(-1)-(-3)][(-1)+(-3)]=2(-4)=-8$
b. $=3x^4-2xy^3+x^3y^2+3x^2y+12xy+15y-12xy-12$

$=3x^4-2xy^3+x^3y^2+3x^2y+15y-12$
=3-2.1(-2)^3+1^3.(-2)^2+3.1^2(-2)+15(-2)-12$
$=-25$
c.

$=2x^4+3x^3y-4x^3y-12xy+12xy=2x^4-x^3y$

$=x^3(2x-y)=(-1)^3[2(-1)-2]=-1.(-4)=4$

d. 

$=2x^2y+4x^2-5xy^2-10x+3xy^2-3x^2y$

$=(2x^2y-3x^2y)+4x^2+(-5xy^2+3xy^2)-10x$

$=-x^2y+4x^2-2xy^2-10x$

$=-3^2.(-2)+4.3^2-2.3(-2)^2-10.3=0$