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\(M=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)
\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-xy-y^2+2y+y+x-2+2019\)
\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(y+x-2\right)+2019\)
\(\Rightarrow M=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)+2019\)
\(\Rightarrow M=\left(x^2-y+1\right)\left(x+y-2\right)+2019\)
\(\Rightarrow M=\left(x^2-y+1\right).0+2019\)
\(\Rightarrow M=0+2019\)
\(\Rightarrow M=2019\)
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
\(M = x^3 + x^2y - 2x^2 - xy - y^2 + 3y + x + 2017.\)
\(M=(x^3+x^2y-2x^2)-(xy-y^2+2y)+(x+y-2)+2019\)
\(M=x^2.(x+y-2)-y.(x-y+2)+(x+y-2)+2019\)
\(M=x^2.0-y.0+0+2019\)
\(M=0-0+0+2019\)
\(M=2019\)
\(A=5x^2y-xy^2+4xy+6\) bậc : 3
a)\(B=-5x^2y+xy^2-4xy-6\)
b)\(=>C=-2xy+1-5x^2y+xy^2-4xy-6\)
\(C=-5x^2y+xy^2-6xy-5\)
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
a: \(A=3\cdot\dfrac{1}{8}\cdot\dfrac{-1}{3}+6\cdot\dfrac{1}{8}\cdot\dfrac{1}{9}+3\cdot\dfrac{1}{2}\cdot\dfrac{-1}{27}\)
\(=-\dfrac{1}{8}+\dfrac{1}{12}-\dfrac{1}{18}\)
\(=-\dfrac{7}{72}\)
b: \(B=\left(-1\cdot3\right)^2+\left(-1\right)\cdot3+\left(-1\right)^3+3^3\)
\(=9-3-1+27=36-4=32\)
c: \(C=-\dfrac{3}{4}xy^2-2x^2y-\dfrac{9}{2}xy\)
\(=\dfrac{-3}{4}\cdot\dfrac{1}{2}\cdot\left(-1\right)^2-2\cdot\dfrac{1}{4}\cdot\left(-1\right)-\dfrac{9}{2}\cdot\dfrac{1}{2}\cdot\left(-1\right)\)
\(=\dfrac{-3}{8}+\dfrac{1}{2}+\dfrac{9}{4}=\dfrac{19}{8}\)
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:
$x+y-3=0\Rightarrow x+y=3$
\(M=x^2(x+y)-(x+y)x^2-y(x+y)+4y+x+2019\)
\(=-3y+4y+x+2019=x+y+2019=3+2019=2022\)
\(A=x^3+x^2y-2x^2-xy-y^2+3y+x+2019\)
\(=x^3+x^2\left(2-x\right)-2x^2-y\left(x+y\right)+3y+x+2019\)
\(=x^3+2x^2-x^3-2x^2-2y+3y+x+2019\)
\(=x+y+2019=2021\)
Lời giải:
\(M=x^3+x^2y-2x^2-xy-y^2+3y+x-1\)
\(=(x^3+x^2y-2x^2)-(xy+y^2-2y)+y+x-1\)
\(=x^2(x+y-2)-y(x+y-2)+(y+x-2)+1\)
\(=x^2.0-y.0+0+1=1\)
\(N=x^3-2x^2-xy^2+2xy+2y-2x-2\)
\(=(x^3-2x^2+x^2y)-(x^2y+xy^2-2xy)+2y+2x-4-4x+2\)
\(=x^2(x-2+y)-xy(x+y-2)+2(y+x-2)-4x+2\)
\(=x^2.0-xy.0+2.0-4x+2=2-4x\) (không tính được giá trị cụ thể, bạn thử xem lại đề)
\(P=(x^4+x^3y-2x^3)+(x^3y+x^2y^2-2x^2y)-x(x+y-2)\)
\(=x^3(x+y-2)+x^2y(x+y-2)-x(x+y-2)\)
\(=x^3.0+x^2y.0-x.0=0\)