Cho x-y-2 = 0 . Tính
A= x3 + x2y - 2x2 - xy - y2 + 3y + x -1
B = x3 + x2y - 2x2 - x2y - xy2 + 2xy + 2y +2x - 2
C = x4 + 2x3y - 2x3 + x2y2 - 2x2y - x(x+y) + 2x + 3
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\(A=5x^2y-xy^2+4xy+6\) 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) x2y+xy+x+1= (x2y+xy)+(x+1)=xy(x+10+(x+1)=(x+1)(xy+1)
b) x2-(a+b)x+ab=x2-ax-bx+ab=(x2-ax)-(bx-ab)=x(x-a)-b(x-a)=(x-a)(x-b)
c) ax2+ay-bx2-by=(ax2+ay)-(bx2+by)=a(x2+y)-b(x2+y)=(a-b)(x2+y)
d) ax-2x-a2+2a=(ax-2x)-(a2-2a)=x(a-2)-a(a-2)=(a-2)(x-a)
e) 2x2+4ax+x+2a=(2x2+4ax)+(x+2a)=2x(x+2a)+(x+2a)=(x+2a)(2x+1)
f) x3+ax2+x+a=(x3+ax2)+(x+a)=x2(x+a)+(x+a)=(x2+1)(x+a)
\(M=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)
\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-xy-y^2+2y+y+x-2+2019\)
\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(y+x-2\right)+2019\)
\(\Rightarrow M=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)+2019\)
\(\Rightarrow M=\left(x^2-y+1\right)\left(x+y-2\right)+2019\)
\(\Rightarrow M=\left(x^2-y+1\right).0+2019\)
\(\Rightarrow M=0+2019\)
\(\Rightarrow M=2019\)
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: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)
\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)
b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)
\(\dfrac{2x-3y}{xy^2}=\dfrac{\left(2x-3y\right)\left(x+y\right)}{xy^2\left(x+y\right)}\)
c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)
\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)
\(A=x^3+x^2y-2x^2-xy-y^2+3y+x+2019\)
\(=x^3+x^2\left(2-x\right)-2x^2-y\left(x+y\right)+3y+x+2019\)
\(=x^3+2x^2-x^3-2x^2-2y+3y+x+2019\)
\(=x+y+2019=2021\)
a) \(=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
\(=\left(x-1\right)^2\left(x^2+x+1\right)\)
b) \(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
c) Đổi đề: \(a^2x+a^2y-7x-7y\)
\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)
d) \(=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)
e) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\left(x^2-x+1\right)\)
g) \(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)
i) \(=\left(x+1\right)^2-4=\left(x+1-2\right)\left(x+1+2\right)=\left(x-1\right)\left(x+3\right)\)
a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)
e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)
i)\(=\left(x-1\right)^2-4=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\)