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Bài làm:
1) Ta có: \(2x^2+5xy+2y^2\)
\(=\left(2x^2+4xy\right)+\left(xy+2y^2\right)\)
\(=2x\left(x+2y\right)+y\left(x+2y\right)\)
\(=\left(2x+y\right)\left(x+2y\right)\)
2) Ta có: \(2x^2+2xy-4y^2\)
\(=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)\)
\(=2x\left(x-y\right)+4y\left(x-y\right)\)
\(=2\left(x+2y\right)\left(x-y\right)\)
\(1)2x^2+5xy+2y^2=2x^2+4xy+xy+2y^2=\left(2x^2+4xy\right)+\left(xy+2y^2\right)=2x\left(x+2y\right)+y\left(x+2y\right)=\left(2x+y\right)\left(x+2y\right)\)\(2)2x^2+2xy-4y^2=2x^2+4xy-2xy-4y^2=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)=2x\left(x-y\right)+4y\left(x-y\right)=\left(2x+4y\right)\left(x-y\right)\)
\(2x^2+y^2-3xy+3x-2y+1=0\)
\(2,25x^2-2.1,5.x\left(y-1\right)+\left(y-1\right)^2-0,25x^2=0\)
\(\left(1,5x-y+1\right)^2-\left(0,5x\right)^2=0\)
\(\left(1,5x-y+1-0,5x\right)\left(1,5x-y+1+0,5x\right)=0\)
\(\left(x-y+1\right)\left(2x-y+1\right)=0\)
Đề bài là j thì b tự lm nhé~
\(x^3-3xy^2-2y^3\)
\(=x^3-xy^2-2xy^2-2y^3\)
\(=x\left(x^2-y^2\right)-2y^2\left(x+y\right)\)
\(=x\left(x-y\right)\left(x+y\right)-2y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy-2y^2\right)\)
\(=\left(x+y\right)\left(x^2-2xy+xy-2y^2\right)\)
\(=\left(x+y\right)\left[x\left(x-2y\right)+y\left(x-2y\right)\right]\)
\(=\left(x+y\right)^2\left(x-2y\right)\)
\(x^3-3xy^2-2y^3\)
\(=x^3-xy^2-2xy^2-2y^3\)
\(=x\left(x^2-y^2\right)-2y^2\left(x+y\right)\)
\(=x\left(x-y\right)\left(x+y\right)-2y^2\left(x+y\right)\)
\(=\left(x+y\right)\left[x\left(x+y\right)-2y^2\right]\)
\(=\left(x+y\right)\left(x^2+xy-2y^2\right)\)
\(=\left(x+y\right)\left(x^2+2xy-xy-2y^2\right)\)
\(=\left(x+y\right)\left[x\left(x-2y\right)-y\left(x-2y\right)\right]\)
\(=\left(x-y\right)\left(x-y\right)\left(x-2y\right)\)
\(=\left(x-y\right)^2\left(x-2y\right)\)
Thiếu y3 nha bạn :
\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
Làm tính nhân
(4x3+3xy2-2y3).(3x2-5xy-6y2)
=12x5+12y5-20x4y-36x2y3-8xy4
Phân tích đa thức thành nhân tử
10x3+5x2y-10x2y-10xy2+5y3
=10x3-5x2y-10xy2+5y3
=5(2x3-x2y-2xy2+y3-)
\(x^4+y^2-2x^2y+x^2+2x-2y\)
\(=\left(y^2-x^2y-xy\right)-\left(x^2y-x^4-x^3\right)+\left(xy-x^3-x^2\right)-\left(2y-2x^2-2x\right)\)
\(=y\left(y-x^2-x\right)-x^2\left(y-x^2-x\right)+x\left(y-x^2-x\right)-2\left(y-x^2-x\right)\)
\(=\left(y-x^2+x-2\right)\left(y-x^2-x\right)\)
Bài làm:
Ta có: \(a^2x^2+b^2y^2-a^2y^2-b^2x^2\)
\(=a^2\left(x^2-y^2\right)-b^2\left(x^2-y^2\right)\)
\(=\left(a^2-b^2\right)\left(x^2-y^2\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(x-y\right)\left(x+y\right)\)
\(a^2x^2+b^2y^2-a^2y^2-b^2x^2\)
\(=\left(a^2x^2-a^2y^2\right)-\left(b^2x^2-b^2y^2\right)\)
\(=a^2\left(x^2-y^2\right)-b^2\left(x^2-y^2\right)\)
\(=\left(a^2-b^2\right)\left(x^2-y^2\right)\)
Bài làm:
Ta có: \(2x^2-3xy-2y^2\)
\(=\left(2x^2-4xy\right)+\left(xy-2y^2\right)\)
\(=2x\left(x-2y\right)+y\left(x-2y\right)\)
\(=\left(2x+y\right)\left(x-2y\right)\)
\(2x^2-3xy-2y^2\)
\(=\left(2x^2-4xy\right)+\left(xy-2y^2\right)\)
\(=2x\left(x-2y\right)+y\left(x-2y\right)\)
\(=2x\left(x-2y\right)+y\left(x-2y\right)\)