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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: \(2x^2+2xy-4y^2\)
\(=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)\)
\(=2x\left(x-y\right)+4y\left(x-y\right)\)
\(=2\left(x+2y\right)\left(x-y\right)\)
\(1)2x^2+5xy+2y^2=2x^2+4xy+xy+2y^2=\left(2x^2+4xy\right)+\left(xy+2y^2\right)=2x\left(x+2y\right)+y\left(x+2y\right)=\left(2x+y\right)\left(x+2y\right)\)\(2)2x^2+2xy-4y^2=2x^2+4xy-2xy-4y^2=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)=2x\left(x-y\right)+4y\left(x-y\right)=\left(2x+4y\right)\left(x-y\right)\)
=(x^5-5x^2)-(2xy+10y)
=(x^5-5x^2)-(2xy-10y)
=x^2.(x-5)-2y.(x-5)
=(x^2-2y).(x-5)
k nha
\(x^3-5x^2-2xy+10y=x^2\left(x-5\right)-2y\left(x-5\right)=\left(x-5\right)\left(x^2-2y\right)=\left(x-5\right)\left(x-2y\right)\left(x+2y\right)\)
\(x^2-2xy+5x-10y\)
\(=x\left(x-2y\right)+5\left(x-2y\right)\)
\(=\left(x+5\right)\left(x-2y\right)\)
\(x^2-2xy+5x-10y\)
\(=\left(x^2-2xy\right)+\left(5x-10y\right)\)
\(=x\left(x-2y\right)+5\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+5\right)\)
\(x-3\sqrt{x}+\sqrt{xy}-3y\)
\(=\left(x-3\sqrt{x}\right)+\left(\sqrt{xy}-3y\right)\)
\(=\sqrt{x}\left(\sqrt{x}-3\right)+y\left(\sqrt{x}-3\right)\)
\(=\left(\sqrt{x}-3\right)\left(\sqrt{x}+y\right)\)
\(x^2-5x+2xy-10y\)
\(=\left(x^2-5x\right)+\left(2xy-10y\right)\)
\(=x\left(x-5\right)+2y\left(x-5\right)\)
\(=\left(x+2y\right)\left(x-5\right)\)
\(=x^2y\left(x-5\right)-2y\left(x-5\right)+0\)
\(=\left(x-5\right)\left(x^2y-2y\right)\)
xong phân tích nốt cái bậc 2 kia để được max điểm :)))
\(2xy-x^2+3y^2-4y+1\)
\(=-\left(x^2-2xy+y^2\right)+4y^2-4y+1\)
\(=-\left(x-y\right)^2+\left(2y-1\right)^2\)
\(=\left(2y-1+x-y\right)\left(2y-1-x+y\right)\)
\(=\left(y+x-1\right)\left(3y-x-1\right)\)
Cách 1: \(x^2-2xy+y^2+4x-4y-5=\left(y^2-xy+y\right)+\left(-xy+x^2-x\right)+\left(-5y+5x-5\right)\)
\(=y\left(y-x+1\right)-x\left(y-x+1\right)-5\left(y-x+1\right)=\left(y-x+1\right)\left(y-x-5\right)\)
Cách 2: \(x^2-2xy+y^2+4x-4y-5=\left(x^2+y^2+2^2-2xy+4x-4y\right)-9\)
\(=\left(y-x-2\right)^2-3^2=\left(y-x-2-3\right)\left(y-x-2+3\right)=\left(y-x-5\right)\left(y-x+1\right)\)
\(g,\left(4x^2+8x+4\right)-x^2\)
\(=\left(2x+2\right)^2-x^2\)
\(=\left(3x+2\right)\left(x+2\right)\)
\(=2xy+5x+4y^2+10y\)
\(=x\left(2y+5\right)+2y\left(2y+5\right)\)
\(=\left(x+2y\right)\left(2y+5\right)\)
\(=2x\left(x+2y\right)+5\left(x+2y\right)=\left(x+2y\right)\left(2x+5\right)\)