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Ta có: \(\left(x-y\right)^3+4y\left(2x^2+y^2\right)\)
\(=x^3-3x^2y+3xy^2-y^3+8x^2y+4y^3\)
\(=x^3+5x^2y+3xy^2+3y^3\)
\(=x^3+3x^2y+3xy^2+y^3+2x^2y+2y^3\)
\(=\left(x+y\right)^3+2y\left(x^2+y^2\right)\)
\(VT=\dfrac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}=\dfrac{\left(x+y\right)\left(x+2y\right)}{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}=\dfrac{1}{x-y}\)
\(\text{a) }\left(x-1\right)\left(x^2+y\right)-\left(x^2-y\right)\left(x-2\right)-x\left(x+2y\right)+3\left(y-5\right)\)
\(=\left(x^3+xy-x^2-y\right)-\left(x^3-2x^2-xy+2y\right)-\left(x^2+2xy\right)+\left(3y-15\right)\)
\(=x^3+xy-x^2-y-x^3+2x^2+xy-2y-x^2-2xy+3y-15\)
\(=\left(x^3+x^3\right)+\left(-x^2+2x^2-x^2\right)+\left(xy+xy-2xy\right)+\left(-y-2y+3y\right)-15\)
\(=0+0+0+0-15\)
\(=-15\)
\(\text{b) }6\left(x^3y+x-3\right)-6x\left(2xy^3+1\right)-3x^2y\left(2x-4y^2\right)\)
\(=\left(6x^3y+6x-18\right)-\left(12x^2y^3+6x\right)-\left(6x^3y-12x^2y^3\right)\)
\(=6x^3y+6x-18-12x^2y^3-6x-6x^3y+12x^2y^3\)
\(=\left(6x^3y-6x^3y\right)+\left(6x-6x\right)+\left(-12x^2y^3+12x^2y^3\right)-18\)
\(=0+0+0-18\)
\(=-18\)
\(\text{c) }\left(x^2+2xy+4y^2\right)\left(x-2y\right)-6\left(\frac{1}{2}-\frac{4}{3}y^3\right)\)
\(=\left(x^3-2x^2y+2x^2y-4xy^2+4xy^2-8y^3\right)-\left(3-8y^3\right)\)
\(=\left(x^3-8y^3\right)-\left(3-8y^3\right)\)
\(=x^3-8y^3-3+8y^3\)
\(=x^3-3\)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
Ta phân tích mẫu:
\(x^3+2x^2y-xy^2-2y^3\)
\(=x^3+3x^2y+2xy^2-x^2y-3xy^2-2y^3\)
\(=x\left(x^2+3xy+2y^2\right)-y\left(x^2+3xy+2y^2\right)\)
\(=\left(x-y\right)\left(x^2+3xy+2y^2\right)\)
Thay vào ta có:
\(\frac{x^2+3xy+2y^2}{\left(x-y\right)\left(x^2+3xy+2y^2\right)}=\frac{1}{x-y}\)
Vậy ta có điều phải chứng minh
\(\left(x-y\right)^3+4y\left(2x^2+y^2\right)=\left(x+y\right)^3+2y\left(x^2+y^2\right)\)
\(\Leftrightarrow x^3-3x^2y+3xy^2-y^3+8x^2y+4y^3=x^3+3x^2y+3xy^2+y^3+2x^2y+2y^3\)
\(\Leftrightarrow\left(-3x^2y+8x^2y\right)+3xy^2+3y^3=\left(3x^2y+2x^2y\right)+3xy^2+3y^2\)
\(\Leftrightarrow5x^2y+3xy^2+3y^2=5x^2y+3xy^2+3y^2\)