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B) Ta có: 2x-2y-x2+2xy-y2
⇔ 2(x-y)-(x2-2xy+y2)
⇔ 2(x-y)-(x-y)2
⇔ (x-y)(2-x+y)
Đúng thì tick nhé
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\(\dfrac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\dfrac{\left(x+y\right)\left(x+2y\right)}{x\left(x^2-y^2\right)+2y\left(x^2-y^2\right)}\)
\(=\dfrac{x+y}{x^2-y^2}\)
\(=\dfrac{1}{x-y}\)
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a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)
\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)
\(=6x^2-3x+\dfrac{5}{2}\)
b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)
\(=3x-y-y-x+2x^2-2x\)
\(=2x^2-2y\)
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a: =xy(1/3+4-2)=7/3xy
b: =xy^2(-1+3/2+4/3)=(1/3+3/2)xy^2=11/6xy^2
c: =4x^2y^2+2/3x^2y^2-4/3x^2y=-4/3x^2y+14/3x^2y^2
d: =3x^2y^2z+4x^2y^2z-8x^2y^2z=-x^2y^2z
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Mình nghĩ là phân tích đa thức
a)\(3x+2y+xy+6\)
\(=x\left(y+3\right)+2\left(y+3\right)\)
\(=\left(x+2\right)\left(y+3\right)\)
b)\(2x^2+3xy-2y^2-10x-5y+12\)
\(=2x^2+\left(3y-10\right)x-\left(2y^2+5y-12\right)\)
\(=\left[2x+\left(y-4\right)\right]\left(x+2y+3\right)\)
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\(B=\dfrac{x^3+2x^2y-xy^2-2y^3}{x^2+3xy+2y^2}\)
\(B=\dfrac{x^2\left(x+2y\right)-y^2\left(x+2y\right)}{x^2+xy+2xy+2y^2}\)
\(B=\dfrac{\left(x+2y\right)\left(x^2-y^2\right)}{x\left(x+y\right)+2y\left(x+y\right)}\)
\(B=\dfrac{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(2y+x\right)}\)
\(B=x-y\)\(\left(\text{Đ}K:x+2y\ne0;x+y\ne0\right)\)
Tham khảo nhé~
\(B=\dfrac{x^3+2x^2y-xy^2-2y^3}{x^2+3xy+2y^2}\)
\(=\dfrac{x^2\left(x+2y\right)-y^2\left(x+2y\right)}{x^2+xy+2xy+2y^2}\)
\(=\dfrac{\left(x^2-y^2\right)\left(x+2y\right)}{x\left(x+y\right)+2y\left(x+y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x+2y\right)}{\left(x+2y\right)\left(x+y\right)}\)
\(=x-y\)