mong mn giải giúp mình ạ
Thực hiện phép nhân đa thức với đa thức
(x^2 - xy + y^2).(x + y)
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TL:
\(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y+xy^2-yx^2-xy^2-y^3\)
\(=x^3-y^3\)
\(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x\left(x^2+xy+y^2\right)-y\left(x^2+xy+y^2\right)\)
\(=\left(x^3+x^2y+xy^2\right)-\left(x^2y+xy^2+y^3\right)\)
\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3\)
\(=x^3-y^3\)
`@` `\text {Ans}`
`\downarrow`
\((x+y)(x-y)+(xy^4-x^3y^2) \div (xy^2) \)
`= x(x-y) + y(x-y) + xy^4 \div xy^2 - x^3y^2 \div xy^2`
`= x^2 - xy + xy - y^2 + y^2 - x^2`
`= (x^2 - x^2) + (-xy + xy) + (-y^2 + y^2)`
`= 0`
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
(5x - 2y)(x2 - xy + 1)
= 5x3 - 5x2y + 5x - 2x2y + 2xy2 - 2y
= 5x3 - 7x2y + 2xy2 + 5x - 2y
(x - 1)(x + 1)(x + 2)
= (x2 - 1)(x + 2)
= x3 + 2x2 - x - 2
1/2x2y2(2x + y)(2x - y)
= 1/2x2y2(4x2 - y2)
= 2x4y2 - 1/2x2y4
h) \(y\left(y-x\right)^3-x\left(x-y\right)^2+xy\left(x-y\right)=y\left(y-x\right)^3-x\left(y-x\right)^2-xy\left(y-x\right)=\left(y-x\right)\left[y\left(y-x\right)^2-x-xy\right]=\left(y-x\right)\left[y\left(y^2-2xy+x^2\right)-x-xy\right]=\left(y-x\right)\left(y^3-2xy^2+x^2y-x-xy\right)\)
i) \(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2=10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(a-2b\right)^2=\left(a-2b\right)^2\left(10x^2-x^2-2\right)=\left(a-2b\right)^2\left(9x^2-2\right)\)
\(\dfrac{xy}{2}-x+\dfrac{x^2}{4}=x\left(\dfrac{y}{2}-1+\dfrac{x}{4}\right)\)
Ta có: \(x^2y-xy^2+y^2z-yz^2+xz^2-x^2z=xy\left(x-y\right)-z\left(x^2-y^2\right)+z^2\left(x-y\right)\)
\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)
\(=\left(x-y\right)\left(x\left(y-z\right)-z\left(y-z\right)\right)=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
\(x\left(x-y\right)^2-y\left(x-y\right)^2+xy^2-x^2y\)
\(=\left(x-y\right)^2\left(x-y\right)-xy\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2xy+y^2-xy\right)\)
\(=\left(x-y\right)\left(x^2-3xy+y^2\right)\)
\(=x^3+x^2y-x^2y-xy^2+xy^2+y^3=x^3+y^3\)
(x2 - xy + y2) . ( x + y )
= ( x + y ).(x2 - xy + y2)
= x3 + y3