Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
`a, (2x-3)^3 = 8x^3 - 36x^2 + 54x - 27`
`b, (a+3b)^3 = a^3 + 9a^2b + 27ab^2 + 27b^3`
`c, (xy-1)^3 = x^3y^3 - 3x^2y^2 + 3xy -1`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(3x-5\right)\left(3x+5\right)\)
\(=\left(3x\right)^2-5^2\)
\(=9x^2-25\)
b) \(\left(x-2y\right)\left(x+2y\right)\)
\(=x^2-\left(2y\right)^2\)
\(=x^2-4y^2\)
c) \(\left(-x-\dfrac{1}{2}y\right)\left(-x+\dfrac{1}{2}y\right)\)
\(=\left(-x\right)^2-\left(\dfrac{1}{2}y\right)^2\)
\(=x^2-\dfrac{1}{4}y^2\)
`a, (3x-5)(3x+5) = 9x^2 - 25`
`b, (x-2y)(x+2y) = x^2 -4y^2`
`c, (-x-1/2y)(-x+1/2y) = x^2 - 1/4y^2`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-5\right)\left(a^2+5a+25\right)\)
\(=a^3-5^3\)
\(=a^3-125\)
b) \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(=x^3+\left(2y\right)^3\)
\(=x^3+8y^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(4x^4-8x^2y^2+12x^5y\right):\left(-4x^2\right)\)
\(=4x^4:-4x^2-8x^2y^2:-4x^2+12x^4y:-4x^2\)
\(=-x^2+2y^2-3x^2y\)
b) \(x^2\left(x-y^2\right)-xy\left(1-xy\right)-x^3\)
\(=x^3-x^2y^2-xy+x^2y^2-x^3\)
\(=-xy\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)
b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)
\(\dfrac{2\left(x-1\right)}{x^2}.\dfrac{x}{\left(x-1\right)}=\dfrac{2\left(x-1\right)x}{x^2\left(x-1\right)}=\dfrac{2}{x}\)
B2:
a)\(2x^2-4x+2=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)
b)\(x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv
`a, (a-1)(a+1)(a^2+1)`
`= (a^2-1)(a^2+1)`
`= a^4-1`
`b, (xy+1)^2 - (xy-1)^2`
`= x^2y^2 + 2xy + 1 - x^2y^2 + 2xy - 1`
`= 4xy`
a) \(\left(a-1\right)\left(a+1\right)\left(a^2+1\right)\)
\(=\left(a^2-1\right)\left(a^2+1\right)\)
\(=a^4-1\)
b) \(\left(xy+1\right)^2-\left(xy-1\right)^2\)
\(=\left[\left(xy+1\right)-\left(xy-1\right)\right]\left[\left(xy+1\right)+\left(xy-1\right)\right]\)
\(=\left(xy+1-xy+1\right)\left(xy+1+xy-1\right)\)
\(=4xy\)