Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,Q=\left(-2x^3y+7x^2y+3xy\right)+P=\left(-2x^3y+7x^2y+3xy\right)+\left(3x^2y-2xy^2-4xy+2\right)\\ =-2x^3y+7x^2y+3xy+3x^2y-3xy^2-4xy+2\\ =-2x^3y^2+10x^2y-3xy^2-xy+2\)
\(b,M=\left(3x^2y^2-5x^2y+8xy\right)-P\\ =\left(3x^2y^2-5x^2y+8xy\right)-\left(3x^2y-2xy^2-4xy+2\right)\\ =3x^2y^2-5x^2y+8xy-3x^2y^2+2xy^2+4xy-2\\ =-3x^2y+12xy-2\)
1a) (x - 2y) (x2 - 2xy + y2)
= (x - 2y) (x - y)2
= x2 - xy - 2xy + 2y2
= (x2 - xy) - (2xy - 2y2)
= x (x - y) - 2y (x - y)
= (x - y) (x - 2y)
2a) x (x - 3) - y (3 - x)
= x (x - 3) + y (x - 3)
= (x - 3) (x + y)
b) 3x2 - 5x - 3xy + 5y
= (3x2 - 3xy) - (5x - 5y)
= 3x (x - y) - 5 (x - y)
= (x - y) (3x - 5)
3) 12x (3 - 4x) + 7 (4x - 3) = 0
12x (3 - 4x) - 7 (3 - 4x) = 0
(3 - 4x) (12x - 7) = 0
=> 3 - 4x = 0 hoặc 12x - 7 = 0
* 3 - 4x = 0 => x = \(\frac{3}{4}\)
* 12x - 7 = 0 => x = \(\frac{7}{12}\)
Vậy x =\(\frac{3}{4}\)hoặc x =\(\frac{7}{12}\)
\(c,3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=3x^3y^2\left(x+1\right)+3y^2\left(x+1\right)\)
\(=3y^2\left(x+1\right)\left(x^3+1\right)\)
\(=3y^2\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
a, \(\left(xy+1\right)^2-\left(x+y\right)^2\)
= \(\left(xy\right)^2+2xy+1-\left(x^2+2xy+y^2\right)\)
= \(x^2y^2+2xy+1-x^2-2xy-y^2\)
= \(x^2y^2-x^2-y^2+1\) = \(\left(x^2y^2-x^2\right)-y^2+1=x^2\left(y^2-1\right)-\left(y^2-1\right)=\left(y^2-1\right)\left(x^2-1\right)\)
b,\(3x^4y^2+3x^3y^2+3xy^2+3y^2\)
= \(\left(3x^4y^2+3x^3y^2\right)+\left(3xy^2+3y^2\right)\)
= \(3x^3y^2\left(x+1\right)+3y^2\left(x+1\right)=3y^2\left(x+1\right)\left(x^3+1\right)\)
Bài làm :
\(\text{a)}9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)
\(\text{b)}3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=\left(3x^4y^2+3xy^2\right)+\left(3x^3y^2+3y^2\right)\)
\(=3xy^2\left(x^3+1\right)+3y^2\left(x^3+1\right)\)
\(=\left(3xy^2+3y^2\right)\left(x^3+1\right)\)
\(=3y^2\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
\(\text{c)}\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2+x+y^2+y+1-xy\right)\)
\(d ) x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
Đây, bản full đây thím, tớ thực sự đã kiên nhẫn lắm đấy ...
a)\(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)=4\left(x^2-y^2-2x+2ay-a^2+1\right)\)
\(=4\left[\left(x^2-2x+1\right)-\left(a^2-2ay+y^2\right)\right]\)
\(=4\left[\left(x-1\right)^2-\left(a-y\right)^2\right]\)
\(=4\left(x-1-a+y\right)\left(x-1+a-y\right)\)
b)\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1\right)-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)
c)\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x^2-1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)\left(5x+5\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)\)
\(=\left(x-1\right)\left(x+3\right)^2\)
d)\(a^5+a^4+a^3+a^2+a+1=a^4\left(a+1\right)+a^2\left(a+1\right)+\left(a+1\right)\)
\(=\left(a+1\right)\left(a^4+a^2+1\right)\)
\(=\left(a+1\right)\left(a^4+2a^2+1-a^2\right)\)
\(=\left(a+1\right)\left[\left(a^2+1\right)^2-a^2\right]\)
\(=\left(a+1\right)\left(a^2-a+1\right)\left(a^2+a+1\right)\)
e)\(x^3-3x^2+3x-1-y^3=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)
\(=\left(x-1-y\right)\left(x^2-2x+1+xy-y+y^2\right)\)
f)\(5x^3-3x^2y-45xy^2+27y^3=5x\left(x^2-9y^2\right)-3y\left(x^2-9y^2\right)\)
\(=\left(x^2-9y^2\right)\left(5x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y\right)\left(5x-3y\right)\)
g)\(3x^2\left(a-b+c\right)+36xy\left(a-b+c\right)+108y^2\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(3x^2+36xy+108y^2\right)\)
\(=3\left(a-b+c\right)\left(x^2+12xy+36y^2\right)\)
\(=3\left(a-b+c\right)\left(x+6y\right)^2\)
a/ \(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)\)
\(=\left(4x^2-8x+4\right)-\left(4y^2-8ay+4a^2\right)\)
\(=\left(2x-2\right)^2-\left(2y-2a\right)^2=\left(2x-2+2y-2a\right)\left(2x-2-2y+2a\right)\)
b/ \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)=\left(x+y-1\right)\left(x^2+y^2+2xy+x+y+1\right)-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)\)
Giải giúp bạn 2 bài tiêu biểu thôi nha
\(\begin{array}{l}A.\left( { - 3xy} \right) = 9{x^3}y + 3x{y^3} - 6{x^2}{y^2}\\ \Rightarrow A = \left( {9{x^3}y + 3x{y^3} - 6{x^2}{y^2}} \right):\left( { - 3xy} \right)\\ = 9{x^3}y:\left( { - 3xy} \right) + 3x{y^3}:\left( { - 3xy} \right) - 6{x^2}{y^2}:\left( { - 3xy} \right)\\ = - 3{x^2} - {y^2} + 2xy\end{array}\)