Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(50x^5-8x^3\)
\(=2x^3\left(25x^2-4\right)\)
\(=2x^3\left(5x-2\right)\left(5x+2\right)\)
b: \(x^4-5x^2-4y^2+10y\)
\(=\left(x^2-2y\right)\left(x^2+2y\right)-5\left(x^2-2y\right)\)
\(=\left(x^2-2y\right)\left(x^2+2y-5\right)\)
c: \(36a^2+12a+1-b^2\)
\(=\left(6a+1\right)^2-b^2\)
\(=\left(6a+1-b\right)\left(6a+1+b\right)\)
d: \(x^3+y^3-xy^2-x^2y\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+y\right)\cdot\left(x-y\right)^2\)
e: Ta có: \(4x^2+4x-3\)
\(=4x^2+6x-2x-3\)
\(=2x\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(2x-1\right)\)
f: Ta có: \(9x^4+16x^2-4\)
\(=9x^4+18x^2-2x^2-4\)
\(=9x^2\left(x^2+2\right)-2\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(9x^2-2\right)\)
g: Ta có: \(-6x^2+5xy+4y^2\)
\(=-6x^2+8xy-3xy+4y^2\)
\(=-2x\left(3x-4y\right)-y\left(3x-4y\right)\)
\(=\left(3x-4y\right)\left(-2x-y\right)\)
h: Ta có: \(\left(x^2+4x\right)^2+8\left(x^2+4x\right)+15\)
\(=\left(x^2+4x\right)^2+3\left(x^2+4x\right)+5\left(x^2+4x\right)+15\)
\(=\left(x^2+4x+3\right)\cdot\left(x^2+4x+5\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+5\right)\)
\(-9x^4+3x^2+2\\ =-9x^4+6x^2-3x^2+2\\ =-3x^2\left(3x^2-2\right)-\left(3x^2-2\right)\\ =-\left(3x^2-2\right)\left(3x^2+1\right)\)
\(-9x^4+3x^2+2\)
\(=-9x^4+6x^2-3x^2+2\)
\(=-3x^2\left(3x^2-2\right)-\left(3x^2-2\right)\)
\(=\left(3x^2-2\right)\left(-3x^2-1\right)\)
a) \(9x^2-16\)
\(=\left(3x\right)^2-4^2\)
\(=\left(3x-4\right)\left(3x+4\right)\)
b) \(x^2+4xy+4y^2-3x-6y\)
\(=\left(x^2+4xy+4y^2\right)-\left(3x+6y\right)\)
\(=\left[x^2+2\cdot x\cdot2y+\left(2y\right)^2\right]-3\left(x+2y\right)\)
\(=\left(x+2y\right)^2-3\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+2y-3\right)\)
#\(Toru\)
2) 9x2+ 12x+ 4
<=>(3x)2+ 2.3x.2+ 22 <=>(3x+ 2)2
3) 4x4+ 20x2+ 25
<=>(2x2)2+ 2.2x2.5+ 52 <=>(2x2+5)2
4) 25x2- 20xy+ 4y2
<=> (5x)2- 2.5x.2y+ (2y)2<=> (5x-2y)2
5) 9x4- 12x2y+ 4y2
<=> (3x2)2- 2.3x2.2.y+ (2y)2<=> (3x2- 2y)2
6) 4x4- 16x2y3+ 16y6
<=> (2x2)2- 2.2x2.4y3+ (4y3)2<=> (2x2- 4y3)2
7) 9x4- 12x5+ 4x6
<=> (3x2)2- 2.3x2.2x3+ (2x3)2<=> (3x2- 2x3)2
\(a^3+a+30\)
\(=a^3+3a^2-3a^2-9a+10a+30\)
\(=\left(a+3\right)\left(a^2-3a+10\right)\)
\(x^3+x^2+100\)
\(=x^3+5x^2-4x^2-20x+20x+100\)
\(=\left(x+5\right)\left(x^2-4x+20\right)\)
a) Ta có: \(x^4+64\)
\(=x^4+16x^2+64-16x^2\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
b) Ta có: \(81x^4+4y^4\)
\(=81x^4+36x^2y^2+4y^4-36x^2y^2\)
\(=\left(9x^2+2y^2\right)^2-\left(6xy\right)^2\)
\(=\left(9x^2-6xy+2y^2\right)\left(9x^2+6xy+2y^2\right)\)
c) Ta có: \(x^5+x+1\)
\(=x^5+x^2-x^2+x-1\)
\(=x^2\left(x^3+1\right)-\left(x^2-x+1\right)\)
\(=x^2\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)
\(9x^2-12xy-20y-25=9x^2-25-4y\left(3x+5\right)\)
\(=\left(3x+5\right)\left(3x-5\right)-4y\left(3x+5\right)=\left(3x+5\right)\left(3x-4y-5\right)\)
\(xy^2-49x^3-28x^2-4x=x\left[y^2-\left(49x^2+28x+4\right)\right]\)
\(=x\left[y^2-\left(7x+2\right)^2\right]=x\left(y+7x+2\right)\left(y-7x-2\right)\)
\(x^2-3x-2019.2022=x^2-3x-2019\left(2019+3\right)\)
\(=x^2-3x-2019^2-3.2019=\left(x-2019\right)\left(x+2019\right)-3\left(x+2019\right)\)
\(=\left(x+2019\right)\left(x-2022\right)\)
a: \(9x^2-12xy-20y-25\)
\(=\left(3x-5\right)\left(3x+5\right)-4y\left(3x+5\right)\)
\(=\left(3x+5\right)\left(3x-5-4y\right)\)
a, 2xy^2 ( x^3 -3xy - 4 )
b, x^2 - 4x - 4x +16
= x(x-4) - 4(x-4)
= (x-4) (x-4)
a, 7x - 14
= 7(x-2)
b, 2x - 2y + \(x^2\)- xy
= (2x-2y) + (\(x^2\)-xy)
= 2(x-y) + x(x-y)
= (x-y)(2+x)
c, 6x + 12
= 6(x+2)
\(a,=7\left(x-2\right)\\ b,=2\left(x-y\right)+x\left(x-y\right)=\left(x+2\right)\left(x-y\right)\\ c,=6\left(x+2\right)\\ d,\text{Sai đề}\)