NY
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NM
1
4 tháng 8 2015
x^4+ 2x^2 + 5x^3 + 10x- 2x^2 - 4
= x^2. [x^2+ 2 ] +5.x[x^2+2] - 2.[x^2+2]
= [x^2+5x-2] . [x^2+2]
HN
2
DN
0
28 tháng 10 2021
\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
10 tháng 9 2021
a: Ta có: \(-3x^4+20x^3-35x^2-10x+48\)
\(=-\left(3x^4-20x^3+35x^2+10x-48\right)\)
\(=-\left(3x^4-9x^3-11x^3+33x^2+2x^2-6x+16x-48\right)\)
\(=-\left(x-3\right)\left(3x^3-11x^2+2x+16\right)\)
\(=-\left(x-3\right)\left(3x^3-6x^2-5x^2+10x-8x+16\right)\)
\(=-\left(x-3\right)\left(x-2\right)\left(3x^2-5x-8\right)\)
\(=-\left(x-3\right)\left(x-2\right)\left(3x-8\right)\left(x+1\right)\)
b: Ta có: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)
\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)
\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)
\(=-\left(x-1\right)\left(2x^3+2x^2+7x^2+7x+3x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\left(2x^2+7x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\cdot\left(x+3\right)\left(2x+1\right)\)
24 tháng 9 2019
a) \(x^4+5x^3+10x-4\)
\(=\left(x^4+2x^2\right)+\left(5x^3+10x\right)-\left(2x^2+4\right)\)
\(=x^2\left(x^2+2\right)+5x\left(x^2+2\right)-2\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2+5x-2\right)\)
\(=\left(x^2+2\right)\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}-2\right)\)
\(=\left(x^2+2\right)\left[\left(x+\frac{5}{2}\right)^2-\frac{33}{4}\right]\)
\(=\left(x^2+2\right)\left[\left(x+\frac{5}{2}\right)^2-\left(\frac{\sqrt{33}}{2}\right)^2\right]\)
\(=\left(x^2+2\right)\left(x+\frac{5}{2}-\frac{\sqrt{33}}{2}\right)\left(x^2+\frac{5}{2}+\frac{\sqrt{33}}{2}\right)\)
b) \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+2xy-zx-zy+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-zx-zy\right)\)
HN
25 tháng 8 2021
\(B=x^8+2x^5-2x^4+x^2-2x-100+10x\left(x^4+x\right)+\left(5x-1\right)^2\)
\(=x^8+2x^5-2x^4+x^2-2x-100+10x^5+25x^2-10x+1\)
\(=x^8+12x^5-2x^4+36x^2-12x-99\)
\(=x^8+6x^5+9x^4+6x^5+36x^2+54x-11x^4-66x-99\)
\(=x^4\left(x^4+6x+9\right)+6x\left(x^4+6x+9\right)-11\left(x^4+6x+9\right)\)
\(=\left(x^4+6x+9\right)\left(x^4+6x-11\right)\)
17 tháng 10 2021
\(x^4+2x^3+10x-25\)
\(=x^4+5x^2+2x^3+10x-5x^2-25\)
\(=\left(x^2+5\right)\left(x^2+2x-5\right)\)
29 tháng 9 2016
a, x^2 + 5x +4
= x^2 + 1x + 4x + 4
= (x^2 + 1x) + (4x + 4)
= x ( x + 1 ) + 4 ( x + 1 )
= (x + 1) (x + 4)
b, x^2 - 6x + 5
= x^2 - 1x - 5x + 5
= (x^2 - 1x) - (5x - 5)
= x (x - 1) - 5 (x - 1)
= (x - 1) (x - 5)
c, x^2 + 7x + 12
= x^2 + 3x + 4x + 12
= (x^2 + 3x) + (4x + 12)
= x (x + 3) + 4 (x + 3)
= (x + 3) (x + 4)
d, 2x^2 - 5x + 3
= 2^x2 - 2x - 3x + 3
= 2x (x - 1) - 3 (x - 1)
= (x-1) (2x - 3)
e, 7x - 3x^2 - 4
= 3x + 4x - 3x^2 - 4
= (3x - 3x^2) + (4x - 4)
= 3x (1 - x) + 4 (x - 1)
= 3x (1-x) - 4 (1 - x)
= (1 - x) (3x - 4)
f, x^2 - 10x + 16
= x^2 - 2x - 8x + 16
= (x^2 - 2x) - (8x - 16)
= x (x - 2) - 8 (x - 2)
= (x - 2) (x - 8)
VD
29 tháng 9 2016
a, (x+1)(x+4)
b,(x-5)(x-1)
c,(x+3)(x+4)
d,(2x-3)(x-1)
e,(-3x+4)(x-1)
f, (x-8)(x-2)
\(x^4+5x^3+10x-4\)
\(=x^4+5x^3-2x^2+2x^2+10x-4\)
\(=x^2\left(x^2+5x-2\right)+2\left(x^2+5x-2\right)\)
\(=\left(x^2+2\right)\left(x^2+5x-2\right)\)
Mình cũng vừa làm được cách 2:
\(x^4+5x^3+10x-4\)
=\(x^4-4+5x^3+10x\)
=\(\left(x^2+2\right)\left(x^2-2\right)+5x\left(x^2+2\right)\)
=\(\left(x^2+2\right)\left(x^2+5x-2\right)\)