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]

k đi rồi giúp cho

k rồi giúp cho

\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

bạn gõ lại công thức cho rõ đi, khó đọc quá

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: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)

\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)

\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)

\(=-\left(x-1\right)\left(2x^3+2x^2+7x^2+7x+3x+3\right)\)

\(=-\left(x-1\right)\left(x+1\right)\left(2x^2+7x+3\right)\)

\(=-\left(x-1\right)\left(x+1\right)\cdot\left(x+3\right)\left(2x+1\right)\)

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) \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+2xy-zx-zy+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-zx-zy\right)\)

\(B=x^8+2x^5-2x^4+x^2-2x-100+10x\left(x^4+x\right)+\left(5x-1\right)^2\)

\(=x^8+2x^5-2x^4+x^2-2x-100+10x^5+25x^2-10x+1\)

\(=x^8+12x^5-2x^4+36x^2-12x-99\)

\(=x^8+6x^5+9x^4+6x^5+36x^2+54x-11x^4-66x-99\)

\(=x^4\left(x^4+6x+9\right)+6x\left(x^4+6x+9\right)-11\left(x^4+6x+9\right)\)

\(=\left(x^4+6x+9\right)\left(x^4+6x-11\right)\)

\(x^4+2x^3+10x-25\)

\(=x^4+5x^2+2x^3+10x-5x^2-25\)

\(=\left(x^2+5\right)\left(x^2+2x-5\right)\)

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)

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)