help me

a)\(x^4+6x^3+11x^2+6x+1\)

\(=x^4+9x^2+1+6x^3+6x+2x^2\)

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

\(x^4+5x^3-12x^2+5x+1\)

\(=\left(x^4-2x^3+x^2\right)+\left(7x^3-14x^2+7x\right)+\left(x^2-2x+1\right)\)

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

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

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

2: =(2x+1)^2-y^2

=(2x+1+y)(2x+1-y)

3: =x^2(x^2+2x+1)

=x^2(x+1)^2

4: =x^2+6x-x-6

=(x+6)(x-1)

5: =-6x^2+3x+4x-2

=-3x(2x-1)+2(2x-1)

=(2x-1)(-3x+2)

6: =5x(x+y)-(x+y)

=(x+y)(5x-1)

7: =2x^2+5x-2x-5

=(2x+5)(x-1)

8: =(x^2-1)*(x^2-4)

=(x-1)(x+1)(x-2)(x+2)

9: =x^2(x-5)-9(x-5)

=(x-5)(x-3)(x+3)

help me

dễ mà bạn xin 20 phút làm ra giấy nhé :)) 

a) \(\left(x^4+6x^3+9x^2\right)+2x^2+6x+1\)

      \(\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)

        \(\left(x^2+3x+1\right)^2\)

b) \(x^4+x^3+x^2+x+1\)

câu b,  chúa sẽ c/m x ko tồn tại , và nó là 1 đa thức bất khả Q . trong R 

vì lớp 8 chưa học đến số phức  

     \(x^4+x^3=-x^2-x-1\)

 \(x^4+x^3+\frac{1}{4}x^2=\left(\frac{1}{4}x^2-x^2\right)-x-1\)

\(\left(x^2+\frac{1}{2}x\right)^2=-\frac{3}{4}x^2-x-1\)

\(4\left(x^2+\frac{1}{2}x\right)^2=-3x^2-4x-4\)

\(\Delta`=\left(-2\right)^2-\left(-4\right).\left(-3\right)=4-12< 0\)

          denta < 0 x vô nghiệm

vậy đa thức trên ko thể phân tích và nó là 1 đa thức bất khả Q

c) ,   

\(\left(6x^4-12x^3\right)+\left(17x^3-34x^2\right)-\left(4x^2-8x\right)-\left(3x-6\right)\)

\(6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)

     \(\left(x-2\right)\left(6x^3+17x^2-4x-3\right)\)

      \(\left(x-2\right)\left\{\left(6x^3+18x^2\right)-\left(x^2+3x\right)-\left(x+3\right)\right\}\)

      \(\left(x-2\right)\left\{6x^2\left(x+3\right)-x\left(x+3\right)-\left(x+3\right)\right\}\)

\(\left(x-2\right)\left(x+3\right)\left(6x^2-x-1\right)\)

 \(\left(x-2\right)\left(x+3\right)\left\{\left(6x^2+\frac{6}{3}x\right)-\left(\frac{9}{3}x+\frac{9}{9}\right)\right\}\)

\(\left(x-2\right)\left(x+3\right)\left\{6x\left(x+\frac{1}{3}\right)-\frac{9}{3}\left(x+\frac{1}{3}\right)\right\}\)

\(\left(X-2\right)\left(X+3\right)\left(X+\frac{1}{3}\right)\left(6x-1\right)\)

d)

\(\left(x^4-x^3\right)+\left(6x^3-6x^2\right)-\left(6x^2-6x\right)-\left(x-1\right)\)

\(x^3\left(x-1\right)+6x^2\left(x-1\right)-6x\left(x-1\right)-\left(x-1\right)\)

\(\left(x-1\right)\left(x^3+6x^2-6x-1\right)\)

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

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

\(\Delta=49-4=45\)

\(x1,2=\frac{-7+\sqrt{45}}{2},\frac{-7-\sqrt{45}}{2}\)

\(\left(x-1\right)^2\left(x-\frac{7+\sqrt{45}}{2}\right)\left(x-\frac{7-\sqrt{45}}{2}\right)\)

1) \(x^3+2x-3\)

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

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

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

2) \(x^3-6x+4\)

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

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

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

3) \(x^3-2x^2+1\)

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

\(=x^2\left(x-1\right)-x\left(x-1\right)-\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-x-1\right)\)

4) \(x^3+5x^2-12\)

\(=\left(x^3+2x^2\right)+\left(3x^2+6x\right)-\left(6x+12\right)\)

\(=x^2\left(x+2\right)+3x\left(x+2\right)-6\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+3x-6\right)\)

5) \(x^3-6x^2+9x\) (chắc đề như vậy)

\(=x\left(x^2-6x+9\right)\)

\(=x\left(x-3\right)^2\)

6) \(4x^3-9x^2+5x\)

\(=x\left(4x^2-9x+5\right)\)

\(=x\left[\left(4x^2-4x\right)-\left(5x-5\right)\right]\)

\(=x\left[4x\left(x-1\right)-5\left(x-1\right)\right]\)

\(=x\left(x-1\right)\left(4x-5\right)\)

x⁴ + x³ + 6x² + 5x + 5

=x⁴+x³+x²+5x²+5x+5

=x².(x²+x+1)+5.(x²+x+1)

=(x²+x+1)(x²+5)

=x^4 +x^3 + x^2 + 5x^2 + 5x + 5

=x^2(x^2+x+1) +5(x^2+x+1)

=(x^2+x+1)(x^2+5)

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

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

\(1.\)

\(4x^2-4x-3\)

\(=4x^2-2x+6x-3\)

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

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

\(2.\)

\(2x^2-5x-3\)

\(=2x^2-6x+x-3\)

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

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

\(3.\)

\(3x^2-5x-2\)

\(=3x^2+x-6x-2\)

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

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

\(4.\)

\(2x^2+5x+2\)

\(=2x^2+4x+x+2\)

\(=2x\left(x+2\right)+\left(x+2\right)\)

\(=\left(2x+1\right)\left(x+2\right)\)

\(5.\)

\(6x^2-x-1\)

\(=6x^2-3x+2x-1\)

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

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

\(6.\)

\(6x^2-6x-3\)

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

\(7.\)

\(15x^2-2x-1\)

\(=15x^2+3x-5x-1\)

\(=3x\left(5x+1\right)-1\left(5x+1\right)\)

\(=\left(5x+1\right)\left(3x-1\right)\)

\(8.\)

\(x^4-13x^2+36\)

\(=\left(x-3\right)\left(x^3+3x^2-4x-12\right)\)

\(=\left(x-3\right)\left(x-2\right)\left(x^2+5x+6\right)\)

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

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)