1)  \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)

2)  \(x^3-9x^2+6x+16\)

\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)

\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)

\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)

3)   \(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

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

\(=\left(x-5\right)\left(x^2-x-1\right)\)

4)  \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

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

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

5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)

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

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

gửi phần này trước còn lại làm sau !!! tk mk nka !!!

nhiều thế

6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)

\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)

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

7)   \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)          (NHÂN x + 2 vs x +  5  và  x + 3 vs x + 4 )

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

ĐẶT   \(x^2+7x+11=y\)   ta được :  

\(\left(y+1\right)\left(y-1\right)-24=y^2-1-24\)

\(=y^2-25=\left(y-5\right)\left(y+5\right)\)

8)  \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)

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

9) sai đề rùi bạn ơi ! đề đúng nè 

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

Ta thấy :  

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

Thay vào biểu thức bài cho ta được : 

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

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

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

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

bài ở trên câu 3 : kết luận là  \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs  

`@` `\text {Ans}`

`\downarrow`

`a)`

Thu gọn:

`P(x)=`\(5x^4 + 3x^2 - 3x^5 + 2x - x^2 - 4 +2x^5\)

`= (-3x^5 + 2x^5) + 5x^4 + (3x^2 - x^2) + 2x - 4`

`= -x^5 + 5x^4 + 2x^2 + 2x - 4`

`Q(x) =`\(x^5 - 4x^4 + 7x - 2 + x^2 - x^3 + 3x^4 - 2x^2\)

`= x^5 + (-4x^4 + 3x^4) - x^3 + (x^2 - 2x^2) + 7x - 2`

`= x^5 - x^4 - x^3 - x^2 + 7x - 2`

`@` Tổng:

`P(x)+Q(x)=`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) + (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 + x^5 - x^4 - x^3 - x^2 + 7x - 2`

`= (-x^5 + x^5) - x^3 + (5x^4 - x^4) + (2x^2 - x^2) + (2x + 7x) + (-4-2)`

`= 4x^4 - x^3 + x^2 + 9x - 6`

`@` Hiệu:

`P(x) - Q(x) =`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) - (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 - x^5 + x^4 + x^3 + x^2 - 7x + 2`

`= (-x^5 - x^5) + (5x^4 + x^4) + x^3 + (2x^2 + x^2) + (2x - 7x) + (-4+2)`

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

`b)`

`@` Thu gọn:

\(H (x) = ( 3x^5 - 2x^3 + 8x + 9) - ( 3x^5 - x^4 + 1 - x^2 + 7x)\)

`= 3x^5 - 2x^3 + 8x + 9 - 3x^5 + x^4 - 1 + x^2 - 7x`

`= (3x^5 - 3x^5) + x^4 - 2x^3 - x^2 + (8x + 7x) + (9+1)`

`= x^4 - 2x^3 - x^2 + 15x + 10`

\(R( x) = x^4 + 7x^3 - 4 - 4x ( x^2 + 1) + 6x\)

`= x^4 + 7x^3 - 4 - 4x^3 - 4x + 6x`

`= x^4 + (7x^3 - 4x^3) + (-4x + 6x) - 4`

`= x^4 + 3x^3 + 2x - 4`

`@` Tổng:

`H(x)+R(x)=` \((x^4 - 2x^3 - x^2 + 15x + 10)+(x^4 + 3x^3 + 2x - 4)\)

`= x^4 - 2x^3 - x^2 + 15x + 10+x^4 + 3x^3 + 2x - 4`

`= (x^4 + x^4) + (-2x^3 + 3x^3) - x^2 + (15x + 2x) + (10-4)`

`= 2x^4 + x^3 - x^2 + 17x + 6`

`@` Hiệu: 

`H(x) - R(x) =`\((x^4 - 2x^3 - x^2 + 15x + 10)-(x^4 + 3x^3 + 2x - 4)\)

`=x^4 - 2x^3 - x^2 + 15x + 10-x^4 - 3x^3 - 2x + 4`

`= (x^4 - x^4) + (-2x^3 - 3x^3) - x^2 + (15x - 2x) + (10+4)`

`= -5x^3 - x^2 + 13x + 14`

`@` `\text {# Kaizuu lv u.}`

1. 2x(3x² - 5x + 3) = 6x³ - 10x² + 6x

2. \(-\dfrac{1}{2}x^2\left(2x^3-4x+3\right)=-x^5+2x^3+\dfrac{-3}{2}x^2\)

3. -2x(x² + 5x - 3) = -2x³ - 10x² + 6x

4. x(3x² - 2x + 5) = 3x³ - 2x² + 5x

5. 3xy²(2x - 4y + 3xy) = 6x²y² - 12xy³ = 9x²y³

a,\(x^3-7x+6\)

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

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

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

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

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

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

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

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

b,\(x^3-9x^2+6x+16\)

\(=x^3-8x^2-x^2+8x-2x+16\)

\(=\left(x^3-8x^2\right)-\left(x^2-8x\right)-\left(2x-16\right)\)

\(=x^2.\left(x-8\right)-x.\left(x-8\right)-2.\left(x-8\right)\)

\(=\left(x-8\right).\left(x^2-x-2\right)\)

\(=\left(x-8\right).\left(x^2+x-2x-2\right)\)

\(=\left(x-8\right).\left[\left(x^2+x\right)-\left(2x+2\right)\right]\)

\(=\left(x-8\right).\left[x.\left(x+1\right)-2.\left(x+1\right)\right]\)

\(=\left(x-8\right).\left(x+1\right).\left(x-2\right)\)

c,\(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

\(=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)

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

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

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

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

\(=\left(x-5\right).\left[x.\left(x+2\right)-3.\left(x+2\right)\right]\)

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

Chúc bạn học tốt!!!

d,\(2x^3-x^2+5x+3\)

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

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

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

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

e, \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=\left(27x^2-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)

\(=9x^2.\left(3x-1\right)-6x.\left(3x-1\right)+4.\left(3x-1\right)\)

\(=\left(3x-1\right).\left(9x^2-6x+4\right)\)

Chúc bạn học tốt!!!

7, \(\left(x+2\right).\left(x+3\right).\left(x+4\right).\left(x+5\right)-24\)

\(=\left[\left(x+2\right).\left(x+5\right)\right].\left[\left(x+3\right).\left(x+4\right)\right]-24\)

\(=\left(x^2+5x+2x+10\right).\left(x^2+4x+3x+12\right)-24\)

\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)(1)

Đặt \(t=x^2+7x+10\Rightarrow t+2=x^2+7x+12\)

\(\Rightarrow\left(1\right)=t.\left(t+2\right)-24\)

\(=t^2+2t-24=t^2-4t+6t-24\)

\(=\left(t^2-4t\right)+\left(6t-24\right)=t.\left(t-4\right)+6.\left(t-4\right)\)

\(=\left(t-4\right).\left(t+6\right)\) (2)

Vì \(t=x^2+7x+10\) nên:

(2) \(=\left(x^2+7x+10-4\right).\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right).\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right).\left(x^2+7x+16\right)\)

\(=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)

\(=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)

\(=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)

Chúc bạn học tốt!!!

Chia nhỏ ra cậu ơi :v

Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((

1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)

b: \(\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)

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

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

c: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=5x^3+14x^2+12x+8\)

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^4-10x^3+6x^2\)

c: \(=-x^5+2x^3-\dfrac{3}{2}x^2\)

d: \(=2x^3+10x^2-8x-x^2-5x+4=2x^3+9x^2-13x+4\)