\(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)

\(=a^4\left(a+b-a-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)

\(=-a^4\left(c-a\right)-a^4\left(a-b\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)

\(=\left(b^4-a^4\right)\left(c-a\right)+\left(c^4-a^4\right)\left(a-b\right)\)

\(=\left(b^2+a^2\right)\left(b^2-a^2\right)\left(c-a\right)+\left(c^2-a^2\right)\left(c^2+a^2\right)\left(a-b\right)\)

\(=\left(b^2+a^2\right)\left(b-a\right)\left(b+a\right)\left(c-a\right)+\left(c-a\right)\left(c+a\right)+\left(c^2+a^2\right)\left(a-b\right)\)

\(=\left(b-a\right)\left(c-a\right)[\left(b^2+a^2\right)\left(a+b\right)-\left(c+a\right)\left(c^2+a^2\right)]\)

\(=\left(b-a\right)\left(c-a\right)\left(ab^2+a^3+b^3+a^2b-c^3-ac^2-a^3-a^2c\right)\)

\(=\left(b-a\right)\left(c-a\right)\left(ab^2+b^3+a^2b-c^3-ac^2-a^2c\right)\)

\(=\left(b-a\right)\left(c-a\right)[\left(ab^2-ac^2\right)+\left(a^2b-a^2c\right)+\left(b^3+c^3\right)]\)

\(=\left(b-a\right)\left(c-a\right)[a\left(b^2-c^2\right)+a^2\left(b-c\right)+\left(b-c\right)\left(b^2+bc+c^2\right)]\)

\(=\left(b-a\right)\left(c-a\right)\left(b-c\right)\left(ab+ac+a^2+b^2+c^2+bc\right)\)

x⁴ - 4x³ - 8x² + 8x 

 = x(x³ - 4x² - 8x + 8) 

= x[x³ + 8 - 4x(x + 2)] 

= x[(x + 2)(x² - 2x + 4) - 4x(x + 2)] 

= x(x + 2)(x² - 6x + 4)

= x(x + 2)(x² - 6x + 9 - 5) 

 = \(x\left(x+2\right)\left[\left(x-3\right)^2-5\right]=x\left(x+2\right)\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)\)

\(x^4-4x^3-8x^2+8x\)

\(=x\left(x^3-4x^2-8x+8\right)\)

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

\(=x\left[\left(x^3-6x^2+4x\right)+\left(2x^2-12x+8\right)\right]\)

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

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

\(=x\left[\left(x-3\right)^2-\left(\sqrt{5}\right)^2\right]\left(x+2\right)\)

\(=x\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)

1, bạn xem lại đề nhé 

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

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

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

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

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

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

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

Phân tích đa thức thành nhân tử:

x^3-x+y^3-y 

= (y+x)(y^2-xy+x^2-1)

x^2+x-6 

= (x-2)(x+3)

2x^2+3x-5

= (x-1)(2x+5)

x^4+3x^3+x+3

= (x+1)(x+3)(x^2-x+1)

\(x^3-x+y^3-y\)

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

\(x^2+x-6\)

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

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

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

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

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

\(x^4+3x^3+x+3\)

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

\(\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

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

Trả lời:

c, \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-9.x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9x^2+18x+9=15\)

\(\Leftrightarrow45x+9=15\)

\(\Leftrightarrow45x=6\)

\(\Leftrightarrow x=\frac{2}{15}\)

Vậy x = 2/15 là nghiệm của pt.

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

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

\(\Leftrightarrow x^3-27+4x-x^3=1\)

\(\Leftrightarrow4x-27=1\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

Vậy x = 7 là nghiệm của pt.

e, \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-19\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-19\)

\(\Leftrightarrow12x-4=-19\)

\(\Leftrightarrow12x=--15\)

\(\Leftrightarrow x=-\frac{5}{4}\).

Vậy x = - 5/4 là nghiệm của pt. 

g, \(3x\left(x-2\right)-x\left(1+3x\right)=14\)

\(\Leftrightarrow3x^2-6x-x-3x^2=14\)

\(\Leftrightarrow-7x=14\)

\(\Leftrightarrow x=-2\)

Vậy x = - 2 là nghiệm của pt.

h, \(\left(3x+1\right)^2+\left(5x-2\right)^2=34\left(x+2\right)\left(x-2\right)\)

\(\Leftrightarrow9x^2+6x+1+25x^2-20x+4=34\left(x^2-4\right)\)

\(\Leftrightarrow34x^2-14x+5=34x^2-136\)

\(\Leftrightarrow34x^2-14x+5-34x^2+136=0\)

\(\Leftrightarrow-14x+151=0\)

\(\Leftrightarrow-14x=-151\)

\(\Leftrightarrow x=\frac{151}{14}\)

Vậy x = 151/14 là nghiệm của pt.

i, \(x^3+3x^2+3x+1=125\)

\(\Leftrightarrow\left(x+1\right)^3=5^3\)

\(\Rightarrow x+1=5\)

\(\Leftrightarrow x=4\)

Vậy x = 4 là nghiệm của pt.

cam on ban nhieu lam

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

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

\(=x\)

\(2x\left(x-3y\right)-\left(8x^3y-12x^2y^2\right):2xy\)

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

\(=-2x^2\)

a, \(A=\left(5x-2\right)\left(x+1\right)-\left(x-3\right)\left(5x+1\right)-17\left(x+3\right)\)

\(=5x^2+3x-2-5x^2+14x+3-17x-51=-50\)

Vậy biểu thức ko phụ thuộc giá trị biến x 

b, \(B=\left(6x-5\right)\left(x+8\right)-\left(3x-1\right)\left(2x+3\right)-9\left(4x-3\right)\)

\(=6x^2+43x-40-6x^2-7x+3-36x+27=-10\)

Vậy biểu thức ko phụ thuộc giá trị biến x 

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

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

Vậy biểu thức ko phụ thuộc giá trị biến x 

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

Vậy biểu thức ko phụ thuộc giá trị biến x