\(x^3\left(x^2-7\right)^2-36x\)

\(=x.\left[x^2.\left(x^2-7\right)^2-36\right]\)

\(=x.\left[\left(x^3-7x\right)^2-6^2\right]\)

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

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

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

Ta có : \(x^3\left(x^2-7\right)^2-36x\)

= \(x^3\left(x^4-14x^2+49\right)-36x\)

= \(x\left(x^6-14x^4+49x^2-36\right)\)

= \(x\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)\)---- chỗ này tắt 

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

\(x\left(x^2\left(x^2-7\right)^2-6^2\right)=x\left(x^3-7-6\right)\left(x^3-7+6\right)=x\left(x^3-13\right)\left(x^3-1\right)\)

5^x+3 là số chẵn, 5^y+4 là số lẻ. Phân tích 516 = 2x2x3x43

do đó, 5^y+4 = 129, vậy y=3

5^x+3 = 4, nên x=0

a)    \(3x^2-7x+2\)

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

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

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

b)   \(a\left(x^2+1\right)-x\left(a^2+1\right)\)

\(=ax^2+a-a^2x-x\)

\(=ax\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(ax-1\right)\)

\(x^5+y^5-\left(x+y\right)^5\)

\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)

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

\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)

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

bạn thu gom 2 đa thức đầu tiên thành 1 nhóm và 2 đa thức sau thành 1 nhóm . sau đó dùng hđt rồi đem chung 

nên nhớ 8=2³

a, = [(x-2).(x+1)]^2+(x-2)^2

    = (x-2)^2.(x+1)^2+(x-2)^2

    = (x-2)^2.[(x+1)^2+1]

    = (x-2)^2.(x^2+2x+2)

Tk mk nha

b)  \(6x^5+15x^4+20x^3+15x^2+6x+1\)

\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)

\(=\left(2x+1\right)\left(3x^4+6x^3+7x^2+4x+1\right)\)

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

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

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

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

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

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

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

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

\(x^4-9x^3+28x^2-36x+16\)

\(=x^4-x^3-8x^3+8x^2+20x^2-20x-16x+16\)

\(=\left(x^4-x^3\right)-\left(8x^3-8x^2\right)+\left(20x^2-20x\right)-\left(16x-16\right)\)

\(=x^3\left(x-1\right)-8x^2\left(x-1\right)+20x\left(x-1\right)-16\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3-8x^2+20x-16\right)\)

\(=\left(x-1\right)\left(x^3-2x^2-6x^2+12x+8x-16\right)\)

\(=\left(x-1\right)[x^2\left(x-2\right)-6x\left(x-2\right)+8\left(x-2\right)]\)

\(=\left(x-1\right)\left(x-2\right)\left(x^2-6x+8\right)\)

\(=\left(x-1\right)\left(x-2\right)\left(x^2-4x-2x+8\right)\)

\(=\left(x-1\right)\left(x-2\right)[x\left(x-4\right)-2\left(x-4\right)]\)

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

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

Để tui cân hết choa 

a, Lời giải 

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

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

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

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

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

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

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