\(4.\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)

\(=4.\left[\left(x+5\right)\left(x+12\right)\right].\left[\left(x+6\right)\left(x+10\right)\right]-3x^2\)

\(=4.\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)

Đặt \(a=x^2+16x+60\) ta có :

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

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

Thay a , ta có ;

\(\left(2a-x\right)\left(2a+3x\right)=\left[2.\left(x^2+16x+60\right)-x\right].\left[2.\left(x^2+16x+60\right)+3x\right]\)

\(=\left(2x^2+32x+120-x\right)\left(2x^2+32x+120+3x\right)\)

\(=\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)

\(=\left(2x^2+16x+15x+120\right)\left(2x^2+35x+120\right)\)

\(=\left[2x.\left(x+8\right)+15.\left(x+8\right)\right]\left(2x^2+35x+120\right)\)

\(=\left(x+8\right)\left(2x+15\right)\left(2x^2+35x+120\right)\)

Ta có: ( 4x + 1)(12x - 1)(3x + 2)(x+1) - 4

= [(4x+1)(3x+2)]. [(12x-1)(x+1)] - 4 = (12x² +11x + 2)(12x² + 11x - 1) - 4

Đặt a = 12x² + 11x - 1. Thay vào biểu thức ta có:

(a+3).a - 4 = a² + 3a - 4 =a² + 4a - a - 4 = a(a+4) - (a+4)

= (a+4)(a-1)

=> (4x+1)(12x-1)(3x+2)(x+1) - 4 = (12x² + 11x + 3)(12x²+11x - 2)

Thanks you bn

Đề sai hả bạn?

\(2x^2-x^2-3x-1\)

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

\(=\frac{1}{4}\left(4x^2-12x-4\right)\)

\(=\frac{-1}{4}\left[13-\left(4x-12x+9\right)\right]\)

\(=-\frac{1}{4}\left[13-\left(2x-3\right)^2\right]\)

\(=-\frac{1}{4}\left(\sqrt{13}-2x+3\right)\left(\sqrt{13}+2x-3\right)\)

Nguyễn Công Tỉnh.Bad maths.^-^

\(=x^7+x^6+x^5-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

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

\(x^2-xy-10x+10y\)

\(=x\left(x-y\right)-10\left(x-y\right)\)

\(=\left(x-y\right)\left(x-10\right)\)

đầu tiên , x^4 + x^3 + 2X^2 +x+1 = (X^2)^2 + 2X^2 + 1 + X^3 + X = (x^2+1)^2 + x(X^2 +1) = ... đoạn này tự lm nha

Mình có cách khác :

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

\(4x-x^3+2x^2y-xy^2\)

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

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

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

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

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

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

\(x^4-3x^2+9\)

\(=x^4+6x^2+9-9x^2\)

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

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

x⁴ - 3x² + 9 = x⁴ - 2.x².\(\frac{3}{2}\) + \(\frac{9}{4}\) + \(\frac{27}{4}\) = ( x² - \(\frac{3}{2}\) ) - \(\frac{27}{4}\) = ......ko biết .....

x⁴ -3x² +9  = x⁴ + 6x² +9 - 9x²

= (x² + 3)² - (3x)²

= (x² +3 - 3x)(x² +3+3x²)

`a)3x-3a+yx-ya`

`=3(x-a)+y(x-a)`

`=(x-a)(y+3)`

`b)x^2-9-4(x+3)`

`=(x-3)(x+3)-4(x+3)`

`=(x+3)(x-3-4)`

`=(x+3)(x-7)`

a)  \(=3x+yx-\left(3a+ya\right)\) \(=x\left(3+y\right)-a\left(3+y\right)\) \(=\left(3+y\right)\left(x-a\right)\)

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

a)   \(=x^4-x^3-2x^3+2x^2+2x^2-2x-x+1\)

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

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

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

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

c)

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

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

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

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

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

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

b)

\(=x^4+1011x^2+1011+\left(1010x^2-2020x+1010\right)\)

\(=x^4+1011x^2+1011+1010\left(x^2-2x+1\right)\)

\(=x^4+1011x^2+1011+1010\left(x-1\right)^2\)

CÓ:   \(x^4+1010\left(x-1\right)^2+1011x^2\ge0\forall x\)

=>   \(x^4+1010\left(x-1\right)^2+1011x^2+1011\ge1011>0\forall x\)

=> ĐA THỨC b > 0 => Ko ph được thành nhân tử.