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

\(=\left(x^7+x^4\right)+\left(x^5+x^2\right)+\left(x^3+1\right)\)

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

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

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

Goi y la them bot

\(x^7+x^2+1\)

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

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

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

a, k ph đc

b,Đặt \(A=...=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt x^2+5x+4=t,ta có:

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

\(=\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)=x\left(x+5\right)\left(x^2+5x+10\right)\)

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

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

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

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

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

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

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

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

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

a)\(x^2-x-12\)

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

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

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

b) \(x^2+8x+15\)

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

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

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

Gợi ý:

Nhóm:\(\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-8\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-8\)

Đặt \(t=x^2+5x+4\) thì biểu thức trở thành:

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

Rồi bạn làm tiếp, nếu còn phân tích được thì phải phân tích, mình bận rồi.

(x + 1)(x + 2)(x + 3)(x + 4) - 8

= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 8

= (x² + 4x + x + 4)(x² + 3x + 2x + 6) - 8

= (x² + 5x + 4)(x² + 5x + 6) - 8

Đặt x² + 5x + 5 = t

⇒ (x² + 5x + 5 - 1)(x² + 5x + 5 + 1) - 8 (1)

Thay t = x² + 5x + 5 vào (1), ta có:

(t - 1)(t + 1) - 8 = t² - 1 - 8 = t² - 9

= (t - 3)(t + 3)

⇔ (x² + 5x + 5 - 3)(x² + 5x + 5 + 3)

= (x² + 5x + 2)(x² + 5x + 8)

Chúc bạn học tốt !!!!!!!!

(x+1)(x+2)(x+3)(x+4)-8

= [(x+1)(x+4)][(x+2)(x+3)]-8

= (x²+4x+x+4)(x²+3x+2x+6)-8

= (x²+5x+5-1)(x²+5x+5+1)-8

= (x²+5x+5)²-1²-8

= (x²+5x+5)²-9

= (x²+5x+5) -3²

= (x²+5x+5-3)(x²+5x+5+3) {HĐT số 3}

= (x²+5x+2)(x²+5x+8)

a)\(x^2+x-6\)

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

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

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

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

a) x² + x - 6

= x² - 2x + 3x - 6

= (x² - 2x) + (3x - 6)

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

= (x + 3)(x - 2)

b) x⁴ + 4

= x⁴ + 4x² + 4 - 4x²

= (x⁴ + 4x² + 4) - 4x²

= (x + 2)² - 4x²

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