đề có sai ko bn?

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

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

K sai dau
giao an truong Tran dai nghia do

\(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left[\left(x-1\right)\left(x-7\right)\right].\left[\left(x-3\right)\left(x-5\right)\right]-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(x^2-8x+11=t\) \(\Rightarrow\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20=\left(t-4\right)\left(t+4\right)-20=t^2-16-20=t^2-36=\left(t-6\right)\left(t+6\right)\)\(\Rightarrow\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left(x^2-8x+11-6\right)\left(x^2-8x+11+6\right)=\left(x^2-8x+17\right)\left(x^2-8x+5\right)\)

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)

\(x^4+2x^2+1=\left(x^2+1\right)^2\) (Nhớ k cho mình với nhé!)

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

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

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

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

\(x^4-6x^3+54x-81\)

\(=x^4+3x^3-9x^3+27x^2-27x^2 +81x-27x-81\)

\(=\left(x^4+3x^3\right)-\left(9x^3+27x^2\right)+\left(27x^2+81x\right)-\left(27+81\right)\)

\(=x^3\left(x+3\right)-9x^2\left(x+3\right)+27x\left(x+3\right)-27\left(x+3\right)\)

\(=\left(x+3\right)\left(x^3-9x^2+27x-27\right)\)

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

nhan tu là j hở bn

ns ik mình sẽ giải cko

cau cau hoi roi do ban

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

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

\(=\left(x+1+\sqrt{x}\right)\left(x+1-\sqrt{x}\right)\)