x⁶+x⁴+x²y²+y⁴-y⁶=(x⁶-y⁶)+(x⁴+x²y²+y⁴)=(x²-y²)(x⁴+x²y²+y⁴)+(x⁴+x²y²+y⁴)=(x⁴+x²y²+y⁴)(x²-y²+1)=((x²+y²)²-x²y²)(x²-y²+1)

                          =(x²+xy+y²)(x²-xy+y²)(x²-y²+1)

x⁴-30x²+31x-30=(x⁴+x)-(30x²-30x+30)=x(x+1)(x²-x+1)-30(x²-x+1)=(x²-x+1)(x²+x-30)=(x²-x+1)(x-5)(x+6)

b)\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-24\)4

\(=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]-24\)

\(=\left(x^2-4x-x+4\right)\left(x^2-3x-2x+6\right)-24\)

\(=\left(x^2-5x+4\right)\left(x^2-5x+4+2\right)-24\)

\(\)Đặt  \(x^2-5x+4\)là a,ta có

\(=a\left(a+2\right)-24\)

\(=a^2+2a-24\)

\(=a^2+6a-4a-24\)

\(=a\left(a+6\right)-4\left(a+6\right)\)

\(=\left(a+6\right)\left(a-4\right)\)

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

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

Câu hỏi của Huỳnh Bảo Nguyên - Toán lớp 8 - Học toán với OnlineMath

Mk làm òi nhé !

\(=\left(x^4+x\right)-30x^2+30x-30\)

\(=x\left(x^3+1\right)-30\left(x^2-x+1\right)\)

\(=x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]\)

\(=\left(x^2-x+1\right)\left(x^2+x-30\right)\)

\(2x^4-8x^3-7x^3+28x^2+7x^2-28x-2x+8\\ =2x^3\left(x-4\right)-7x^2\left(x-4\right)+7x\left(x-4\right)-2\left(x-4\right)\\ =\left(x-4\right)\left(2x^3-7x^2+7x-2\right)\\ =\left(x-4\right)\left(2x^3-4x^2-3x^2+6x+x-2\right)\\ =\left(x-4\right)\left[2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)\right]\\ =\left(x-4\right)\left(x-2\right)\left(2x^2-2x-x+1\right)\\ =\left(x-4\right)\left(x-2\right)\left(2x-1\right)\left(x-1\right)\)