Question

Tính:

a) (x-1)^3-(x+1)^3+6.((x+1).(x-1)

b) (x-1)^3-(x-1).(x^2+x+1)-3.(1-x).x

Answer 1

a. Câu hỏi của Nguyễn Thị Anh Thư - Toán lớp 8 - Học toán với OnlineMath

Answer 2

a, \(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

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

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

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

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

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

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

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

\(=\left(x-1\right).0=0\)