Question

tinh gia tri bieu thuc (x-1)^3-4x(x+1)(x-1)+3(x-1)(x^2+x+1) voi x=-2

Answer 1

Đặt \(A\) , ta có :

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

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

Thay \(x=2\) vào biểu thức , ta có :

\(A=\left(-2-1\right)^3-4.\left(-2\right).\left[\left(-2\right)^2-1\right]+3.\left[\left(-2\right)^3-1\right]\)

\(A=\left(-3\right)^3+8.3+3.\left(-9\right)\)

\(A=-27+24-27\)

\(A=-30\)