Question

Giúp mik bài 3 chỉ cần lm cau K ,I ,F

Answer 1

i) Ta có: \(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(=18^8-18^8+1\)

=1

Answer 2

Bài 4: 

a) Ta có: \(\left(x+3\right)^2-\left(x-3\right)^2-12x\)

\(=x^2+6x+9-x^2+6x-9-12x\)

=0

b) Ta có: \(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\)

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

=1

c) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x^3\)

\(=x^3-27-x^3\)

=-27

d) Ta có: \(\left(3x+2\right)\left(9x^2-6x+4\right)-9x\left(3x^2+1\right)+9x\)

\(=27x^3+8-27x^3-9x+9x\)

=8