Question

(3x^2)^3-7x^3[x^3-x(4x^2+1]+(-x^2)^2,其中x=2分之1

Answer 1

原式=3³×(x²)³-7x³(x³-x×4x²-x×1)+x^4

=27x^6-7x³(x³-4x³-x)+x^4

=27x^6-7x³(-4x³-x)+x^4

=27x^6-7x³×(-4x³)-7x³×(-x)+x^4

=27x^6+28x^6+7x^4+x^4

=55x^6+8x^4

当x=1/2时

原式=55×(1/2)^6+8×(1/2)^4

=55×1/64+8×1/16

=55/64+1/2

=87/64