Question

a) (4x-1)^3-(4x-3)(16x^2+3)

b) 2(x^3+y^3)-3(x^2+y^2 với x+y=1

Answer 1

a) \((4x-1)^3-(4x-3)(16x^2+3)\)

\(=(4x)^3-3.(4x)^2.1+3.4x.1^2-1^3-(4x-3) (16x^2+3)\)

\(=64x^3-48x^2+12x-1-64x^3-12x-48x^2-9\)

\(=9\)

Answer 2

\(a,\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-48x^2+12x-1-64x^3-4x+48x^2+9\)\(=12x+8\)

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

\(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left[\left(x+y\right)^2-2xy\right]\)\(=2x^2-2xy+2y^2+6xy\) ( vì x + y =1 )

\(=2x^2+4xy+2y^2\)

\(=2\left(x^2+2xy+y^2\right)=2\left(x+y\right)^2=2\)