Question

1) Chứng minh bt sau ko phụ thuộc vào biến

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

b) (2x+3y) ( 4x\(^2\)- 6xy + 9y\(^2\)) - ( 2x - 3y ) ( 4x\(^2\)+ 6xy + 9y\(^2\)) - 27 ( 2y\(^3\)- 1 )

c) y( x\(^2\)- y\(^2\)) ( x\(^2\)+ y\(^2\)) - y( x\(^4\)- y\(^4\))

d) ( x-1)\(^3\)- ( x-1) ( x\(^2\)+ x + 1 ) - 3 ( 1-x).x

Answer 1

a: \(=x^3-3x^2+3x-1-x^3-64+3x^2-3x\)

=-65

b \(=8x^3+27y^3-8x^3+27y^3-54y^3+27\)

=27

c: \(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)=0\)

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

\(=-3x^2+3x-3x+3x^2=0\)