Question

Chứng minh giá trị các biểu thức sau không phụ thuộc vào biến 

a, ( 2x + 3 ) . ( 4x² - 6x + 9 ) - 2 ( 4x³ - 1 )

b, ( 4x - 1 )³ - ( 4x - 3 ) . ( 16x² + 3 )

c, 2( x³ + y³ ) - 3 ( x² + y² ) với x + y = 1

d, ( x + 1 )³ - ( x - 1 )³ - 6 ( x + 1 ) (x - 1 )

Answer 1

\(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)=8x^3+27-8x^3+2=29\)

\(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)=64x^3-48x^2+12x-1-\left(64x^3+12x-48x^2-9\right)=8\)

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

\(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x^2+y^2\right)\)

\(=2\left(x^2-xy+y^2\right)-3x^2-3y^2\)

\(=-2xy-x^2-y^2\)

\(=-\left(x^2+2xy+y^2\right)=-\left(x+y\right)^2=-1^2=-1\)

        \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x+1\right)\left(x-1\right)\)

\(=x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-1\right)\)

\(=x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+6=8\)

Chúc bạn học tốt.