Question

[Nguyên Hoàng]

chúng minh giá trị biểu thức sau ko phụ thuộc vào giá trị biến

a) (x-1)(x-2)(1+x+x²)(4+2x+x²)-x⁶+9x³

b) (\(\dfrac{1}{3}\)+2x)(4x²-\(\dfrac{2}{3}\)x+\(\dfrac{1}{9}\))-(2x-\(\dfrac{1}{3}\))(4x²+\(\dfrac{2}{3}\)x+\(\dfrac{1}{4}\))

c) (x-1)³-(x-1)(x²+x+1)-3(1-x)x

d) (x²-y²)(x²+xy+y²)(x²-xy+y²)+y⁶-x⁶

Mọi người giúp em vs ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

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

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

\(=\left(x^3-1\right)\left(x^3-8\right)-x^6+9x^3\)

\(=x^6-9x^3+8-x^6+9x^3=8\)

b) Ta có: \(\left(\dfrac{1}{3}+2x\right)\left(\dfrac{1}{9}-\dfrac{2}{3}x+4x^2\right)-\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2}{3}x+\dfrac{1}{4}\right)\)

\(=\dfrac{1}{27}+8x^3-8x^3+\dfrac{1}{27}\)

\(=\dfrac{2}{27}\)

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

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

=0

d) Ta có: \(\left(x^2-y^2\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)-x^6+y^6\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)-x^6+y^6\)

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

\(=x^6-y^6-x^6+y^6=0\)