Trả lời:

( x + 2 )³ + ( x - 2 )³ - 2x ( x² + 10 )

= x³ + 6x² + 12x + 8 + x³ - 6x² + 12x - 8 - 2x³ - 20x 

= 4x

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\(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+10\right)\)

\(=\left(x+2+x-2\right)\left[\left(x+2\right)^2+\left(x+2\right)\left(x-2\right)-\left(x-2\right)^2\right]-2x^3-20x\)

\(=2x\left(x^2+4x+4+x^2-4-x^2+4x-4\right)-2x^3-20x\)

\(=2x\left(x^2+8x-4\right)-2x^3-20x=2x^3+16x^2-8x-2x^3-20x\)

\(=16x^2-28x\)

Đặt biểu thức đã cho là A.

Ta có: 2A = (3 - 1) * (3 + 1) * (3^2 + 1) * .... * (3^64 + 1)

= (3^2 - 1) * (3^2 + 1) * ... * (3^64 + 1) (hằng đẳng thức a^2 - b^ 2 = (a+b)(a-b))

Rút gọn triệt tiêu ta được 2A=3^64 - 1

=> A = (3^64 - 1)/2

P=1+5²+5⁴+5⁶+...+5⁹⁸

=>5².P=P=5²+5⁴+5⁶+...+5⁹⁸+5¹⁰⁰

=>25P-P=24P=P=5²+5⁴+5⁶+...+5⁹⁸+5¹⁰⁰-(1+5²+5⁴+5⁶+...+5⁹⁸)

=>24P=5¹⁰⁰-1

=>P=\(\frac{5^{100}-1}{24}.\)

a: Ta có: \(\left(x+y\right)^2+\left(x-y\right)^2-2x^2\)

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

\(=2y^2\)

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

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

=2

c) \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)-\left(x-2y\right)\left(x^2+2xy+4y^2\right)+2y^3\)

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

\(=x^3+8y^3-x^3+8y^3+2y^3\)

\(=8y^3+8y^3+2y^3\)

\(=18y^3\)

\(=x^6-6x^4+12x^2-8-x^3+x+6x^2-18x\\ =x^6-6x^4-x^3+18x^2-17x-8\)

\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)

a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)

b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)