Khi x=1 và y=-2 thì A=1^2+3*1*(-2)+2

=3-6=-3

1; \(x^2\) + 3\(x^2\) + 3\(x\) = 4\(x^2\) + 3\(x\) (1) 

Thay \(x=99\) vào (1) ta có:

4.99² + 3.99 = 4.9801 + 297 = 39204 + 297 = 39501

a) \(\left(x+2\right)\left(x^2-24+4\right)\left(x^3+8\right)\)

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

\(=\left(x^3-20x+2x^2-40\right)\left(x^3+8\right)\)

\(=x^6+8x^3-20x^4+160x+2x^5+16x^2-40x^3-120\)

\(=x^6+2x^5-20x^4-32x^3+16x^2+160x-120\)

b) \(\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

\(=8x^3+2x^2+\dfrac{1}{2}x-2x^2-\dfrac{1}{2}x-\dfrac{1}{8}\)

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

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

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

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

\(=2x^4\)

d) \(\left(x+3\right)\left(x^2-3x+9\right)-x^3\)

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

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

\(=27\)

e) \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-26x^3\)

\(=\left(3x+y\right)\left[\left(3x\right)^2-3x\cdot y+y^2\right]-26x^3\)

\(=\left(3x\right)^3+y^3-26x^3\)

\(=27x^3+y^3-26x^3\)

\(=x^3+y^3\)

g) \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=\left(x+3y\right)\left[x^2-x\cdot3y+\left(3y\right)^2\right]+\left(3x-y\right)\left[\left(3x\right)^2+3x\cdot y+y^2\right]\)

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

\(=x^3+27y^3+27x^3-y^3\)

\(=28x^3+26y^3\)

Yêu cầu của đề là gì ?

a) Sửa đề:

(x + 2)(x² - 2x + 4)(x³ + 8)

= (x³ + 8)(x³ + 8)

= (x³ + 8)²

b) (2x - 1/2)(4x² + x + 1/4)

= (2x)³ - (1/2)³

= 8x³ - 1/8

c) (x² + y)(x² - y) + y² + x⁴

= (x²)² - y² + y² + x⁴

= 2x⁴

d) (x + 3)(x² - 3x + 9) - x³

= x³ + 3³ - x³

= 27

e) (3x + y)(9x² - 3xy + y²) - 26x³

= (3x)³ + y³ - 26x³

= 27x³ + y³ - 26x³

= x³ + y³

g) (x + 3y)(x² - 3xy + 9y²) + (3x - y)(9x² + 3xy + y²)

= x³ + (3y)³ + (3x)³ - y³

= x³ + 27y³ + 27x³ - y³

= 28x³ + 26y³