a) x⁶ + y⁶ = (x²)³ + (y²)³

= (x² + y²)(x⁴ - x²y² + y⁴)

b) x⁶ - y⁶

= (x³)² - (y³)²

= (x³ - y³)(x³ + y³)

= (x - y)(x² + xy + y²)(x + y)(x² - xy + y²)

1A:

a: \(x^3+2x=x\left(x^2+2\right)\)

b: \(3x-6y=3\left(x-2y\right)\)

c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

\(=5\left(x+3y\right)\left(1-3x\right)\)

d: \(3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+3\right)\)

1A. a. x(x²+2) 

b. 3(x-2y)

c. 5(x+3y)(1-3x) 

d. (x-y) (3-5x)

1B. a. 2x(2x-3)

b.xy(x²-2xy+5)

c. 2x(x+1)(x+2)

d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)

1B:

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(x^3y-2x^2y^2+5xy\)

\(=xy\left(x^2-2xy+5\right)\)

\(a,=\left(3x+1\right)^2-y^2=\left(3x-y+1\right)\left(3x+y+1\right)\\ b,=x\left(x^2-5x+6\right)=x\left(x^2-2x-3x+6\right)=x\left(x-2\right)\left(x-3\right)\)

x 6 - y 6 = x 3 2 - y 3 2 = x 3 + y 3 x 3 - y 3 = x + y x 2 - x y + y x - y x 2 + x y + y 2

ủa mũ hay để nguyên z?? 

hđt =x^2(x+1)^2(x^2-2x+2)

x⁶ - x⁴ + 2x³ + 2x²

= x²(x⁴ - x² + 2x + 2)

= x²[(x⁴ - x²) + (2x + 2)]

= x²[x²(x² - 1) +2(x + 1)]

= x²[x²(x - 1)(x + 1) + 2(x + 1)]

= x²(x + 1)[x²(x - 1) + 2]

= x²(x + 1)(x³ - x² + 2)

= x²(x + 1)(x³ + x² - 2x² - 2x + 2x + 2)

= x²(x + 1)[(x³ + x²) - (2x² + 2x) + (2x + 2)]

= x²(x + 1)[x²(x + 1) - 2x(x + 1) + 2(x + 1)]

= x²(x + 1)²(x² - 2x + 2)

a,

\(A=4(x-2)(x+1)+(2x-4)^2+(x+1)^2\\=[2(x-2)]^2+2\cdot2(x-2)(x+1)+(x+1)^2\\=[2(x-2)+(x+1)]^2\\=(2x-4+x+1)^2\\=(3x-3)^2\)

Thay $x=\dfrac12$ vào $A$, ta được:

\(A=\Bigg(3\cdot\dfrac12-3\Bigg)^2=\Bigg(\dfrac{-3}{2}\Bigg)^2=\dfrac94\)

Vậy $A=\dfrac94$ khi $x=\dfrac12$.

b,

\(B=x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\\=(x^9-1)-(x^7-x^4)-(x^6-x^3)-(x^5-x^2)\\=[(x^3)^3-1]-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1)-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1-x^4-x^3-x^2)\\=(x^3-1)(x^6-x^4-x^2+1)\)

Thay $x=1$ vào $B$, ta được:

\(B=(1^3-1)(1^6-1^4-1^2+1)=0\)

Vậy $B=0$ khi $x=1$.

$Toru$

=362880 nha

chúc bạn học tốt nha!!!!!

Kết quả:362880

362880

anh chị xinh gái đẹp ơi hãy giúp em giải bài toán này

Câu 1: D

Câu 2: A

\(a,=\sqrt{xy}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{xy}+1\right)\left(\sqrt{x}-1\right)\\ b,=\sqrt{xy}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{xy}+1\right)\)