Ta có : \(x2-y2=2\Rightarrow\left(x-y\right)2=2\Rightarrow x-y=1\)

\(A=2\left(x6-y6\right)-6\left(x4+y4\right)\)

\(\Rightarrow2\left[\left(x-y\right)6\right]-6\left[\left(x+y\right)4\right]\)

Mà \(x-y=1\Rightarrow A=2.6-6\left[\left(x+y\right)4\right]\)

\(\Rightarrow A=6\left[2-\left(x+y\right)4\right]\)

\(\Rightarrow A=6\left[2-4x-4y\right]=6\left[2-4\left(x-y\right)\right]\)

\(\Rightarrow A=6\left[2-4.1\right]=6.\left[2-4\right]=6.\left(-2\right)=-12\)

Vậy A = -12

Khi x = - 1; y = 1 thì xy = (-1).1= -1

Ta có: xy – x2y2 + x3y3 – x4y4 + x5y5 – x6.y6

= xy – (xy)2 + (xy)3 – (xy)4 + (xy)5 – (xy)6

= -1 – (-1)2 + (-1)3 – (-1)4 + (-1)5 - (-1)6

= -1 – 1 + (-1) – 1 + (-1) – 1

= - 6

Chọn đáp án D

D đúng nha!

P = x⁶ + y⁶ = (x² + y²)(x⁴ - x² y² + y⁴) 

= (x² + y²)² - 3x² y² \(\ge1-3×\frac{\left(x^2+y^2\right)^2}{4}=1-\frac{3}{4}=\frac{1}{4}\)

Đạt được khi x² = y² = \(\frac{1}{2}\)

Bài 3: 

\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)+15\)

\(=\left(x^2-9\right)\left(x^2-1\right)+15\)

\(=x^4-10x^2+9+15\)

\(=x^4-10x^2+24\)

\(=\left(x^2-4\right)\left(x^2-6\right)\)

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

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

Ta có: \(1+xy=x^2+y^2\ge2xy\Rightarrow xy\le1\)

\(1+xy=x^2+y^2\ge-2xy\Rightarrow xy\ge-\dfrac{1}{3}\)

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

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

Đặt \(a=xy\Rightarrow P=f\left(a\right)=-2a^2+2a+1\)

Xét hàm \(f\left(a\right)=-2a^2+2a+1\) trên \(\left[-\dfrac{1}{3};1\right]\)

\(-\dfrac{b}{2a}=\dfrac{1}{2}\in\left[-\dfrac{1}{3};1\right]\)

\(f\left(-\dfrac{1}{3}\right)=\dfrac{1}{9}\) ; \(f\left(\dfrac{1}{2}\right)=\dfrac{3}{2}\) ; \(f\left(1\right)=1\)

\(\Rightarrow M=\dfrac{3}{2}\) ; \(m=\dfrac{1}{9}\) \(\Rightarrow Mm=\dfrac{1}{6}\)

a, \(8^3yz+12^2yz+6xyz+yz\)

\(=512yz+144yz+6xyz+yz\)

\(=yz\left(512+14+6x+1\right)\)

\(=yz\left(527+6x\right)\)

$---$

b, \(81x^4\left(z^2-y^2\right)-z^2+y^2\)

\(=81x^4\left(z^2-y^2\right)-\left(z^2-y^2\right)\)

\(=\left(z^2-y^2\right)\left(81x^4-1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left[\left(9x^2\right)^2-1^2\right]\)

\(=\left(z-y\right)\left(z+y\right)\left(9x^2-1\right)\left(9x^2+1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left[\left(3x\right)^2-1^2\right]\left(9x^2+1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left(3x-1\right)\left(3x+1\right)\left(9x^2+1\right)\)

$---$

c, \(\dfrac{x^3}{8}-\dfrac{y^3}{27}+\dfrac{x}{2}-\dfrac{y}{3}\)

\(=\left[\left(\dfrac{x}{2}\right)^3-\left(\dfrac{y}{3}\right)^3\right]+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)

\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}\right)+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)

\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}+1\right)\)

$---$

d, \(x^6+x^4+x^2y^2+y^4-y^6\)

\(=\left(x^6-y^6\right)+\left(x^4+x^2y^2+y^4\right)\)

\(=\left[\left(x^2\right)^3-\left(y^2\right)^3\right]+\left(x^4+x^2y^2+y^4\right)\)

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

\(=\left(x^4+x^2y^2+y^4\right)\left(x^2-y^2+1\right)\)

$Toru$

 từ x^2 - y^2 = 2, bình phương và lập phương 2 vế để tìm x^4 + y^4 = ??? và x^6 + y^6 = ???
sau đó thế vào A là tìm dc giá trị của A