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!

Bài 3: 

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

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

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

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

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

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

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, \(x^6+x^4+x^2y^2+y^4-y^6\)

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

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

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

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

$Toru$

a) \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)=0\)

b) \(B=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x=x^3-3x^2+3x-1-x^3-x^2-x+x^2+x+1-3x+3x^2=0\)

a: Ta có: \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)

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

=0

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

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

=0

a) \(\left(x-5\right)^2=\left(3+2x\right)^2\)

\(\Rightarrow\left(3+2x\right)^2-\left(x-5\right)^2=0\)

\(\Rightarrow\left(3+2x+x-5\right)\left(3+2x-x+5\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(x+8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)

b) \(27x^3-54x^2+36x=9\)

\(\Rightarrow27x^3-54x^2+36x-9=0\)

\(\Rightarrow27x^3-54x^2+36x-8+8-9=0\)

\(\Rightarrow\left(3x-2\right)^3-1=0\)

\(\Rightarrow\left(3x-2-1\right)\left[\left(3x-2\right)^2+3x-2+1\right]=0\)

\(\Rightarrow\left(3x-3\right)\left[\left(3x-2\right)^2+3x-2+\dfrac{1}{4}-\dfrac{1}{4}+1\right]=0\)

\(\Rightarrow\left(3x-3\right)\left[\left(3x-2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)

\(\Rightarrow\left(3x-3\right)\left[\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\left(1\right)\)

mà \(\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0,\forall x\)

\(\left(1\right)\Rightarrow3x-3=0\Rightarrow3x=3\Rightarrow x=1\)

(\(x-5\))² = (3 +2\(x\))² ⇒ \(\left[{}\begin{matrix}x-5=3+2x\\x-5=-3-2x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=-8\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x\in\){-8; \(\dfrac{2}{3}\)}

  27\(x^3\) - 54\(x^2\) + 36\(x\) = 9

27\(x^3\) - 54\(x^2\) + 36\(x\) - 8 = 1

(3\(x\) - 2)³ = 1 ⇒ 3\(x\) - 2 = 1 ⇒ \(x\) = 1

1

\(\left\{{}\begin{matrix}x-y=4\\xy=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\y\left(y+4\right)=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\y^2+4y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y+4\\\left[{}\begin{matrix}y=-2+\sqrt{5}\\y=-2-\sqrt{5}\end{matrix}\right.\end{matrix}\right.\)

Với \(y=-2+\sqrt{5}\Rightarrow x=2+\sqrt{5}\)

Với \(y=-2-\sqrt{5}\Rightarrow x=2-\sqrt{5}\)

\(\Rightarrow A=x^2+y^2=\left(-2+\sqrt{5}\right)^2+\left(2+\sqrt{5}\right)^2=\left(2-\sqrt{5}\right)^2+\left(-2-\sqrt{5}\right)^2=18\)

\(B=x^3+y^3\Rightarrow\left[{}\begin{matrix}B=\left(2+\sqrt{5}\right)^3+\left(-2+\sqrt{5}\right)^3=34\sqrt{5}\\B=\left(2-\sqrt{5}\right)^3+\left(-2-\sqrt{5}\right)^3=-34\sqrt{5}\end{matrix}\right.\)

\(\Rightarrow C=x^4+y^4=\left(-2+\sqrt{5}\right)^4+\left(2+\sqrt{5}\right)^4=\left(2-\sqrt{5}\right)^4+\left(-2-\sqrt{5}\right)^4=322\)

(x - 5)² = (3 + 2x)²

(x - 5)² - (3 + 2x)² = 0

[(x - 5) - (3 + 2x)][(x - 5) + (3 + 2x)] = 0

(x - 5 - 3 - 2x)(x - 5 + 3 + 2x) = 0

(-x - 8)(3x - 2) = 0

-x - 8 = 0 hoặc 3x - 2 = 0

*) -x - 8 = 0

-x = 8

x = -8

*) 3x - 2 = 0

3x = 2

x = 2/3

Vậy x = -8; x = 2/3

--------------------

27x³ - 54x² + 36x = 9

27x³ - 54x² + 36x - 9 = 0

27x³ - 27x² - 27x² + 27x + 9x - 9 = 0

(27x³ - 27x²) - (27x² - 27x) + (9x - 9) = 0

27x²(x - 1) - 27x(x - 1) + 9(x - 1) = 0

(x - 1)(27x² - 27x + 9) = 0

x - 1 = 0 hoặc 27x² - 27x + 9 = 0

*) x - 1 = 0

x = 1

*) 27x² - 27x + 9 = 0

Ta có:

27x² - 27x + 9

= 27(x² - x + 1/3)

= 27(x² - 2.x.1/2 + 1/4 + 1/12)

= 27[(x - 1/2)² + 1/12] > 0 với mọi x ∈ R

⇒ 27x² - 27x + 9 = 0 (vô lí)

Vậy x = 1

A = x² + y²

= x² - 2xy + y² + 2xy

= (x - y)² + 2xy

= 4² + 2.1

= 16 + 2

= 18

B = x³ - y³

= (x - y)(x² + xy + y²)

= (x - y)(x² - 2xy + y² + xy + 2xy)

= (x - y)[(x - y)² + 3xy]

= 4.(4² + 3.1)

= 4.(16 + 3)

= 4.19

= 76

C = x⁴ + y⁴

= (x²)² + (y²)²

= (x²)² + 2x²y² + (y²)² - 2x²y²

= (x² + y²)² - 2x²y²

= (x² - 2x²y² + y² + 2x²y²)² - 2x²y²

= [(x - y)² + 2x²y²]² - 2x²y²

= (4² + 2.1²)² - 2.1²

= (16 + 2)² - 2

= 18² - 2

= 324 - 2

= 322

a: =>(2x+3)^2-(x-5)^2=0

=>(2x+3+x-5)(2x+3-x+5)=0

=>(x+8)(3x-2)=0

=>x=2/3 hoặc x=-8

b: =>27x^3-54x^2-36x-9=0

=>3x^3-6x^2-4x-1=0

=>\(x\simeq2,57\)

c: A=x^2+y^2=(x-y)^2+2xy=4^2+2=18

B=x^3-y^3=(x-y)^3+3xy(x-y)

=4^3+3*1*4

=64+12=76

C=(x^2+y^2)^2-2x^2y^2

=18^2-2*1^2=322

Bài 1:

$2xy=(x+y)^2-(x^2+y^2)=4^2-10=6\Rightarrow xy=3$ 

$M=x^6+y^6=(x^3+y^3)^2-2x^3y^3$

$=[(x+y)^3-3xy(x+y)]^2-2(xy)^3=(4^3-3.3.4)^2-2.3^3=730$

Bài 2:
$8x^3-32y-32x^2y+8x=0$

$\Leftrightarrow (8x^3+8x)-(32y+32x^2y)=0$

$\Leftrightarrow 8x(x^2+1)-32y(1+x^2)=0$

$\Leftrightarrow (8x-32y)(x^2+1)=0$
$\Rightarrow 8x-32y=0$ (do $x^2+1>0$ với mọi $x$)

$\Leftrightarrow x=4y$

Khi đó:

$M=\frac{3.4y+2y}{3.4y-2y}=\frac{14y}{10y}=\frac{14}{10}=\frac{7}{5}$

Bài cuối $x^21$ không rõ. Bạn xem lại.

\(\text{a) x^2 + y^2 = (x+y)^2 - 2xy = a^2 - 2b}\)

\(\text{b) x^3 + y^3 = (x+y)^3 - 3xy(x+y) = a^3 - 3ab}\)

\(\text{c) x^4 + y^4 = (x^2+y^2)^2 - 2x^2y^2 = (a^2-2b)^2 - 2b^2 = a^4 - 4a^2b + 2b^2}\)

\(\text{d) x^5 + y^5 = (x^3+y^3)(x^2+y^2) - x^2y^2(x+y) = a^5 - 5a^3b + 5ab^2}\)