thêm x;y thuộc z nhé

\(x^8-y^8=\left(x^4\right)^2-\left(y^4\right)^2=\left(x^4-y^4\right)\left(x^4+y^4\right)=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\)

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

vì \(x-y⋮x-y;x,y\in Z\Rightarrow\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)⋮x-y\Rightarrow x^8-y^8⋮x-y\)

\(x+y⋮x+y;x,y\in Z\Rightarrow\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)⋮x+y\Rightarrow x^8-y^8⋮x+y\)

Cảm ơn bạn nha! Bài bạn làm đúng rồi nhé!

Đặt \(\left(x-1;y-1\right)=\left(a;b\right)\Rightarrow\left(x;y\right)=\left(a+1;b+1\right)\)

\(VT=\dfrac{\left(a+1\right)^3+\left(b+1\right)^3-\left(a+1\right)^2-\left(b+1\right)^2}{ab}=\dfrac{a^3+a+b^3+b+2\left(a^2+b^2\right)}{ab}\)

\(VT\ge\dfrac{2a^2+2b^2+2\left(a^2+b^2\right)}{ab}=\dfrac{4\left(a^2+b^2\right)}{ab}\ge\dfrac{8ab}{ab}=8\)

a. Do \(x=y-1\Rightarrow x-y=1\)

Ta có:

\(A=x^3-y^3-3xy=\left(x-y\right)^3+3xy\left(x-y\right)-3xy=1^3+3xy.1-3xy=1\left(đpcm\right)\)

b. \(B=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

(Do \(x-y=1\))

(Bạn áp dụng hằng đẳng thức \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)vào bài toán)

Kết quả, \(B=x^{16}-y^{16}\left(đpcm\right)\)

a)\(x=y+1\Rightarrow x-y=1\Rightarrow\left(x-y\right)^3=1\)

Hay x³- 3xy(x-y) -  y³=1  => x³- y³ -3xy =1

b) 1.(x+y)(x²+y²)(x⁴+y⁴)(x⁸+y⁸) = (x-y)(x+y)......................=(x²-y²)(x²+y²)..........=(x^4-y⁴)(x⁴+y⁴)......=(x⁸-y⁸)(x⁸+y⁸) =x¹⁶-y¹⁶

Áp dụng tc dtsbn:

\(\dfrac{x}{2013}=\dfrac{y}{2014}=\dfrac{z}{2015}=\dfrac{x-z}{-2}=\dfrac{y-z}{-1}=\dfrac{x-y}{-1}\\ \Leftrightarrow\dfrac{x-z}{2}=\dfrac{y-z}{1}=\dfrac{x-y}{1}\\ \Leftrightarrow x-z=2\left(y-z\right)=2\left(x-y\right)\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)

a, Ta có: \(2\left(x^8+y^8\right)\ge\left(x^3+y^3\right)\left(x^5+y^5\right)\)

\(\Leftrightarrow x^8+y^8\ge x^5y^3+x^3y^5\)

Ta CM: \(\Leftrightarrow x^8+y^8\ge x^5y^3+x^3y^5\)

Áp dụng bđt Cô si:

\(x^8+x^8+x^8+x^8+x^8+y^8+y^8+y^8\ge8x^5y^3\) (*)

Tương tự, \(5y^3+3x^3\ge8x^3y^5\) (**)

Từ (*), (**) \(\Rightarrowđpcm\)

Bài 3:

x=y+1 nên x-y=1

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

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

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

=x^8-y^8

Ta có \(x-y=1\)

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

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

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

\(A=x^8-y^8\)

= \(-\left[\left(x-y\right)\left(x^2-y^2\right)\left(x^4-y^4\right)\left(x^8-y^8\right)\left(x^{16}-y^{16}\right)\right]\)

= \(-\left[\left(x-y\right)\left(x-y\right)^2\left(x-y\right)^4\left(x-y\right)^8\left(x-y\right)^{16}\right]\)

= \(-\left(1\cdot1^2\cdot1^4\cdot1^8\cdot1^{16}\right)\)

= -1

1. \(\left\{{}\begin{matrix}3x^2+y^2+4xy=8\left(1\right)\\\left(x+y\right)\left(x^2+xy+2\right)=8\end{matrix}\right.\)

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

<=> \(\left(x+y\right)\left(3x+y\right)=\left(x+y\right)\left(x^2+xy+2\right)=0\)

<=> \(\left(x+y\right)\left(x^2+xy+2-3x-y\right)=0\)

<=> \(\left[{}\begin{matrix}x=-y\\x^2+xy+2-3x-y=0\end{matrix}\right.\)

TH1: x = -y thay vào pt (1), ta được:

3y² + y² - 4y² = 8

<=> 0y = 8 (vô lí)

TH2: \(x^2+xy+2-3x-y=0\)

<=> x (x + y) - (x + y) - 2(x - 1) = 0

<=> (x - 1)(x + y) - 2(X - 1) = 0

<=> (x - 1)(x + y - 2) = 0

<=> \(\left[{}\begin{matrix}x=1\\x+y-2=0\end{matrix}\right.\)

Với x =  1 thay vào pt (1) -> 3 + y² + 4y = 8

<=> y² + 4y - 5 = 0 <=> (y + 5)(y - 1) = 0

<=> \(\left[{}\begin{matrix}y=-5\\y=1\end{matrix}\right.\)

Với x + y - 2 = 0 => x = 2 - y thay vào pt (1)

=> 3(2 - y)² + y² + 4(2 - y)y = 8

<=> 3y² - 12y + 12 + y² + 8 - 4y² = 8

<=> 12 = 12y <=> y= 1 => x = 2 - 1 = 1

Vậy ....