Question

Cho x=y+1. Chứng minh rằng:

a)\(x^3-y^3-3xy=1\)

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

 

Answer 1

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

 

 

Answer 2

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