\(2^x+2^y=72\)

\(2^x+2^y=64+8\)

\(2^x+2^y=2^6+2^3\)

\(\Rightarrow x=6;y=3\)

Giả sử x>y, ta có:

2^x + 2^y = 72

=> 2^y (1 + 2^x-y) = 2³. 3²

Vì 1 + 2^x-y là số lẻ nên 1 + 2^x-y = 1;3;9

Vậy x = 3 và y = 0

x=4 ; y=6 ngược lại

lam cach nao vay

mk lai ra 3 va 6 co

\(\Leftrightarrow2x^2+x+2=y\left(2x-1\right)\)

\(\Leftrightarrow y=\dfrac{2x^2+x+2}{2x-1}=x+1+\dfrac{3}{2x-1}\)

\(y\in Z\Rightarrow\dfrac{3}{2x-1}\in Z\)

Mà x nguyên dương \(\Rightarrow2x-1>0\)

\(\Rightarrow2x-1=Ư\left(3\right)\Rightarrow x=\left\{1;2\right\}\) 

\(\Rightarrow\left(x;y\right)=\left(1;5\right);\left(2;4\right)\)

Ta có: \(2x^2+2y^2-x-y-2xy+\frac{1}{2}=0\)

\(\Leftrightarrow\left(x^2+y^2-2xy\right)+\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}^2\right)=0\)

Nhận xét \(\left(x-y\right)^2\ge0;\left(x-\frac{1}{2}\right)^2\ge0;\left(y-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-\frac{1}{2}\right)^2=0\\\left(y-\frac{1}{2}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-y=0\\x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Leftrightarrow}x=y=\frac{1}{2}}\)