LINK THAM KHẢO
https://olm.vn/hoi-dap/detail/101095140158.html
\(x^2+x+3=y^2\)
\(\Leftrightarrow4x^2+4x+12=4y^2\)
\(\Leftrightarrow\left(4x^2+4x+1\right)-4y^2=-11\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2y\right)^2=-11\)
\(\Leftrightarrow\left(2x+1-2y\right)\left(2x+y+2y\right)=-11=\left(-1\right)\cdot11=11\cdot\left(-1\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1\)
Đến đây bạn tự làm nốt nhá.tớ làm thử cho 1 TH tham khảo nhé !
\(\hept{\begin{cases}2x+1-2y=-1\\2x+1+2y=11\end{cases}}\Rightarrow\hept{\begin{cases}x-y=-1\\x+y=5\end{cases}}\Rightarrow x=2\Rightarrow y=3\)
Còn lại tương tự:3
he equation you’ve provided is 2^x + 3 = y^2. Let’s explore some ways to approach this equation:
Algebraic Solution:
Rearrange the equation to isolate the exponential term: [2^x = y^2 - 3]Take the natural logarithm (base e) of both sides: [x \ln(2) = \ln(y^2 - 3)]Solve for (x): [x = \frac{\ln(y^2 - 3)}{\ln(2)}]Graphical Approach:
We can plot the graphs of both sides of the equation and find the points of intersection.Visit the Desmos Graphing Calculator and input the following equations:Left side: (y_1 = 2^x)Right side: (y_2 = y^2 - 3)The points where (y_1) and (y_2) intersect correspond to solutions.Integer Solutions:
For integer solutions, we can explore the Diophantine equation (1 + x + x^2 + x^3 = y^2).According to Ribenboim’s book on Catalan’s conjecture, for (k = 3), only (x = 1) and (x = 7) are possible solutions 1.Feel free to explore these approaches further! 🤓
