Tìm nghiệm nguyên của phương trình: \(^{x^4+x^2+1=y^2}\)
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\(\left(1+x\sqrt{x^2+1}\right)\left(\sqrt{x^2+1}-x\right)=1\)
\(\Rightarrow\dfrac{1+x\sqrt{x^2+1}}{\sqrt{x^2+1}+x}=1\)
\(\Rightarrow1+x\sqrt{x^2+1}=\sqrt{x^2+1}+x\)
\(\Rightarrow1+x\sqrt{x^2+1}-\sqrt{x^2+1}-x=0\)
\(\Rightarrow-\left(x-1\right)+\left(x-1\right)\sqrt{x^2+1}=0\)
\(\Rightarrow\left(x-1\right)\left(\sqrt{x^2+1}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\sqrt{x^2+1}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\\sqrt{x^2+1}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x^2+1=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
\(a,2y^2-x+2xy=y+4\\ \Leftrightarrow2y\left(x+y\right)-\left(x+y\right)=4\\ \Leftrightarrow\left(2y-1\right)\left(x+y\right)=4=4\cdot1=\left(-4\right)\left(-1\right)=\left(-2\right)\left(-2\right)=2\cdot2\)
Vì \(x,y\in Z\Leftrightarrow2y-1\) lẻ
\(\left\{{}\begin{matrix}2y-1=1\\x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2y-1=-1\\x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=0\end{matrix}\right.\)
Vậy PT có nghiệm \(\left(x;y\right)=\left\{\left(3;1\right);\left(4;0\right)\right\}\)
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-Tham khảo:
https://hoc24.vn/hoi-dap/tim-kiem?id=45441263315&q=T%C3%ACm%20nghi%E1%BB%87m%20nguy%C3%AAn%20c%E1%BB%A7a%20ph%C6%B0%C6%A1ng%20tr%C3%ACnh%20sau%C2%A0%5C%28x%5E6%203x%5E2%201%3Dy%5E4%5C%29
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the bon may co biet
Ta có \(x^4+x^2+1\le x^4+2x^2+1=\left(x^2+1\right)^2\)
Mà \(\left(x^2\right)^2=x^4< x^4+x^2+1\)nên \(\left(x^2\right)^2< x^4+x^2+1\le\left(x^2+1\right)^2\)
\(\Leftrightarrow x^4+x^2+1=\left(x^2+1\right)^2\)\(\Leftrightarrow y^2=\left(x^2+1\right)^2\)
Thay vào phương trình đã cho, ta có: \(x^4+x^2+1=\left(x^2+1\right)^2\)
\(\Leftrightarrow x^4+x^2+1=x^4+2x^2+1\)\(\Leftrightarrow x^2=0\)\(\Leftrightarrow x=0\)
Khi đó \(y^2=\left(x^2+1\right)^2=\left(0^2+1\right)^2=1\)\(\Leftrightarrow y=\pm1\)
Vậy phương trình đã cho có hai nghiệm nguyên là \(\left(0;1\right)\)và \(\left(0;-1\right)\)