1.
Với \(n=0;1\) không thỏa mãn
Với \(n>1\)
\(A=\left(n^2+n\right)^2+n^2+3n+7>\left(n^2+n\right)^2\)
\(A=\left(n^2+n+2\right)^2-\left[3\left(n^2-1\right)+n\right]< \left(n^2+n+2\right)^2\)
\(\Rightarrow\left(n^2+n\right)^2< A< \left(n^2+n+2\right)^2\)
\(\Rightarrow A=\left(n^2+n+1\right)^2\)
\(\Rightarrow n^4+2n^3+2n^2+3n+7=\left(n^2+n+1\right)^2\)
\(\Rightarrow n^2-n-6=0\Rightarrow\left[{}\begin{matrix}n=-2\left(loại\right)\\n=3\end{matrix}\right.\)
3.
TH1:
\(x>y\Rightarrow x^2y^2+x-y>x^2y^2\)
Mặt khác x; y nguyên dương \(\Rightarrow y\ge1\Rightarrow xy-\left(x-y\right)=x\left(y-1\right)+y>0\Rightarrow xy>x-y\)
\(\Rightarrow2xy+1>x-y\Rightarrow x^2y^2+x-y< x^2y^2+2xy+1\)
\(\Rightarrow x^2y^2< x^2y^2+x-y< \left(xy+1\right)^2\)
\(\Rightarrow x^2y^2+x-y\) nằm giữa 2 SCP liên tiếp nên ko thể là SCP (trái giả thiết) \(\Rightarrow\) loại
TH2: \(x< y\Rightarrow x^2y^2+x-y< x^2y^2\)
\(x-y-\left(-2xy+1\right)=\left(x-1\right)+y\left(2x-1\right)>0\Rightarrow x-y>-2xy+1\)
\(\Rightarrow x^2y^2+x-y>x^2y^2-2xy+1=\left(xy-1\right)^2\)
\(\Rightarrow\left(xy-1\right)^2< x^2y^2+x-y< x^2y^2\)
\(\Rightarrow x^2y^2+x-y\) nằm giữa 2 SCP liên tiếp \(\Rightarrow\) ko thể là SCP => trái giả thiết => loại
Vậy \(x=y\)
2.
\(C=n^4-2n^3+3n^2-6n+8\)
\(C=\left(n^2-n\right)^2+2n^2-6n+8=\left(n^2-n\right)^2+2\left(n-\dfrac{3}{2}\right)^2+\dfrac{7}{2}>\left(n^2-n\right)^2\)
\(C=\left(n^2-n+3\right)^2-\left(4n^2+1\right)< \left(n^2-n+3\right)^2\)
\(\Rightarrow\left(n^2-n\right)^2< C< \left(n^2-n+3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}C=\left(n^2-n+1\right)^2\\C=\left(n^2-n+2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4n-7=0\left(vn\right)\\2n^2+2n-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}n=1\\n=-2\end{matrix}\right.\)
2b.
\(4B=4n^4+4n^3+4n^2+4n+4=\left(2n^2+n\right)^2+3n^2+4n+4>\left(2n^2+n\right)^2\)
\(4B=\left(2n^2+n+2\right)^2-5n^2\le\left(2n^2+n+2\right)^2\)
\(\Rightarrow\left(2n^2+n\right)^2< B\le\left(2n^2+n+2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}4B=\left(2n^2+n+1\right)^2\\4B=\left(2n^2+n+2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4n^4+4n^3+4n^2+4n+4=\left(2n^2+n+1\right)^2\\4n^4+4n^3+4n^2+4n+4=\left(2n^2+n+2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}n^2-2n-3=0\\5n^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}n=3\\n=0\end{matrix}\right.\)
mn ơi ai thấy bài này giúp em được ko ạ,e đang cần gấp lắm
