\(x^2+7x+12=x^2+3x+4x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)

=> \(B=\left(x+3\right)\left(x+4\right)\left(x-2\right)\left(x-1\right)\)

\(=\left[\left(x+3\right)\left(x-1\right)\right]\left[\left(x+4\right)\left(x-2\right)\right]+2013\)

\(=\left[x^2+2x-3\right]\left[x^2+2x-8\right]+2013\)

Đặt : \(t=x^2+2x-3\)

Ta có: \(B=t\left(t-5\right)+2013=t^2-5t+2013=t^2-2.t.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+2013\)

\(=\left(t-\frac{5}{2}\right)^2+\frac{8027}{4}\ge\frac{8027}{4}\)

"=" xảy ra <=> \(t=\frac{5}{2}\Leftrightarrow x^2+2x-3=\frac{5}{2}\Leftrightarrow\left(x+1\right)^2=\frac{13}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{13}{2}}-1\\x=-\sqrt{\frac{13}{2}}-1\end{cases}}\)(tm)

Vậy min B = 8027/4 tại x =....

\(16x^4+8x^2+1-8x^2\)

\(=\left(4x^2+1\right)^2-8x^2\)

\(=\left(4x^2+1-x\sqrt{8}\right)\left(4x^2+1+x\sqrt{8}\right)\)

\(A=\frac{3x+1}{2x^2-x+3}\)

\(A=\frac{2x^2-x+3-2x^2+4x-2}{2x^2-x+3}\)

\(A=\frac{\left(2x^2-x+3\right)-2\left(x^2-2x+1\right)}{2x^3-x+3}\)

\(A=1-\frac{2\left(x-1\right)^2}{2x^2-x+3}\)

\(A=1-\frac{2\left(x-1\right)^2}{2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+\frac{23}{8}}\)

\(A=1-\frac{2\left(x-1\right)^2}{2\left(x-\frac{1}{4}\right)^2+\frac{23}{8}}\le1\)

Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(x-\frac{1}{4}\right)^2\ge0\forall x\end{cases}\Rightarrow\frac{2\left(x-1\right)^2}{2\left(x-\frac{1}{4}\right)^2+\frac{23}{8}}\ge0\forall x}\)

Dấu '' = '' xảy ra khi x = 1 

Vậy Max A =1 khi x = 1 .

cách 2 nếu chưa học bezout

Mà \(A\left(x\right):\left(x-1\right)\)dư 4\(\Rightarrow m+n+1=4\)

                                                 \(\Rightarrow m+n=3\left(1\right)\)

Mà \(A\left(x\right):\left(x+1\right)\)dư 6\(\Rightarrow n-m-1=6\)

                                               \(\Rightarrow n-m=7\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\Rightarrow\hept{\begin{cases}n+m=3\\n-m=7\end{cases}\Rightarrow\hept{\begin{cases}n=5\\m=-2\end{cases}}}\)

Vậy n=5 và m=-2

Áp dụng định lý Bezout ta có:

\(A\left(x\right)\)chia x-1 dư 4 \(\Rightarrow A\left(1\right)=4\)

                                    \(\Rightarrow1+m+n=4\)

                                     \(\Rightarrow m+n=3\left(1\right)\)

\(A\left(x\right)\)chia x+1 dư 6 \(\Rightarrow A\left(-1\right)=6\)

                                       \(\Rightarrow-1-m+n=6\)

                                      \(\Rightarrow-m+n=7\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\Rightarrow\hept{\begin{cases}m+n=3\\-m+n=7\end{cases}\Rightarrow}\hept{\begin{cases}n=5\\m=-2\end{cases}}\)

Vậy n=5 và m=-2 

Làm hơi tắt nhé

• Nếu \(y=0\Rightarrow x^2=65\Rightarrow x\notin Z\)
• Nếu \(y>1\Rightarrow x^2+y^3-3y^2=65-3y\Leftrightarrow x^2+\left(y^3-3y^2+3y-1\right)=64\Leftrightarrow x^2-\left(y-1\right)^3=64\)
• Mà \(x;y-1\in N;64=0^2+4^3=8^2+0^3\)
• \(Th1:\hept{\begin{cases}x=0\\y-1=4\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=5\end{cases}}}\)
• \(Th2:\hept{\begin{cases}x=8\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=8\\y=1\end{cases}}}\)
• Thử lại ta có nghiệm nguyên là : \(\left(0;5\right),\left(8;1\right)\)

<=> x²  = 64 - (y-1)³ \(\ge0< =>4\ge y-1< =>y\le5.\)

y=5 => x=0 (thỏa mãn); y=4 => x² = 37 (loại); y=3 => x² =56 (loại); y= 2 => x² = 63 loại; y=1 => x= 8; y=0 => x= 65 loại

vậy các nghiệm (x;y) = (0;5); (1;8)

Ta có:

M = (a + 1)(a + 2)(a + 3)(a + 4)  + 1

M = [(a + 1)(a + 4)][(a + 2)(a + 3)] + 1

M = (a² + 4a + a + 4)(a² + 3a + 2a + 6) + 1

M = (a² + 5a + 4)(a² + 5a + 6) + 1

M = (a² + 5a + 4)² + 2(a² + 5a + 4) + 1

M = (a² + 5a + 4 + 1)²

M = (a² + 5a + 5)²

=> M là bình phương của 1 số nguyên 

\(M=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1\)

\(M=\left[\left(a+1\right)\left(a+4\right)\right]\left[\left(a+2\right)\left(a+3\right)\right]+1\)

\(M=\left[a\left(a+4\right)+\left(a+4\right)\right]\left[a\left(a+3\right)+2\left(a+3\right)\right]+1\)

\(M=\left(a^2+4a+a+4\right)\left(a^2+3a+2a+6\right)+1\)

\(M=\left(a^2+5a+5\right)^2-1+1\)

\(M=a^2+5a+5\)

Mà \(a\inℤ\Rightarrow\left(a^2+5a+5\right)\inℤ\)

Vậy,..................