Đặt \(f\left(x\right)=\left(x-a\right)\left(x-10\right)+1=x^2-\left(a+10\right)x+10a+1\).

Theo đề bài, ta đặt \(f\left(x\right)=\left(x-m\right)\left(x-n\right)\) với \(m,n\inℤ\). 

\(f\left(x\right)=x^2-\left(m+n\right)x+mn\)

Khi đó, ta thu được hệ pt:

\(\left\{{}\begin{matrix}m+n=a+10\\mn=10a+1\end{matrix}\right.\) 

Ta thấy nếu \(\left(a+10\right)^2-4\left(10a+1\right)< 0\) 

\(\Leftrightarrow\left(a-12\right)\left(a-8\right)< 0\)

\(\Leftrightarrow8< a< 12\) 

thì sẽ không tồn tại \(m,n\) thỏa mãn. Vậy \(\left[{}\begin{matrix}a\le8\\a\ge12\end{matrix}\right.\)

 Khi đó \(m,n\) là nghiệm nguyên của pt \(X^2-\left(a+10\right)X+10a+1=0\)         (*)

 Pt này có \(\Delta=\left(a+10\right)^2-4\left(10a+1\right)\) \(=\left(a-10\right)^2-4\) mà (*) lại có 2 nghiệm nguyên nên \(\left(a-10\right)^2-4\) phải là số chính phương.

 Đặt \(\left(a-10\right)^2-4=k^2\) (với \(k\inℕ\))

\(\Leftrightarrow\left(a-10\right)^2-k^2=4\)

\(\Leftrightarrow\left(a-10-k\right)\left(a-10+k\right)=4\)

Vì \(a-10-k\le a-10+k\) nên ta xét các TH sau:

TH1: \(\left\{{}\begin{matrix}a-10+k=2\\a-10-k=2\end{matrix}\right.\), khi đó \(k=0\) và \(a=12\)

\(\Rightarrow f\left(x\right)=x^2-22x+121=\left(x-11\right)^2\) thỏa ycbt.

TH2: \(\left\{{}\begin{matrix}a-10-k=1\\a-10+k=4\end{matrix}\right.\Rightarrow2k=3\), vô lí.

TH3: \(\left\{{}\begin{matrix}a-10-k=-2\\a-10+k=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}k=0\\a=8\end{matrix}\right.\). 

Thử lại, ta có \(f\left(x\right)=x^2-18x+81=\left(x-9\right)^2\) thỏa ycbt.

TH4; \(\left\{{}\begin{matrix}a-10-k=-4\\a-10+k=-1\end{matrix}\right.\Rightarrow2k=3\), vô lí.

Vậy \(a\in\left\{8;12\right\}\) thỏa ycbt.

\(3x^2+3xy-17=7x-2y\)

\(\Leftrightarrow3x\left(x+y\right)+2x+2y-9x-17=0\)

\(\Leftrightarrow3x\left(x+y\right)+2\left(x+y\right)-9x-6-11=0\)

\(\Leftrightarrow\left(x+y\right)\left(3x+2\right)-3\left(3x+2\right)=11\)

\(\Leftrightarrow\left(3x+2\right)\left(x+y-3\right)=11\)

\(\Leftrightarrow\left(3x+2\right);\left(x+y-3\right)\in\left\{-1;1;-11;11\right\}\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(-1;-7\right);\left(-\dfrac{1}{3};\dfrac{43}{3}\right);\left(-\dfrac{11}{3};\dfrac{17}{3}\right);\left(3;1\right)\right\}\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(-1;-7\right);\left(3;1\right)\right\}\left(x;y\inℤ\right)\)

\(x^2+y^2=3-xy\)

\(\Leftrightarrow\left(x-y\right)^2+2xy=3-xy\)

\(\Leftrightarrow\left(x-y\right)^2=3-3xy\)

\(\Leftrightarrow\left(x-y\right)^2=3\left(1-xy\right)\)

mà \(\left(x-y\right)^2\ge0,\forall x;y\inℤ\)

PT\(\Leftrightarrow\left\{{}\begin{matrix}x-y=3\\1-xy=3\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-y=0\\1-xy=0\end{matrix}\right.\)

\(TH1:\left\{{}\begin{matrix}x-y=3\\1-xy=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y+3\\xy=-2\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(1;-2\right);\left(2;-1\right);\left(-1;2\right);\left(-2;1\right)\right\}\)

\(TH2:\left\{{}\begin{matrix}x-y=0\\1-xy=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y\\xy=1\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)

Vậy \(\Leftrightarrow\left(x;y\right)\in\left\{\left(1;-2\right);\left(2;-1\right);\left(-1;2\right);\left(-2;1\right);\left(1;1\right);\left(-1;-1\right)\right\}\)

Gọi \(a;b;c\) là các cạnh tam vuông

Theo đề bài ta có :

 \(\left\{{}\begin{matrix}a^2+b^2=c^2\\\dfrac{1}{2}ab=\left(a+b+c\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=c^2\left(1\right)\\ab=2\left(a+b+c\right)\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow c^2=\left(a+b\right)^2-2ab\)

\(\Leftrightarrow c^2=\left(a+b\right)^2-4\left(a+b+c\right)\) (do (2))

\(\Leftrightarrow c^2+4=\left(a+b\right)^2-4\left(a+b\right)-4c+4\)

\(\Leftrightarrow\left(a+b\right)^2-4\left(a+b\right)+4=c^2+4c+4\)

\(\Leftrightarrow\left(a+b-2\right)^2=\left(c+2\right)^2\)

\(\Leftrightarrow a+b-2=c+2\left(đk:a+b\ge2\right)\)

\(\Leftrightarrow c=a+b-4\)

Thay vào \(\left(2\right)\) ta được

\(ab=2\left(a+b+a+b-4\right)\)

\(\Leftrightarrow ab=4a+4b-8\)

\(\Leftrightarrow ab-4a-4b+16=8\)

\(\Leftrightarrow a\left(b-4\right)-4\left(b-4\right)=8\)

\(\Leftrightarrow\left(a-4\right)\left(b-4\right)=8\)

\(\Leftrightarrow\left(a-4\right);\left(b-4\right)\in\left\{1;2;4;8\right\}\)

\(\Leftrightarrow\left(a;b\right)\in\left\{\left(5;12\right);\left(6;8\right);\left(8;6\right);\left(12;5\right)\right\}\)

\(\Leftrightarrow\left(a;b;c\right)\in\left\{\left(5;12;13\right);\left(6;8;10\right);\left(8;6;10\right);\left(12;5;13\right)\right\}\) thỏa đề bài

\(\Leftrightarrow x^2+2x+1=y^2+11\)

\(\Leftrightarrow\left(x+1\right)^2-y^2=11\)

\(\Leftrightarrow\left(x+1-y\right)\left(x+1+y\right)=11\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1-y=-1\\x+1+y=-11\end{matrix}\right.\\\left\{{}\begin{matrix}x+1-y=-11\\x+1+y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x+1-y=1\\x+1+y=11\end{matrix}\right.\\\left\{{}\begin{matrix}x+1-y=11\\x+1+y=1\end{matrix}\right.\end{matrix}\right.\)

Giải các hệ PT tìm x; y

 Nếu \(x< -3\) thì \(x^2+x+3< x^2\) và \(x^2+x+3>\left(x+1\right)^2\), vô lý.

 Nếu \(x>2\) thì \(x^2+x+3>x^2\) và \(x^2+x+3< \left(x+1\right)^2\), cũng vô lý.

 Do đó \(x\in\left\{-3;-2;-1;0;1;2\right\}\)

 Thử từng giá trị, ta thấy \(\left(x;y\right)\in\left\{\left(-3;3\right);\left(-3;-3\right)\right\}\) là các cặp số thỏa ycbt.

Ta có \(\left[\dfrac{34x+19}{11}\right]=\left[\dfrac{33x+11}{11}+\dfrac{x+8}{11}\right]=\left[x+1+\dfrac{x+8}{11}\right]\)

Nếu \(x< -19\) thì \(\left[\dfrac{34x+19}{11}\right]< 2x+1\) , vô lí.

Nếu \(-19\le x< -8\) thì \(-1\le\dfrac{x+8}{11}< 0\) nên \(\left[x+1+\dfrac{x+8}{11}\right]=x\), suy ra \(x=2x+1\) \(\Rightarrow x=-1\), loại.

Nếu \(-8\le x< 3\) thì \(0\le\dfrac{x+8}{11}< 1\) nên \(\left[x+1+\dfrac{x+8}{11}\right]=x+1\), suy ra \(x+1=2x+1\Leftrightarrow x=0\) (thỏa mãn)

Nếu \(x\ge3\) thì \(\dfrac{34x+19}{11}>2x+2\) hay \(\left[\dfrac{34x+19}{11}\right]\ge2x+2>2x+1\), vô lí.

Vậy \(x=0\)

\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\)

\(\Leftrightarrow x^2+\dfrac{1}{x^2}+x^2+\dfrac{y^2}{4}=4\left(1\right)\)

Theo Bất đẳng thức Cauchy cho các cặp số \(\left(x^2;\dfrac{1}{x^2}\right);\left(x^2;\dfrac{y^2}{4}\right)\)

\(\left\{{}\begin{matrix}x^2+\dfrac{1}{x^2}\ge2\\x^2+\dfrac{y^2}{4}\ge2.\dfrac{1}{2}xy\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+\dfrac{1}{x^2}\ge2\\x^2+\dfrac{y^2}{4}\ge xy\end{matrix}\right.\)

Từ \(\left(1\right)\Leftrightarrow x^2+\dfrac{1}{x^2}+x^2+\dfrac{y^2}{4}\ge2+xy\)

\(\Leftrightarrow4\ge2+xy\)

\(\Leftrightarrow xy\le2\left(x;y\inℤ\right)\)

\(\Leftrightarrow Max\left(xy\right)=2\)

Dấu "=" xảy ra khi

\(xy\in\left\{-1;1;-2;2\right\}\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(-1;-2\right);\left(1;2\right);\left(-2;-1\right);\left(2;1\right)\right\}\) thỏa mãn đề bài

hình như dấu "=" xảy ra khi x^2 = 1/x^2 với x^2 = y^2/4 mà bạn nhỉ

\(x-6\sqrt{x-3}+6\text{=}x-3-6\sqrt{x-3}+9\)

\(\text{=}\left(\sqrt{x-3}\right)^2-2.3.\sqrt{x-3}+\left(3\right)^2\)

\(\text{=}\left(\sqrt{x-3}-3\right)^2\)

A =  \(x-6\)\(\sqrt{x-3}\) + 6 (đkxd \(x>3\))

A = (\(x\) - 3) - 2.3.\(\sqrt{x-3}\) + 9 

A = (\(\sqrt{x-3}\))² - 2.3.\(\sqrt{x-3}\) + 3²

A = (\(\sqrt{x-3}\)- 3)²