\(2x^2+5x=12\)

\(\Leftrightarrow2x^2+5x-12=0\)

\(\Leftrightarrow2x^2+8x-3x-12=0\)

\(\Leftrightarrow2x\left(x+4\right)-3\left(x+4\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+4=0\\2x-3=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-4\\x=\frac{3}{2}\end{array}\right.\)

Vậy ............

-4/11 = 8/22 = 40/-110

x = 8

y = -110

\(\left(x,y\right)\in\left\{\left(-1;7\right)\left(-7;1\right)\right\}\)

\(\left(x,y\right)\in\left\{\left(-1;7\right);\left(-7;1\right)\right\}\)

(x,y)∈{(−1;7);(−7;1)}

=>5x-8y=0 và x-2y=12

=>5x-8y=0 và 5x-10y=60

=>2y=-60 và 5x=8y

=>y=-30 và 5x=-240

=>x=-48; y=-30

Đặt \(\left\{{}\begin{matrix}x+\sqrt{1+x^2}=a>0\\y+\sqrt{1+y^2}=b>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}1+x^2=a^2+x^2-2ax\\1+y^2=b^2+y^2-2by\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a^2-1}{2a}\\y=\dfrac{b^2-1}{2b}\end{matrix}\right.\)

Giả thiết trở thành: \(ab=2018\)

\(P=\dfrac{a^2-1}{2a}+\dfrac{b^2-1}{2b}=\dfrac{1}{2}\left(a+b\right)-\dfrac{a+b}{2ab}\)

\(P=\dfrac{1}{2}\left(a+b\right)\left(1-\dfrac{1}{ab}\right)=\dfrac{1}{2}\left(a+b\right).\dfrac{2017}{2018}\ge\sqrt{ab}.\dfrac{2017}{2018}=\dfrac{2017}{\sqrt{2018}}\)

\(P_{min}=\dfrac{2017}{\sqrt{2018}}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{2017}{2\sqrt{2018}}\)

2) Ta có: \(\left(2x+1\right).\left(3y-2\right)=-55=\left(-1\right).55=1.\left(-55\right)=\left(-5\right).11=5.\left(-11\right)\)

- Ta có bảng giá trị: 

Vậy \(\left(x,y\right)\in\left\{\left(-28,1\right);\left(-1,19\right);\left(2,-3\right);\left(5,-1\right)\right\}\)

3) Ta có: \(\left(x-2\right).\left(y+3\right)=5=\left(-1\right).\left(-5\right)=1.5\)

- Ta có bảng giá trị:

Vậy \(\left(x,y\right)\in\left\{\left(1,-8\right);\left(3,2\right);\left(-3,-4\right);\left(7,-2\right)\right\}\)

4) Ta có: \(\left(2x+3\right).\left(y-5\right)=10=\left(-1\right).\left(-10\right)=1.10=\left(-2\right).\left(-5\right)=2.5\)

- Vì \(x\in Z\)mà \(2x+3\)là số lẻ \(\Rightarrow\)\(2x+3\in\left\{-1,1,-5,5\right\}\)

- Ta có bảng giá trị:

Vậy \(\left(x,y\right)\in\left\{\left(-2,-5\right);\left(-1,16\right);\left(-4,3\right);\left(1,7\right)\right\}\)

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

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

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

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

\(=\left(y-1\right)\left[-x\left(x-1\right)-y\left(x-1\right)\left(x+1\right)\right]\)

\(=\left(y-1\right)\left(x-1\right)\left(-x-xy-y\right)\)