Ta có:

\(\left|x+3\right|+\left|x-1\right|=\left|x+3\right|+\left|1-x\right|\ge\left|x+3+1-x\right|=4\)

\(3-y^2-2y=4-\left(y^2+2y+1\right)=4-\left(y+1\right)^2\le4\)

\(\Rightarrow\left|x+3\right|+\left|x-1\right|\ge3-y^2-2y\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\left(x+3\right)\left(1-x\right)\ge0\\y+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le1\\y=-2\end{matrix}\right.\)

Các cặp số nguyên thỏa mãn là:

\(\left(x;y\right)=\left(-3;-2\right);\left(-2;-2\right);\left(-1;-2\right);\left(0;-2\right);\left(1;-2\right)\)

(x-3)(y-5)=-7

=>\(\left(x-3;y-5\right)\in\left\{\left(1;-7\right);\left(-7;1\right);\left(-1;7\right);\left(7;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(4;-2\right);\left(-4;6\right);\left(2;12\right);\left(10;4\right)\right\}\)

\(5x-y\left(x-3\right)=8\)

\(\Leftrightarrow5x-15-y\left(x-3\right)=8-15\)

\(\Leftrightarrow5\left(x-3\right)-y\left(x-3\right)=-7\)

\(\Leftrightarrow\left(5-y\right)\left(x-3\right)=-7\)

Bảng giá trị:

Vậy các cặp số nguyên thỏa mãn là:

\(\left(x;y\right)=\left(4;12\right);\left(10;6\right);\left(-4;4\right);\left(2;-2\right)\)

theo bài ra ta có: x/-3=7/y

=>x.y=(-3).7=-21=(-3).7=3.(-7)=21.(-1)=1.(-21)

vì x,y là số nguyên

=>(x;y) thuộc (3;-7);(-7;3);(-3;7);(7;-3);(-1;21);(21;-1);(1;-21);(-21;1)

Ta có: \(\frac{x}{-3}=\frac{7}{y}\Rightarrow x.y=-3.7=-21\)

Vậy bạn tìm các cặp số có tích là -21 nhé

\(y\left(x+3\right)-5x-15=2\\ \Rightarrow y\left(x+3\right)-\left(5x+15\right)=2\\ \Rightarrow y\left(x+3\right)-5\left(x+3\right)=2\\ \Rightarrow\left(y-5\right)\left(x+3\right)=2\)

Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}y-5,x+3\in Z\\y-5,x+3\inƯ\left(2\right)\end{matrix}\right.\)

Ta có bảng:

Vậy \(\left(x,y\right)\in\left\{\left(-2;7\right);\left(-1;6\right);\left(-4;3\right);\left(-5;4\right)\right\}\)

=>y.(x+3)-5(x+3)=2

=>(y-5).(x+3)=2