\(a,\frac{x+3}{5}\in\Leftrightarrow x+3\in B5\Leftrightarrow x\in B5-3\)

\(b,\frac{7}{x-1}\in Z\Leftrightarrow x-1\inƯ7\Leftrightarrow x-1\in\left\{\pm1;\pm7\right\}\Leftrightarrow x\in\left\{-6;0;2;8\right\}\)

\(c,\frac{x+2}{x-1}\in Z\Leftrightarrow\frac{x-1+3}{x-1}\in Z\Leftrightarrow1+\frac{3}{x-1}\in Z\Leftrightarrow\frac{3}{x-1}\in Z\)

\(\Leftrightarrow x-1\inƯ3\Leftrightarrow x-1\in\left\{\pm1;\pm3\right\}\Leftrightarrow x\in\left\{-2;0;2;4\right\}\)

a, Ta có: \(\frac{x}{5}=\frac{-3}{7}\Rightarrow xy=-15\Rightarrow xy=-1.15=1.\left(-15\right)=-15.1=15.\left(-1\right)=3.\left(-5\right)=-3.5=-5.3=5.\left(-3\right)\) Vì \(x,y\in Z\)

Vậy \(\left(x;y\right)\in\left\{\left(-1;15\right);\left(1;-15\right);\left(15;-1\right);\left(-15;1\right);\left(3;-5\right);\left(-5;3\right);\left(5;-3\right);\left(-3;5\right)\right\}\)

b, \(\frac{-11}{x}=\frac{y}{-3}\Rightarrow xy=33\Rightarrow xy=3.11=11.3=-3.\left(-11\right)=-11.\left(-3\right)\)

Vậy \(\left(x;y\right)\in\left\{\left(3;11\right);\left(11;3\right);\left(-3;-11\right);\left(-11;-3\right)\right\}\)

a, Ta có: \(\frac{3n+9}{n-4}\in Z\Leftrightarrow\frac{3n-12+21}{n-4}\in Z\Leftrightarrow\frac{3\left(n-4\right)}{n-4}+\frac{21}{n-4}\in Z\Leftrightarrow3+\frac{21}{n-4}\in Z\)

\(\Leftrightarrow\frac{21}{n-4}\in Z\Leftrightarrow n-4\inƯ21\Leftrightarrow n-4\in\left\{\pm1;\pm3;\pm7;\pm21;\right\}\)

\(\Leftrightarrow n\in\left\{-17;-3;1;3;4;7;11;25\right\}\)

b, Ta có: \(\frac{6n+5}{2n-1}\in Z\Leftrightarrow\frac{6n-3+8}{2n-1}\in Z\Leftrightarrow\frac{3\left(2n-1\right)}{2n-1}+\frac{8}{2n-1}\in Z\Leftrightarrow3+\frac{8}{2n-1}\in Z\Leftrightarrow\frac{8}{2n-1}\in Z\)

\(\Leftrightarrow2n-1\inƯ8\Leftrightarrow2n-1\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Leftrightarrow n\in\left\{1;0\right\}\)  Vì \(n\in Z\)

Ta có: \(\frac{87^3+37^3}{124}-87.37=\frac{\left(87+37\right)\left(87^2-87.37+37^2\right)}{124}-87.37\)

\(=\frac{124\left(87^2-87.37+37^2\right)}{124}-87.37\)

\(=87^2-87.37+37^2-87.37\)

\(=87^2-2.87.37+37^2\)

\(=\left(87-37\right)^2=50^2=2500\)

1, 180

2, 6,25

3, 7,32

Ta có: \(\left(27x^3-1\right):\left(3x-1\right)=\left[\left(3x\right)^3-1^3\right]:\left(3x-1\right)=\left(3x-1\right)\left(9x^2+3x+1\right):\left(3x-1\right)=9x^2+3x+1\)

Ta có: \(\frac{4}{13};\frac{2}{13};\frac{1}{18};\frac{4}{5}< 1\)

\(\frac{7}{5};\frac{7}{3};\frac{7}{4}>1\)

Ta có: \(\frac{7}{5}< \frac{7}{4}< \frac{7}{3}\) \(\Rightarrow\) Hai số lớn nhất là \(\frac{7}{4};\frac{7}{3}\)

Có:  \(\frac{2}{13}< \frac{4}{13}< \frac{4}{5}\)  Quy đồng: \(\frac{1}{18}=\frac{2}{36}< \frac{2}{13}\) 

\(\Rightarrow\frac{2}{13};\frac{1}{18}\) là hai số bé nhất

Vậy tổng 4 số Bình và Hoàng chọn là: \(\frac{7}{4}+\frac{7}{3}+\frac{2}{13}+\frac{1}{18}=\frac{819+1092+72+26}{468}=\frac{2009}{468}\)