a: \(\dfrac{5}{7}+\dfrac{-3}{11}+\dfrac{2}{7}+\dfrac{-8}{11}\)

\(=\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\left(\dfrac{-3}{11}+\dfrac{-8}{11}\right)\)

\(=\dfrac{7}{7}-\dfrac{11}{11}=1-1=0\)

b: \(\dfrac{5}{11}-\dfrac{3}{7}-\dfrac{4}{7}+\dfrac{6}{11}\)

\(=\left(\dfrac{5}{11}+\dfrac{6}{11}\right)-\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)

\(=\dfrac{11}{11}-\dfrac{7}{7}=1-1=0\)

c: \(\dfrac{9}{13}-\dfrac{3}{8}-\dfrac{5}{8}-\dfrac{22}{13}\)

\(=\left(\dfrac{9}{13}-\dfrac{22}{13}\right)+\left(-\dfrac{3}{8}-\dfrac{5}{8}\right)\)

\(=-\dfrac{13}{13}-\dfrac{8}{8}=-1-1=-2\)

d: \(\dfrac{3}{16}-\dfrac{19}{16}+\dfrac{2}{3}+\dfrac{-8}{3}\)

\(=\left(\dfrac{3}{16}-\dfrac{19}{16}\right)+\left(\dfrac{2}{3}-\dfrac{8}{3}\right)\)

\(=-\dfrac{16}{16}+\dfrac{-6}{3}=-1-2=-3\)

\(2^{x+1}=640-2^{x+3}\)

=>\(2^{x+1}+2^{x+3}=640\)

=>\(2^x\cdot2+2^x\cdot8=640\)

=>\(10\cdot2^x=640\)

=>\(2^x=64=2^6\)

=>x=6

511+−37+611+23+−47=(511+611)+(−37+−47)+23=23

=0

\(\dfrac{18}{8}+\dfrac{-3}{8}-\dfrac{2}{5}+\dfrac{-3}{5}\)

\(=\left(\dfrac{18}{8}-\dfrac{3}{8}\right)+\left(-\dfrac{2}{5}-\dfrac{3}{5}\right)\)

=\(-\dfrac{5}{5}+\dfrac{15}{8}=-1+\dfrac{15}{8}=\dfrac{7}{8}\)

\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)

\(=\dfrac{6}{5}+\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{97\cdot99}+\dfrac{3}{99\cdot101}\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)=\dfrac{6}{5}+\dfrac{3}{2}\cdot\dfrac{96}{505}\)

\(=\dfrac{6}{5}+\dfrac{3\cdot48}{505}=\dfrac{150}{101}\)

Lời giải:
Gọi $d=ƯCLN(3n+2, 7n+1)$
$\Rightarrow 3n+2\vdots d; 7n+1\vdots d$

$\Rightarrow 7(3n+2)-3(7n+1)\vdots d$

$\Rightarrow 11\vdots d$
Để phân số trên rút gọn được thì $ƯCLN(3n+2, 7n+1)>1$

Hay $d>1$

$\Rightarrow d=11$

Điều này xảy ra khi: 

$3n+2\vdots 11$

$\Rightarrow 3n+2-11\vdots 11$

$\Rightarrow 3n-9\vdots 11$

$\Rightarrow 3(n-3)\vdots 11\Rightarrow n-3\vdots 11$

Đặt $n=11k+3$ với $k$ tự nhiên

$100< n< 150$

$\Rightarrow 100< 11k+3< 150$

$\Rightarrow 8,8< k< 13,3$

Mà $k$ là stn nên $k\in\left\{9; 10; 11; 12;13\right\}$

$\Rightarrow n\in \left\{102; 113; 124; 135; 146\right\}$

\(A=\dfrac{n+5}{2n-7}=\dfrac{1}{2}\cdot\dfrac{2n+10}{2n-7}\)

\(=\dfrac{1}{2}\cdot\dfrac{2n-7+17}{2n-7}\)

\(=\dfrac{1}{2}\cdot\left(1+\dfrac{17}{2n-7}\right)\)

Để A lớn nhất thì \(1+\dfrac{17}{2n-7}\) max

=>2n-7=1

=>2n=8

=>n=4

=>\(A_{max}=\dfrac{4+5}{2\cdot4-7}=9\)

\(\dfrac{x}{7}+\dfrac{1}{y}=\dfrac{-1}{14}\)

=>\(\dfrac{xy+7}{7y}=\dfrac{-1}{14}\)

=>\(14\left(xy+7\right)=-7y\)

=>2(xy+7)=-y

=>2xy+y=-14

=>y(2x+1)=-14

mà 2x+1 lẻ(do x nguyên)

nên \(\left(2x+1\right)\cdot y=1\cdot\left(-14\right)=\left(-1\right)\cdot14=7\cdot\left(-2\right)=\left(-7\right)\cdot2\)

=>\(\left(2x+1;y\right)\in\left\{\left(1;-14\right);\left(-1;14\right);\left(7;-2\right);\left(-7;2\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(0;-14\right);\left(-1;14\right);\left(3;-2\right);\left(-4;2\right)\right\}\)