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

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

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

Chọn A

A nha

ngu

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