\(3n+1⋮n-1\)
\(\Rightarrow\left(3n+1\right)-3\left(n-1\right)⋮n-1\)

\(\Rightarrow3n+1-3n+3⋮n-1\)

\(\Rightarrow4⋮n-1\)

\(\Rightarrow n-1\in\left\{1;2;-1;-2\right\}\)

\(\Rightarrow n\in\left\{2;3;0;-1\right\}\)

\(\left(3n-1\right)⋮n-1\\ =>\left(3n-1\right)-3\cdot\left(n-1\right)⋮n-1\)

\(\left(3n-1\right)-3\cdot\left(n-1\right)\\ =3n-1-3n+3\\ =\left(3n-3n\right)+\left(-1+3\right)\\ =0+2\\ =2\)

\(=>2⋮n\\ =>n\inƯ\left(2\right)\\ Ư\left(2\right)=\left\{1;-2;2;-1\right\}\)

\(Th1:3n-1=1\\ 3n=2\\ n=\dfrac{2}{3}\)

\(Th2:3n-1=-2\\ 3n=-1\\ =n=-\dfrac{1}{3}\)

\(Th3:3n-1=2\\ 3n=3\\ n=1\)

\(Th4:3n-1=-1\\ 3n=0\\ n=0\)

\(=>n\in\left\{0;1;-\dfrac{1}{3};\dfrac{2}{3}\right\}\)

1,3+(3+4,2:1,4)+-2^2

=1,3+(3+3)+4

=1,3+6+4

=11,3

\(1,3+\left(3+4,2\div1,4\right)+\left(-2\right)^2\)

\(=1,3+\left(3+3\right)+\left(-2\right)^2\)

\(=1,3+6+\left(-4\right)\)

\(=1,3+2\)

\(=3,3\)

1 giờ người thứ nhất làm được: 1 : 4 = \(\dfrac{1}{4}\) ( công việc).

1 giờ người thứ hai làm một mình được:  1 : 6 = \(\dfrac{1}{6}\) ( công việc)

1 giờ người thứ nhất và người thứ hai cùng làm được:

                          \(\dfrac{1}{4}\) + \(\dfrac{1}{6}\) = \(\dfrac{1}{3}\) (công việc)

Nếu hai người làm chung trong 2 giờ thì hai người làm được:

                         \(\dfrac{1}{3}\) \(\times\) 2 = \(\dfrac{2}{3}\) ( công việc)

vì \(\dfrac{2}{3}\) công việc < 1 công việc

Vậy trong hai giờ nếu hai người cùng làm thì không thể xong công việc được 

\(5,34\approx5,3\)
\(-3,65\approx-3,7\)
\(\Rightarrow5,34+\left(-3,65\right)\approx5,3+\left(-3,7\right)\approx1,6\)
\(\dfrac{2^{10}+2^9.9^5}{2^9+2^8.3^{10}}\)
\(=\dfrac{2^{10}+2^9.3^{10}}{2^9+2^8.3^{10}}\)
\(=\dfrac{2^9\left(2+3^{10}\right)}{2^8\left(2+3^{10}\right)}\)
\(=\dfrac{2^9}{2^8}\)
\(=2\)
#Đạt Đang Bận Thở

2¹⁰+2⁹.9⁵/2⁹+2⁸.3¹⁰

=2¹⁹.3¹⁰/2¹⁷.3¹⁰

=2¹⁹/2¹⁷

=2²/1

=4

5.34+(-3.65)

=5.34-3.65

=1.69

\(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{201\cdot203}\)

= \(\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{201\cdot203}\right)\)

= \(\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)

= \(\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)

= \(\dfrac{5}{2}\cdot\dfrac{202}{203}=\dfrac{505}{203}\)

Ta có :

  \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{201.203}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{201.203}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)

\(=\dfrac{5}{2}.\dfrac{202}{203}\)

\(=\dfrac{505}{203}\)

\(\dfrac{2x-1}{12}=\dfrac{5}{3}\)
\(\Rightarrow\left(2x-1\right)3=12\cdot5\)
\(\Rightarrow6x-3=60\)
\(\Rightarrow6x=63\)
\(\Rightarrow x=\dfrac{21}{2}\)
\(\dfrac{2x-3}{15}=\dfrac{3}{5}\)
\(\Rightarrow\left(2x-3\right)5=15\cdot3\)
\(\Rightarrow10x-15=45\)
\(\Rightarrow10x=60\)
\(\Rightarrow x=6\)
#Đạt Đang Bận Thở

\(\dfrac{2x-1}{12}=\dfrac{20}{12}\)

\(\Leftrightarrow2x-1=20\)

\(\Leftrightarrow2x=21\)

\(\Leftrightarrow x=\dfrac{21}{2}\)

b. \(\dfrac{2x-3}{15}=\dfrac{9}{15}\)

\(\Leftrightarrow2x-3=9\)

\(\Leftrightarrow x=6\)

\(\dfrac{x}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{y+3}\)  Đk (\(y\ne-3\))⇒ \(\dfrac{2x+3}{6}\) = \(\dfrac{1}{y+3}\) ⇒ (2\(x\)+3)(y+3) = 6

Ư(6) = { -6; -3; -2; -1; 1; 2; 3; 6}

Lập bảng ta có:

Từ bảng trên ta có các cặp \(x\), y nguyên thỏa mãn đề bài là:

(\(x\), y) = ( -3; -5); ( -2; -9); ( -1; 3); (0; -1); 

\(\dfrac{x-7}{3}+\dfrac{x-6}{4}=2\)

\(\Leftrightarrow\dfrac{4\left(x-7\right)}{12}+\dfrac{3\left(x-6\right)}{12}=\dfrac{2.12}{12}\)

\(\Leftrightarrow\dfrac{4x-28+3x-18}{12}=\dfrac{24}{12}\)

\(\Leftrightarrow\dfrac{7x-46}{12}=\dfrac{24}{12}\)

\(\Leftrightarrow7x-46=24\)

\(\Leftrightarrow7x=70\)

\(\Leftrightarrow x=10\)

\(\dfrac{4\left(x-7\right)}{12}+\dfrac{3\left(x-6\right)}{12}=\dfrac{24}{12}\)

\(4x-28+3x-18=24\)

\(7x-46=24\)

\(7x=70\)

\(x=10\)

A = \(\dfrac{3}{3\times7}\)+ \(\dfrac{3}{7\times11}\)+ \(\dfrac{3}{11\times15}\)+...+\(\dfrac{3}{107\times111}\)

A = \(\dfrac{3}{4}\) \(\times\)( \(\dfrac{4}{3\times7}\)+ \(\dfrac{4}{7\times11}\)+ \(\dfrac{4}{11\times15}\)+...+\(\dfrac{4}{107\times111}\))

A = \(\dfrac{3}{4}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\)+ \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\)+...+ \(\dfrac{1}{107}\)- \(\dfrac{1}{111}\))

A = \(\dfrac{3}{4}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{111}\))

A = \(\dfrac{9}{37}\) > \(\dfrac{9}{45}\) = \(\dfrac{1}{5}\) 

Vậy  \(\dfrac{3}{3\times7}\) + \(\dfrac{3}{7\times11}\)+ \(\dfrac{3}{11\times15}\) + ...+ \(\dfrac{3}{107\times111}\) > \(\dfrac{1}{5}\) ( đpcm)

Bạn ơi thế này thì đúng hơn chứ:

\(\dfrac{3}{3.7}+\dfrac{3}{7.11}+\dfrac{3}{11.15}+...+\dfrac{3}{107.111}>\dfrac{1}{5}\)