Ta có: \(6xy+4x+15y+18=0\\ \Leftrightarrow\left(6xy+15y\right)+\left(4x+18\right)=0\\ \Leftrightarrow3y\left(2x+5\right)+2\left(2x+6\right)=0\Leftrightarrow3y\left(2x+5\right)+2\left(2x+5\right)+2=0\\ \Leftrightarrow\left(2x+5\right)\left(3y+2\right)=-2\)

Vì \(x,y\inℤ\) nên \(2x+5\inℤ;3y+2\inℤ\) 

\(\Rightarrow\left(2x+5;3y+2\right)\inƯ\left(-2\right)\\\Rightarrow\left(2x+5;3y+2\right)\in \left\{\pm1;\pm2\right\}\)

Ta có bảng sau:

\(\Rightarrow\left(x;y\right)\in\left\{\left(\dfrac{8}{11};-\dfrac{1}{11}\right);\left(\dfrac{1}{11};\dfrac{4}{11}\right);\left(-\dfrac{8}{11};\dfrac{1}{11}\right);\left(-\dfrac{1}{11};-\dfrac{4}{11}\right)\right\}\)(Loại)

Vậy không có nghiệm \(x,y\inℤ\)

\(-\dfrac{2}{7}+\dfrac{5}{12}\cdot\dfrac{18}{35}\)

\(=\dfrac{-2}{7}+\dfrac{18}{12}\cdot\dfrac{5}{35}\)

\(=-\dfrac{2}{7}+\dfrac{3}{2}\cdot\dfrac{1}{7}=\dfrac{-2}{7}+\dfrac{3}{14}=\dfrac{-4+3}{14}=-\dfrac{1}{14}\)

-2/7 + 5/12 . 18/35

= -2/7 + 1/2 . 3/7

= -2/7 + 3/14

= -4/14 + 3/14

= 1/14

\(\dfrac{x}{3}-\dfrac{2}{y}=\dfrac{1}{5}\)

=>\(\dfrac{xy-6}{3y}=\dfrac{1}{5}\)

=>5(xy-6)=3y

=>5xy-30=3y

=>5xy-3y=30

=>y(5x-3)=30

mà 5x-3>=-3 và y>=0(vì x và y là số tự nhiên)

nên \(\left(5x-3\right)\cdot y=30\cdot1=15\cdot2=10\cdot3=6\cdot5=5\cdot6=3\cdot10=2\cdot15=1\cdot30\)

=>\(\left(x;y\right)\in\left\{\left(\dfrac{33}{5};1\right);\left(\dfrac{18}{5};2\right);\left(\dfrac{13}{5};3\right);\left(\dfrac{9}{5};5\right);\left(\dfrac{8}{5};6\right);\left(\dfrac{6}{5};10\right);\left(1;15\right);\left(\dfrac{4}{5};30\right)\right\}\)

mà x,y là các số tự nhiên

nên \(\left(x;y\right)\in\left(1;15\right)\)

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

=>\(x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=-\dfrac{10}{3}+1\)

=>\(x\cdot\dfrac{7}{6}=\dfrac{-7}{3}\)

=>\(x=-\dfrac{7}{3}:\dfrac{7}{6}=-\dfrac{7}{3}\cdot\dfrac{6}{7}=-2\)

toán thống kê 

ý bạn là bảng thống kê à?

\(#Sayaa-chan\)

dễ bạn chỉ cần lập bảng theo đề bài yêu cầu.

Vậy nha!

\(#HT\)

\(#Sayaa-chan\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}=\dfrac{1}{n-1}-\dfrac{1}{n}\)

Do đó: \(C=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

=>\(C< 1-\dfrac{1}{n}< 1\)

\(600=2^3\cdot3\cdot5^2\)

=>Số các ước dương của 600 là (3+1)*(1+1)*(2+1)=4*2*3=24(ước)

\(\dfrac{-2}{5}x+\dfrac{2}{3}=\dfrac{7}{15}\)

\(\dfrac{-2}{5}x=\dfrac{7}{15}-\dfrac{2}{3}\)

\(\dfrac{-2}{5}x=\dfrac{-1}{5}\)

\(x=\dfrac{-1}{5}:\dfrac{-2}{5}\)

\(x=\dfrac{-1}{5}\cdot\dfrac{5}{-2}\)

\(x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

Em nói rõ yêu cầu của đề bài ra em nhé.

\(-\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{7}{15}\)

=>\(-\dfrac{2}{5}x=\dfrac{7}{15}-\dfrac{2}{3}=\dfrac{7}{15}-\dfrac{10}{15}=-\dfrac{3}{15}=-\dfrac{1}{5}\)

=>\(x=\dfrac{-1}{5}:\dfrac{-2}{5}=\dfrac{1}{2}\)

\(\dfrac{-2}{5}x+\dfrac{2}{3}=\dfrac{7}{15}\\ \Rightarrow\dfrac{-2}{5}x=\dfrac{7}{15}-\dfrac{2}{3}\\ \Rightarrow\dfrac{-2}{5}x=\dfrac{-1}{5}\\ \Rightarrow x=\dfrac{-1}{5}:\dfrac{-2}{5}\\ \Rightarrow x=\dfrac{-1}{5}.\dfrac{5}{-2}\\ \Rightarrow x=\dfrac{1}{2}\)

Vậy x = \(\dfrac{1}{2}\)