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a,\(\dfrac{x}{3}-\dfrac{1}{y}=\dfrac{1}{2}\)
=> \(\dfrac{1}{y}=\dfrac{x}{3}-\dfrac{1}{2}=>\dfrac{1}{y}=\dfrac{2x-3}{6}\)
=> y(2x-3)=6.1=6
=> y và 2x-3 là Ư (6)= {+-1,+-2,+-3,+-6}
| 2x-3 | -1 | 1 | 2 | -2 | 3 | -3 | 6 | -6 |
| x | 1 | 2 | 2,5 | 1/2 | 3 | 0 | 9/2 | -3/2 |
| y | -6 | 6 | 3 | -3 | 2 | -2 | 1 |
-1 |
vậy (x;y)= .......................
b,c làm tương tự
chúc bn học tốt 
Ta có :
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.................+\dfrac{2}{n\left(n+1\right)}=\dfrac{2003}{2004}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+............+\dfrac{2}{n\left(n+1\right)}=\dfrac{2003}{2004}\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+............+\dfrac{2}{2\left(n+1\right)}=\dfrac{2003}{2004}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{2003}{2004}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{2003}{2004}\)
\(\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{2003}{4008}\)
\(\dfrac{1}{n+1}=\dfrac{1}{4008}\)
\(\Rightarrow n+1=4008\)
\(\Rightarrow n=4007\) (Thỏa mãn \(n\in N\))
Vậy \(n=4007\) là giá trị cần tìm
~~Chúc bn học tốt~~
a) Để phân số \(\dfrac{3}{n-2}\) là số nguyên thì n - 2 \(⋮\) 3
\(\Rightarrow\) n - 2 \(\in\) Ư(3)
\(\Rightarrow\) n - 2 \(\in\){3; -3; 1;-1}
n \(\in\){5; -1; 3; 2}
c) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)
\(=\dfrac{1}{3}-\dfrac{1}{30}\)
\(=\dfrac{10}{30}-\dfrac{1}{30}\)
\(=\dfrac{9}{30}\)
=\(\dfrac{3}{10}\)
2, ta thấy:
\(\dfrac{2008}{2009}< \dfrac{2008}{2009+2010}\left(1\right)\)
\(\dfrac{2009}{2010}< \dfrac{2009}{2009+20010}\left(2\right)\)
từ (1) và (2) cộng vế với vế ta đc :\(\dfrac{2008}{2009}+\dfrac{2009}{20010}< \dfrac{2008}{2009+2010}+\dfrac{2009}{2009+2010}=\dfrac{2008+2009}{2009+2010}\)
2)\(x+y+z=9^2=81\)
Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{x+y+z}{15+20+28}=\dfrac{81}{63}=\dfrac{9}{7}\)
\(\Rightarrow x=\dfrac{135}{7};y=\dfrac{180}{7};z=36\)
Có: \(\dfrac{1}{21}+\dfrac{1}{77}+\dfrac{1}{165}+...+\dfrac{1}{n^2+4n}=\dfrac{56}{673}\)
\(\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+...+\dfrac{1}{n\left(n+4\right)}=\dfrac{56}{673}\)
\(\Leftrightarrow\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{n\left(n+4\right)}=\dfrac{4.56}{673}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{n}-\dfrac{1}{n+4}=\dfrac{224}{673}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{n+4}=\dfrac{224}{673}\)
\(\Leftrightarrow\dfrac{1}{n+4}=\dfrac{1}{2019}\)
\(\Leftrightarrow n=2015\)
a, \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
ta có: \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}=>\dfrac{3}{x}=\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{5-2y}{6}\)
=>\(\dfrac{3}{x}=\dfrac{5-2y}{6}=>x.\left(5-2y\right)=3.6=18\)
=> x và 5-2y thuộc Ư của 18={1,-1,2,-2,3,-3,6,-6}
vì 5-2y là số lẻ=> 5-2y= +-1 hoặc 5-2y=+-3
xét bảng
| 5-2y | 1 | -1 | 3 | -3 |
| y | 2 | 3 | 1 | 4 |
| x | 18 | -18 | 6 | -6 |
vậy giá trị x,y cần tìm là: {x=18.y=2}
{x=-18.y=3}
{x=6, y=1}Ư
{x=-6,y=4}
Bài 1 :
Sửa đề :
Tìm \(n\in Z\) để những phân số sau đồng thời có giá trị nguyên
\(\dfrac{-12n}{n};\dfrac{15}{n-2};\dfrac{8}{n+1}\)
Làm
Ta có :
\(\dfrac{-12n}{n}=-12\)
\(\Leftrightarrow\) Với mọi \(n\) thì \(\dfrac{-12n}{n}\) đều có giá trị nguyên \(\left(1\right)\)
Để \(\dfrac{15}{n-2}\in Z\) \(\Leftrightarrow n-2\inƯ\left(15\right)=\left\{\pm1;\pm15;\pm3;\pm5\right\}\)
\(\Leftrightarrow n\in\left\{-13;\pm3;\pm1;5;7;17\right\}\left(1\right)\)
Để \(\dfrac{8}{n+1}\in Z\Leftrightarrow n+1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Leftrightarrow n\in\left\{-9;-5;\pm3;-2;0;1;7\right\}\left(3\right)\)
Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow n\in\left\{\pm3;1;7\right\}\)
Giải:
Ta có: \(\dfrac{1}{m}+\dfrac{n}{6}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{m}=\dfrac{1}{2}-\dfrac{n}{6}\)
\(\Leftrightarrow\dfrac{1}{m}=\dfrac{3}{6}-\dfrac{n}{6}=\dfrac{3-n}{6}\)
\(\Leftrightarrow1.6=6=m\left(3-n\right)\)
Mà \(6=1.6=2.3=\left(-1\right).\left(-6\right)=\left(-2\right).\left(-3\right)\)
Ta có bảng sau:
Vậy...
Ta có \(\dfrac{1}{m}+\dfrac{n}{6}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{m}=\dfrac{1}{2}-\dfrac{n}{6}\)
\(\Rightarrow\dfrac{1}{m}=\dfrac{3}{6}-\dfrac{n}{6}\)
\(\Rightarrow\dfrac{1}{m}=\dfrac{3-n}{6}\)
\(\Rightarrow1\times6=\left(3-n\right)\times m\)
\(\Rightarrow6=\left(3-n\right)\times m\)
\(\Rightarrow\left(3-n\right);m\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(\Rightarrow\left(3-n\right)\times m=6=(-1)\times\left(-6\right)=(-6)\times\left(-1\right)=\left(-2\right)\times\left(-3\right)=\left(-3\right)\times\left(-2\right)=1\times6=6\times1=2\times3=3\times2\)
Ta có bảng sau
Vậy các cặp m,n thỏa mãn là