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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{2}+\dfrac{-1}{3}+\dfrac{-2}{3}\le x< \dfrac{-3}{5}+\dfrac{1}{6}+\dfrac{-2}{5}+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}+\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)\le x< \left(\dfrac{-3}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{1}{6}+\dfrac{3}{2}\right)\)
\(\Leftrightarrow\dfrac{1}{2}+\left(-1\right)\le x< -1+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{-1}{2}\le x< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{-3}{6}\le x< \dfrac{4}{6}\)
\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x}{6}-\dfrac{5}{2y+1}=\dfrac{2}{3}\)
\(\dfrac{x}{6}-\dfrac{5.2}{2y.2+1.2}=\dfrac{4}{6}\)(vì 2y + 1 là số lẻ)
\(\dfrac{x}{6}-\dfrac{10}{4y+2}=\dfrac{4}{6}\)
Để \(\dfrac{x}{6}-\dfrac{10}{4y+2}=\dfrac{4}{6}\)thì y = 1 để cùng mẫu số
Khi đó ta có\(\dfrac{x}{6}-\dfrac{10}{4y+2}=\dfrac{4}{6}\) = \(\dfrac{x}{6}-\dfrac{10}{4+2}=\dfrac{4}{6}\) = \(\dfrac{x}{6}-\dfrac{10}{6}=\dfrac{4}{6}\)
Vì 4+10 = 14 => x = 14
Vậy y = 1; x = 14
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>x-3=9
=>x=12
b: =>10-x=-26
=>x=36
c: =>x:4-1=2
=>x:4=3
=>x=12
d: =>x^2=4
=>x=2 hoặc x=-2
e: =>(x-2)^2=100
=>x-2=10 hoặc x-2=-10
=>x=12 hoặc x=-8
![](https://rs.olm.vn/images/avt/0.png?1311)
a, `2/(x-1) in ZZ`.
`=> 2 vdots x - 1`
`=> x-1 in Ư(2)`
`=> x - 1 in {+-1, +-2}`.
`=> x - 1 = 1 => x = 2`.
`=> x - 1 = -1 => x = 0`.
`=> x - 1 = -2 => x = -1`.
`=> x - 1 = 2 => x = 3`.
Vậy `x = 2, 0, - 1, 3`.
b, `4/(2x-1) in ZZ`
`=> 4 vdots 2x - 1`.
`=> 2x - 1 in Ư(4)`
Vì `2x vdots 2 => 2x - 1 cancel vdots 2`
`=> 2x - 1 in {+-1}`
`=> 2x - 1 = -1 => x = 0`.
`=> 2x - 1 = 1 => x = 1`
Vậy `x = 0,1`.
c, `(x+3)/(x-1) in ZZ`.
`=> x + 3 vdots x - 1`
`=> x - 1 + 4 vdots x - 1`.
`=> 4 vdots x-1`
`=> x -1 in Ư(4)`
`=> x - 1 in{+-1, +-2, +-4}`
`x - 1 = 1 => x = 2`.
`x - 1 = -1 => x = 0`.
`x - 1 = 2 =>x = 3`.
`x - 1 = -2 => x = -1`.
`x - 1 = 4 => x = 5`.
`x - 1 = -4 => x = -3`.
Vậy `x = 2, 0 , +-1, 5, -3`.
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Trước hết ta chứng minh \(a^2-1=\left(a-1\right)\left(a+1\right)\text{tự chứng minh }\)
Áp dụng bổ đề trên ta có:
\(-A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right) =\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot...\cdot\dfrac{100^2-1}{100^2}=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}=\dfrac{1\cdot2\cdot3^2\cdot...\cdot99^2\cdot100\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\dfrac{1\cdot101}{2\cdot100}>\dfrac{1}{2}\\ \Rightarrow A< -\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
x | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
y | -10 | 10 | -5 | 5 | -2 | 2 | -1 | 1 |
c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
x - 1 | 1 | -1 | 3 | -3 |
y + 1 | 3 | -3 | 1 | -1 |
x | 2 | 0 | 4 | -2 |
y | 2 | -4 | 0 | -2 |
b: =>xy=12
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2 - 1/y = x/3
3y - 6 = 2xy
3y - 2xy = 6
y(3 - 2x) = 6
Do x là số nguyên nên 2x là số chẵn
3 - 2x là số lẻ
* TH1: 3 - 2x = -3 và y = -2
+) 3 - 2x = -3
2x = 3 + 3
2x = 6
x = 6 : 2
x = 3
* TH2: 3 - 2x = -1 và y = -6
+) 3 - 2x = -1
2x = 3 + 1
2x = 4
x = 4 : 2
x = 2
* TH3: 3 - 2x = 1 và y = 6
+) 3 - 2x = 1
2x = 3 - 1
2x = 2
x = 2 : 2
x = 1
* TH4: 3 - 2x = 3 và y = 2
+) 3 - 2x = 3
2x = 3 - 3
2x = 0
x = 0
Vậy ta tìm được các cặp giá trị (x; y) thỏa mãn:
(3; -2); (2; -6); (1; 6); (0; 2)
\(\dfrac{1}{2}\) - \(\dfrac{1}{y}\) = \(\dfrac{x}{3}\) (đk y ≠ 0)
\(\dfrac{x}{3}\) + \(\dfrac{1}{y}\) - \(\dfrac{1}{2}\) = 0
\(\dfrac{2xy+6-3y}{6y}\) = 0
2\(xy\) + 6 - 3y = 0
6 - y.(3 - 2\(x\)) = 0
y.(3 - 2\(x\)) = 6
Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}
lập bảng ta có:
y | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
3 - 2\(x\) | -1 | -2 | -3 | -6 | 6 | 3 | 2 | 1 |
\(x\) | 2 | \(\dfrac{5}{2}\) | 3 | \(\dfrac{9}{2}\) | -\(\dfrac{3}{2}\) | 0 | \(\dfrac{1}{2}\) | 1 |
Vì \(x;y\) nguyên theo bảng trên ta có:
Các cặp \(x;y\) nguyên thỏa mãn đề bài là:
(\(x;y\)) = (2; -6); (3; -2); (0;2); (1;6)
Lời giải:
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+....+\frac{1}{\frac{x(x+1)}{2}}\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x(x+1)}\right)\)
\(=1+2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{(x+1)-x}{x(x+1)}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)
\(=1+2(\frac{1}{2}-\frac{1}{x+1})=2-\frac{2}{x+1}\)
Ta có: $2-\frac{2}{x+1}=2$
$\Leftrightarrow \frac{2}{x+1}=0$ (vô lý)
Vậy không tồn tại $x$ nguyên dương thỏa mãn.