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x=\(\dfrac{52}{63}\) x -\(\dfrac{9}{26}\)
x=-1\(\dfrac{5}{7}\) =-\(\dfrac{2}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{7}{78}:x=\dfrac{35}{52}\)
\(x=\dfrac{7}{78}:\dfrac{35}{52}\)
\(x=\dfrac{7}{78}.\dfrac{52}{35}\)
\(x=\dfrac{2}{15}\)
\(\dfrac{7}{78}=\dfrac{35}{52}x\)
\(x=\dfrac{7.52}{78.35}\)
\(x=\dfrac{2}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left|x\right|\cdot\dfrac{-31}{7}-2\left|x\right|=\dfrac{-8}{7}\)
\(\Leftrightarrow\left|x\right|\cdot\dfrac{-45}{7}=\dfrac{-8}{7}\)
=>|x|=8/45
=>x=8/45 hoặc x=-8/45
b: \(\Leftrightarrow\left(\dfrac{83}{15}-\dfrac{4}{17}\right):x=\dfrac{365}{2002}\)
\(\Leftrightarrow\dfrac{1351}{255}:x=\dfrac{365}{2002}\)
hay \(x\simeq29,06\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}>\dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{25}{100}=\dfrac{1}{4}\)
Do đó: \(A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)(1)
Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{25}{50}=\dfrac{1}{2}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}< \dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
Do đó: \(A< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)(2)
Từ (1) và (2) ta suy ra ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1\cdot3\cdot5\cdot...\cdot99=\dfrac{\left(1\cdot3\cdot5\cdot...\cdot99\right)\cdot\left(2\cdot4\cdot6\cdot...\cdot100\right)}{2\cdot4\cdot6\cdot...\cdot100}\)
\(=\dfrac{1\cdot3\cdot5\cdot...\cdot2\cdot4\cdot6\cdot...\cdot100}{1\cdot2\cdot3\cdot...\cdot50\cdot2\cdot2\cdot...\cdot2}=\dfrac{51}{2}\cdot\dfrac{52}{2}\cdot...\cdot\dfrac{100}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\dfrac{9}{18}-\dfrac{-7}{12}+\dfrac{13}{32}\)
\(=\dfrac{1}{2}+\dfrac{7}{12}+\dfrac{13}{32}\)
\(=\dfrac{13}{12}+\dfrac{13}{32}=\dfrac{143}{96}\)
b, \(\dfrac{5}{-8}+\dfrac{14}{39}-\dfrac{6}{10}\)
\(\dfrac{-5}{8}+\dfrac{14}{39}-\dfrac{3}{5}\)
\(=\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{14}{39}\)
\(=\dfrac{-49}{40}+\dfrac{14}{39}=\dfrac{-1351}{1560}\)
Lời giải:
$\frac{52-x}{7^2}=\frac{9}{52-x}$
$(52-x)^2=7^2.9=7^2.3^2=21^2=(-21)^2$
$\Rightarrow 52-x=21$ hoặc $52-x=-21$
$\Rightarrow x=52-21$ hoặc $x=52+21$
$\Rightarrow x=31$ hoặc $x=73$