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Giải:
Theo bài ra ta có:
\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{12}\)
\(\Rightarrow-3\le x\le\frac{23}{12}\)
\(\Rightarrow x\varepsilon\left\{-2;-1;0;1\right\}\)
\(\frac{-5}{6}+\frac{16}{6}+-\frac{29}{6}\le x\le\frac{-6}{12}+\frac{24}{12}+\frac{5}{12}\)
=>-3\(\le\) x\(\le\) 23/12
=> x thuộc{-2-1;0;1}
Bài 1
a.\(\frac{-3}{4}\)-y:\(\frac{1}{5}\)=\(\frac{9}{28}\)
y:\(\frac{1}{5}\)=\(\frac{-15}{14}\)
y= \(\frac{-3}{14}\)
b.5x + 5x+2=650
5x . 1 + 5x + 52=650
5x(1+25)=650
5x.26=650
5x=25
x=2
\(a,\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left[2x-2\right]\cdot2x}=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left[\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{\left[2x-2\right]\cdot2x}\right]=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right]=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left[\frac{1}{2}-\frac{1}{2x}\right]=\frac{1}{8}\)
\(\Rightarrow\left[\frac{1}{2}-\frac{1}{2x}\right]=\frac{1}{8}:\frac{1}{2}\)
\(\Rightarrow\left[\frac{1}{2}-\frac{1}{2x}\right]=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow2x=4\Leftrightarrow x=2\)
Vậy x = 2
Mun ảnh đại diện cute
<3
À tk mk nhé. giờ mk tk bn trước
\(\frac{15}{41}+\frac{-138}{41}< x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\Leftrightarrow\frac{-123}{41}< x< \frac{1.3+1.2+1}{6}\)
\(\Leftrightarrow-3< x< 1\)
\(\Rightarrow x\in\left\{-2;-1;0\right\}\)
\(\frac{x}{5}=\frac{15}{2}-\frac{51}{10}\)
\(\frac{x}{5}=\frac{15.5-51}{10}\)
\(\frac{x}{5}=\frac{24}{10}\)
\(\frac{x}{5}=\frac{12}{5}\)
\(x=12\)
x-1/9 = 8/3
=> (x-1).3 = 9.8
=> (x-1).3 = 72
x-1 = 72:3
x-1 = 24
x = 24+1
=> x = 25
\(\frac{x-1}{9}=\frac{8}{3}\Rightarrow\frac{x-1}{9}=\frac{24}{9}\)\(\Leftrightarrow x-1=24\Rightarrow x=24+1=25\)
Vậy x=25