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1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
\(1.a,\frac{18}{24}:\frac{5}{2}+\frac{7}{-10}\)
\(=\frac{18}{24}:\frac{5}{2}+\frac{-7}{10}\)
\(=\frac{18}{24}\cdot\frac{2}{5}+\frac{-7}{10}\)
\(=\frac{3}{4}\cdot\frac{2}{5}+\frac{-7}{10}\)
\(=\frac{3}{2}\cdot\frac{1}{5}+\frac{-7}{10}\)
\(=\frac{3}{10}+\frac{-7}{10}=\frac{-4}{10}=\frac{-2}{5}\)
\(\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right]\)
\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right]\)
\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{3-2-1}{6}\right]\)
\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{301}\right]\cdot0=0\)
2. \(a,x+5=15\Leftrightarrow x=15-5=10\)
\(b,x-\frac{5}{3}=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{1}{9}+\frac{5}{3}\)
\(\Leftrightarrow x=\frac{1}{9}+\frac{15}{9}=\frac{16}{9}\)
c, Sửa lại đề một xíu :
\(\frac{x}{15}=\frac{-2}{3}+\frac{3}{5}\)
\(\Leftrightarrow\frac{x}{15}=\frac{-10}{15}+\frac{9}{15}\)
\(\Leftrightarrow\frac{x}{15}=\frac{-1}{15}\)
\(\Leftrightarrow x=-1\)
\(d,\frac{x}{9}< \frac{7}{x}< \frac{x}{6}(x\inℤ)\)
\(\frac{x\cdot x}{9\cdot x}< \frac{7\cdot9}{9\cdot x}< \frac{7\cdot6}{6\cdot x}\)
\(\Leftrightarrow\frac{x^2}{9x}< \frac{63}{9x}< \frac{42}{6x}\)
Tự làm nốt :>