Question

[phạm thị hà phương]

1.thực hiện phép tính

a,\(\frac{18}{24}:\frac{5}{2}+\frac{7}{-10}\)                            \((\frac{12}{199}-\frac{23}{200}+\frac{34}{201}).(\frac{1}{2}-\frac{1}{3}-\frac{1}{6})\)

2.tìm x

a, x+5=15                                                                   b,\(x-\frac{5}{3}=\frac{1}{9}\)

\(c,\frac{x}{15}-\frac{-2}{3}+\frac{3}{5}\)                                             \(d,\frac{x}{9}< \frac{7}{x}< \frac{x}{6}(x\in z)\)

Answer 1

[Huỳnh Quang Sang]

\(1.a,\frac{18}{24}:\frac{5}{2}+\frac{7}{-10}\)

\(=\frac{18}{24}:\frac{5}{2}+\frac{-7}{10}\)

\(=\frac{18}{24}\cdot\frac{2}{5}+\frac{-7}{10}\)

\(=\frac{3}{4}\cdot\frac{2}{5}+\frac{-7}{10}\)

\(=\frac{3}{2}\cdot\frac{1}{5}+\frac{-7}{10}\)

\(=\frac{3}{10}+\frac{-7}{10}=\frac{-4}{10}=\frac{-2}{5}\)

\(\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right]\)

\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right]\)

\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{3-2-1}{6}\right]\)

\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{301}\right]\cdot0=0\)

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 :>