Bài làm:

1) \(\frac{3}{5}\div\frac{2x}{15}=\frac{1}{2}\div\frac{4}{5}\)

\(\Leftrightarrow\frac{9}{2x}=\frac{5}{8}\)

\(\Rightarrow10x=72\)

\(\Leftrightarrow x=\frac{36}{5}\)

2) \(-\frac{4}{2,5}\div\frac{3}{5}=\frac{1}{5}\div x\)

\(\Leftrightarrow\frac{1}{5}\div x=-\frac{8}{3}\)

\(\Rightarrow x=-\frac{3}{40}\)

3) \(0,12\div3=2x\div\frac{3}{5}\)

\(\Leftrightarrow\frac{1}{25}=\frac{10}{3}x\)

\(\Rightarrow x=\frac{3}{250}\)

\(=\dfrac{3}{2}-\dfrac{2}{21}-\dfrac{7}{12}+\left[\dfrac{15}{21}-\dfrac{1}{3}+\dfrac{5}{4}-\dfrac{2}{7}-\dfrac{1}{3}\right]\)

=11/12-2/21+5/7-2/3+5/4-2/7

=11/12-2/3+5/4-2/21+3/7

=11/12-8/12+15/12-2/21+9/21

=18/12+7/21

=3/2+1/3

=9/6+2/6=11/6

\(B=\dfrac{3}{2}-\dfrac{2}{21}-\left\{\dfrac{7}{12}-\left[\dfrac{15}{21}-\left(\dfrac{1}{3}-\dfrac{5}{4}\right)-\left(\dfrac{2}{7}+\dfrac{1}{3}\right)\right]\right\}\)

\(B=\dfrac{3}{2}-\dfrac{2}{21}-\left\{\dfrac{7}{12}-\left[\dfrac{15}{21}-\left(-\dfrac{11}{12}\right)-\dfrac{13}{21}\right]\right\}\)

\(B=\dfrac{3}{2}-\dfrac{2}{21}-\left\{\dfrac{7}{12}-\dfrac{85}{84}\right\}\)

\(B=\dfrac{3}{2}-\dfrac{2}{21}-\left(-\dfrac{3}{7}\right)\)

\(B=\dfrac{11}{6}\)

16:Sửa đề: \(\frac12x+\frac16\left(x-2\right)=\frac34-2x\)

=>\(\frac12x+\frac16x-\frac26=\frac34-2x\)

=>\(\frac46x-\frac13=\frac34-2x\)

=>\(\frac23x+2x=\frac34+\frac13\)

=>\(\frac83x=\frac{9}{12}+\frac{4}{12}=\frac{13}{12}\)

=>\(x=\frac{13}{12}:\frac83=\frac{13}{12}\times\frac38=\frac{13}{4\times8}=\frac{13}{32}\)

19: \(\frac{5}{12}x+3=\frac13-\frac{7}{12}x\)

=>\(\frac{5}{12}x+\frac{7}{12}x=\frac13-3\)

=>\(\frac{12}{12}x=\frac13-\frac93=-\frac83\)

=>\(x=-\frac83\)

20: \(\frac12x+\frac52=\frac72x-\frac34\)

=>\(\frac12x-\frac72x=-\frac34-\frac52\)

=>\(-3x=-\frac34-\frac{10}{4}=-\frac{13}{4}\)

=>\(3x=\frac{13}{4}\)

=>\(x=\frac{13}{4}:3=\frac{13}{12}\)

Ta có :

\(\frac{1}{5^2}>\frac{1}{5.6}\)

\(\frac{1}{6^2}>\frac{1}{6.7}\)

\(..............\)

\(\frac{1}{100^2}>\frac{1}{100.101}\)

\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(\Rightarrow A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow A>\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\left(1\right)\)

Lại có :

\(\frac{1}{5^2}< \frac{1}{4.5}\)

\(\frac{1}{6^2}< \frac{1}{5.6}\)

\(...............\)

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\left(2\right)\)

Từ (1) và (2) => Điều phải chứng minh