* Chứng minh \(\frac16

Ta có: \(F=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots+\frac{1}{100^2}\)

\(F=\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\cdots+\frac{1}{100\cdot100}\)

\(\Rightarrow F<\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\cdots+\frac{1}{99\cdot100}\)

\(\) \(\Rightarrow F<\frac14-\frac15+\frac15-\frac16+\frac16-\frac17+\cdots+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow F<\frac14-\frac{1}{100}\)

\(\Rightarrow F<\frac{12}{25}\)

Mà \(\frac16=\frac{12}{72}<\frac{12}{25}\)

\(\Rightarrow\frac16 (1)

* Chứng minh \(F<\frac14\)

Lại có: \(\) \(F=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots+\frac{1}{100^2}\)

\(F=\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\cdots+\frac{1}{100\cdot100}\)

\(\Rightarrow F>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\cdots+\frac{1}{100\cdot101}\)

\(\Rightarrow F>\frac15-\frac16+\frac16-\frac17+\frac17-\frac18+\cdots+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow F=\frac15-\frac{1}{101}\)

\(\Rightarrow F>\frac{96}{505}\)

Mà \(\frac14=\frac{96}{384}<\frac{96}{505}\)

\(\Rightarrow F<\frac14\) (2)

Từ (1) và (2) suy ra: \(\frac16

Vậy \(\frac16

a) \(x+2x+3x+...+100x=-213\)

\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)

\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)

\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)

d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)

                                                                    \(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)

c) \(3\left(x-2\right)+2\left(x-1\right)=10\)

\(\Rightarrow3x-6+2x-2=10\)

\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)

a) \(x+2x+3x+4x+...+100x=-213\)

\(x.\left(1+2+3+4+...+100\right)=-213\)

\(x.5050=-213\)

\(x=-\frac{213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)

\(\frac{1}{2}x=-\frac{43}{12}\)

\(x=\frac{-43}{6}\)

a) $\frac{7}{9} \times \frac{{15}}{{28}} \times \frac{9}{7} = (\frac{7}{9} \times \frac{9}{7}) \times \frac{{15}}{{28}} = 1 \times \frac{{15}}{{28}} = \frac{{15}}{{28}}$

b) $\frac{9}{{32}} \times \left( {\frac{2}{3} \times \frac{{14}}{{21}}} \right) = \frac{9}{{32}} \times \left( {\frac{2}{3} \times \frac{2}{3}} \right) = \frac{9}{{32}} \times \frac{4}{9} = \frac{{36}}{{288}} = \frac{1}{8}$

1/6 nha

\(\frac{4}{9}:\frac{8}{9}-\frac{7}{13}:\frac{21}{13}\)

\(=\frac{4}{9}.\frac{9}{8}-\frac{7}{13}.\frac{13}{21}\)

\(=\frac{1}{2}-\frac{1}{3}\)

\(=\frac{3}{6}-\frac{2}{6}=\frac{1}{6}\)

Ủng hộ tui nha! :) 

What?

    \(\dfrac{-4}{12}+\dfrac{18}{45}+\dfrac{-6}{9}+\dfrac{-21}{35}+\dfrac{6}{30}\)
\(=\dfrac{-1}{3}+\dfrac{2}{5}+\dfrac{-2}{3}+\dfrac{-3}{5}+\dfrac{1}{5}\)
\(=\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)+\left(\dfrac{2}{5}+\dfrac{-3}{5}+\dfrac{1}{5}\right)\)
\(=-1+0\)
\(=-1\)

\(\#PeaGea\)

chắc thế này :

\(\frac{9}{17}.\frac{21}{13}+\frac{5}{17}.\frac{9}{13}-\frac{8}{17}.2\)

\(=\frac{9}{17}.\frac{21}{13}+\frac{5}{13}.\frac{9}{17}-\frac{8}{17}.2\)

\(=\frac{9}{17}.\left(\frac{21}{13}+\frac{5}{13}\right)-\frac{8}{17}.2\)

\(=\frac{9}{17}.2-\frac{8}{17}.2\)

\(=2.\left(\frac{9}{17}-\frac{8}{17}\right)\)

\(=2.\frac{1}{17}\)

\(=\frac{2}{17}\)

1/17.(189/13+45/13)-16/17=1/17.18-16/17=18/17-16/17=2/17

Bài 1:

a) \(\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)

\(=\left(\frac{9}{25}-36\right):\left(\frac{19}{5}+\frac{1}{5}\right)\)

\(=\left(\frac{9}{25}-\frac{900}{25}\right):4\)

\(=-\frac{891}{25}.\frac{1}{4}\)

\(=-\frac{891}{100}\)

b) \(\frac{3}{8}.19\frac{1}{3}-\frac{3}{8}.33\frac{1}{3}\)

\(=\frac{3}{8}.\frac{58}{3}-\frac{3}{8}.\frac{100}{3}\)

\(=\frac{3}{8}\left(\frac{58}{3}-\frac{100}{3}\right)\)

\(=\frac{3}{8}\left(-\frac{42}{3}\right)\)

\(=\frac{3}{8}.\left(-14\right)\)

\(=-\frac{21}{4}\)

c) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)

\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{1}{2}+\frac{16}{21}\)

\(=\frac{27}{23}+\frac{5}{21}+\left(-\frac{4}{23}\right)+\frac{1}{2}+\frac{16}{21}\)

\(=\left[\frac{27}{23}+\left(-\frac{4}{23}\right)\right]+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac{1}{2}\)

\(=1+1=2\)

d) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)

\(=\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{36}{45}\)

\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)\)

\(=1+1=2\)