Chứng minh rằng (9^2n+1994^93)chia heets cho 5

a) \(x\cdot\frac{1}{2}+x\cdot\frac{1}{4}+x\cdot\frac{1}{8}=\frac{21}{24}\)

\(x\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)=\frac{7}{8}\)

\(x\cdot\frac{7}{8}=\frac{7}{8}\)

\(\Rightarrow x=\frac{7}{8}\div\frac{7}{8}=1\)

b) \(\left(x+4\right)+\left(x+9\right)+\left(x+14\right)+.....+\left(x+44\right)+\left(x+49\right)=1430\)

\(\left(x+x+x+....+x+x\right)+\left(4+9+14+...+44+49\right)=1430\)

\(10x+265=1430\)

\(10x=1430-265\)

\(10x=1165\)

\(\Rightarrow x=\frac{1165}{10}=116,5\)

c) \(x\cdot0,25-0,5=1\)

\(x\cdot0,25=1+0,5\)

\(x\cdot0,25=1,5\)

\(\Rightarrow x=1,5\div0,25=6\)

\(\frac{8}{23}\cdot\frac{46}{24}-x=\frac{1}{3}\)

=> \(\frac{2}{3}-x=\frac{1}{3}\)

=> \(x=\frac{1}{3}\)

\(\frac{10}{12}\div x=\frac{28}{9}\cdot\frac{3}{56}\)

=> \(\frac{10}{12}\div x=\frac{1}{6}\)

=> \(x=\frac{60}{12}=5\)

\(\frac{x-12}{4}=\frac{1}{2}\)

=> \(\left(x-12\right)\cdot2=4\cdot1\)

=> \(2x-24=4\)

=> \(2x=28\)

=> \(x=14\)

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

bây h thì bạn giải đc chưa

Cảm ơn bạn rất nhiều mình đã hiểu rồi 

Chúc bạn học tốt nhé

Ta có:

\(\frac{x}{2013}\)-\(\frac{1}{10}\)-\(\frac{1}{15}\)-\(\frac{1}{21}\)-...-\(\frac{1}{120}\)=\(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- (\(\frac{2}{20}\)+\(\frac{2}{30}\)+\(\frac{2}{42}\)+...+\(\frac{2}{240}\)) = \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{15.16}\)) = \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4}\)-\(\frac{1}{10}\)) = \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- 2.\(\frac{3}{10}\)= \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)= \(\frac{5}{8}\)+\(\frac{6}{10}\)= 1

=> \(x=2013\)

Vậy \(x=2013\)

2013 nha

khong hieu

a Đ

b S

c S

d Đ

a ) S 

b ) Đ

c ) S

d ) Đ

k cho mk nhé 

Mai Hồng Ngọc? Vũ Thị Thu Hằng? Ai đúng dzậy -_-*

\(\frac{-5}{x}=\frac{-y}{8}=\frac{18}{72}\)

\(\Leftrightarrow\frac{-5}{x}=\frac{-y}{8}=\frac{1}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}-\frac{5}{x}=\frac{1}{4}\\-\frac{y}{8}=\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5.4:1\\-y=8.1:4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-20\\-y=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-20\\y=-2\end{cases}}}\)

vậy x=-20 và y=-2

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

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

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

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

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

\(x=-\frac{4}{12}-\frac{9}{12}\)

\(x=-\frac{13}{12}\)

\(x=-\frac{18}{24}+\frac{15}{21}\)

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

\(x=-\frac{21}{28}+\frac{20}{28}\)

\(x=-\frac{1}{28}\)

\(\left(3x+1\right)^2=25\)

\(\Rightarrow\left(3x+1\right)^2=5^2=\left(-5\right)^2\)

\(\Rightarrow\orbr{\begin{cases}3x+1=5\\3x+1=-5\end{cases}\Rightarrow\orbr{\begin{cases}3x=5-1=4\\3x=-5-1=-6\end{cases}}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-2\end{cases}}\)

\(\left[x-\frac{1}{2}\right]+\frac{1}{2}=\frac{5}{8}\)

\(\Rightarrow x-0=\frac{5}{8}\)

\(x=\frac{5}{8}\)

\(\left[x+\frac{3}{4}\right]-\frac{1}{3}=0\)

\(x+\frac{3}{4}=0+\frac{1}{3}=\frac{1}{3}\)

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

\(x=\frac{-5}{12}\)

\(\frac{1}{9}.3^4.3x=3^7\)

\(\Rightarrow\frac{1}{9}.3^5x=3^7\)

\(x=3^7:\frac{1}{9}:3^5\)

\(x=3^2.9\)

\(x=81\)

\(\frac{x}{5}=\frac{4}{21}\)

\(\Rightarrow21x=4.5\)

\(\Rightarrow x=\frac{4.5}{21}\)

\(x=\frac{20}{21}\)

Nhân hết ra,giải phương trình bậc cao đi