\(a,-\dfrac{12}{16}-\left(\dfrac{3}{4}-x\right)=-\dfrac{5}{3}\)

\(\dfrac{3}{4}-x=-\dfrac{12}{16}-\left(-\dfrac{5}{3}\right)\)

\(\dfrac{3}{4}-x=\dfrac{11}{12}\)

\(x=\dfrac{3}{4}-\dfrac{11}{12}\)

\(x=-\dfrac{1}{6}\)

\(b,x-\dfrac{3}{7}:\dfrac{9}{14}=-\dfrac{7}{3}\)

\(x-\dfrac{3}{7}=-\dfrac{7}{3}\times\dfrac{9}{14}\)

\(x-\dfrac{3}{7}=-\dfrac{3}{2}\)

\(x=-\dfrac{3}{2}+\dfrac{3}{7}\)

\(x=-\dfrac{15}{14}\)

\(c,-\dfrac{3}{4}x+\dfrac{5}{8}x=\dfrac{1}{3}\)

\(\left(-\dfrac{3}{4}+\dfrac{5}{8}\right)x=\dfrac{1}{3}\)

\(-\dfrac{1}{8}x=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}:\left(-\dfrac{1}{8}\right)\)

\(x=-\dfrac{8}{3}\)

`-12/16 - ( 3/4 - x ) = -5/3`

`=> 3/4 - x=-12/16 -(-5/3)`

`=> 3/4 - x=-12/16 +5/3`

`=> 3/4 - x=11/12`

`=> x= 3/4-11/12`

`=> x=9/12 -11/12`

`=> x=-1/6`

__

`x - 3/7 : 9/14 = - 7/3`

`=>x - 3/7=-7/3 xx 9/14`

`=>x - 3/7=-3/2`

`=>x= -3/2 + 3/7`

`=>x=-15/14`

__

`-3/4 x + 5/8 x =1/3`

`=> (-3/4 + 5/8) x =1/3`

`=> ( -6/8+ 5/8) x =1/3`

`=> -1/8 x=1/3`

`=>x= 1/3 : (-1/8)`

`=> x= 1/3 xx (-8)`

`=>x= -8/3`

Có: 7-3.\(\frac{-1}{4}^2\)

= 7-3. \(\frac{1}{16}\)

= 7- \(\frac{3}{16}\)

= \(\frac{112}{16}\)-\(\frac{3}{16}\)

= \(\frac{109}{16}\)

`@`Bảng tần số:

\begin{array}{|c|c|c|}\hline \text{Giá trị (x)}&2&3&4&5&6&7&8&9&10&\\\hline \text{Tần số (n)}&3&5&4&4&4&3&3&2&1&N=29\\\hline\end{array}

`@` Mốt của dấu hiệu là: `3 ( n = 5)`

\(M_0=3\)

There are many traffic problem in the city where I live. Firstly, there are too many people using the roads. Secondly, there are too many vehicles on the road that leads to traffic jams. Finally, many roads are narrow and bumpy. Therefore, I think we should obey the traffic rules and traffic signs as well. Moreover, carefulness is very necessary.

Bạn nhớ tick cho mình và like mình nhé! Chúc bạn học tốt!

a,\(\frac{1}{2}+\frac{3}{4}-\frac{3}{4}+\frac{4}{5}\)

=\(\frac{1}{2}+\frac{3}{4}+\frac{-3}{4}+\frac{4}{5}\)

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

= \(0+\frac{1}{2}+\frac{4}{5}=\frac{13}{10}\)

Nhiều quá bạn ơi

A=\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{4^{99}}\)

4A=1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{4^{98}}\)

4A-A=(1+\(\dfrac{1}{4}\)+..+\(\dfrac{1}{4^{98}}\))-(\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{4^{99}}\))

3A=1-\(\dfrac{1}{4^{99}}\)

A=

A=\(\dfrac{1-\dfrac{1}{4^{99}}}{3}\)