\(E=\dfrac{98:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right)\cdot\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}\\ E=\dfrac{98}{\dfrac{3}{5}}+\dfrac{\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\\ E=\dfrac{490}{3}+\dfrac{\dfrac{7}{4}}{7}=\dfrac{490}{3}+\dfrac{1}{4}=\dfrac{1963}{12}\)

Bài 1:

a.

$|x+\frac{7}{4}|=\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)

b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$

$|2x+1|=\frac{11}{15}$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)

c.

$3x(x+\frac{2}{3})=0$

\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)

d.

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

$\Leftrightarrow x=\frac{2}{5}$

Bài 2:

$\frac{1}{100}-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}$

$=\frac{99}{100}$

$\Rightarrow A=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=\frac{-49}{50}$

Bài 1: 

a) Ta có: \(\left|x+\dfrac{7}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{7}{4}=\dfrac{1}{2}\\x+\dfrac{7}{4}=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)

b) Ta có: \(\left|2x+1\right|-\dfrac{2}{5}=\dfrac{1}{3}\)

\(\Leftrightarrow\left|2x+1\right|=\dfrac{11}{15}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{11}{15}\\2x+1=\dfrac{-11}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-4}{15}\\2x=\dfrac{-26}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{15}\\x=\dfrac{-13}{15}\end{matrix}\right.\)

c) Ta có: \(3x\left(x+\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)

`( 1 + 1/12 ) xx ( 1 + 1/13 ) xx ( 1 + 1/14 ) xx ( 1 + 1/15 )`

`= 13/12 xx14/13 xx 15/14 xx 16/15`

`= ( 13 xx 14 xx 15 xx 16 )/( 12xx13xx14xx15 )`

`= 16/12`

`=4/3` 

`x : 4/7 xx 2/3 = 2`

` x  : 4/7 = 2 : 2/3`

` x : 4/7 = 3`

`x = 3 xx 4/7`

`x=12/7` 

Đổi `: 17` tấn `5` tạ `= 175` tạ 

Ô tô thứ nhất chở là `:`

`175 : ( 3 + 4 ) xx 3 = 75` `(` tạ `)`

Ô tô thứ hai chở là `:`

`175-75=100` `(` tạ `)`

Đ/s : Ô tô thứ nhất `: 75` tạ 

Ô tô thứ hai `: 100` tạ 

a) Ta có: \(7\cdot\dfrac{3}{14}-\dfrac{1}{14}\)

\(=\dfrac{3}{2}-\dfrac{1}{14}\)

\(=\dfrac{21}{14}-\dfrac{1}{14}\)

\(=\dfrac{10}{7}\)

b) Ta có: \(\dfrac{3}{2}+\dfrac{7}{4}:\dfrac{5}{2}\)

\(=\dfrac{3}{2}+\dfrac{7}{4}\cdot\dfrac{2}{5}\)

\(=\dfrac{3}{2}+\dfrac{7}{10}\)

\(=\dfrac{15}{10}+\dfrac{7}{10}=\dfrac{22}{10}=\dfrac{11}{5}\)

Lời giải:
\(7\times \frac{3}{14}-\frac{1}{14}=\frac{7\times 3}{14}-\frac{1}{14}=\frac{21}{14}-\frac{1}{14}=\frac{21-1}{14}=\frac{20}{14}=\frac{2\times 10}{2\times 7}=\frac{10}{7}\)

\(\frac{3}{2}+\frac{7}{4}:\frac{5}{2}=\frac{3}{2}+\frac{7}{4}\times \frac{2}{5}=\frac{3}{2}+\frac{7\times 2}{4\times 5}=\frac{3}{2}+\frac{7\times 2}{2\times 2\times 5}\)

\(=\frac{3}{2}+\frac{7}{2\times 5}=\frac{3\times 5}{2\times 5}+\frac{7}{2\times 5}=\frac{3\times 5+7}{2\times 5}=\frac{22}{2\times 5}=\frac{2\times 11}{2\times 5}=\frac{11}{5}\)

a. 7 x \(\dfrac{3}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{7}{1}\) x \(\dfrac{3}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{98}{14}\) x \(\dfrac{3}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{3}{2}\) - \(\dfrac{1}{14}\)

= \(\dfrac{21}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{10}{7}\)

b. \(\dfrac{3}{2}\) + \(\dfrac{7}{4}\) : \(\dfrac{5}{2}\)

= \(\dfrac{6}{4}\) + \(\dfrac{7}{4}\) : \(\dfrac{5}{2}\)

= \(\dfrac{13}{4}\) : \(\dfrac{5}{2}\)

= \(\dfrac{13}{4}\) . \(\dfrac{2}{5}\)

= \(\dfrac{13}{10}\)

viết p/s như nào vậy bạn

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{5}>\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{3}{5}< -\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x>1\\\dfrac{1}{2}x< \dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< \dfrac{2}{5}\end{matrix}\right.\)

a, \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)
\(=\dfrac{5}{7}+\dfrac{-5}{7}+\dfrac{3}{5} =0+\dfrac{3}{5}=\dfrac{3}{5}\)

b, \(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2=\dfrac{-3}{4}-\dfrac{-3}{2}+1=\dfrac{-3}{4}-\dfrac{-6}{4}+1=\dfrac{3}{4}+1=\dfrac{7}{4}\)

 c, \(\dfrac{-5}{9}+\left(\dfrac{-2}{3}\right)^2.\left(20\%-1.2\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(\dfrac{1}{5}-\dfrac{6}{5}\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(-1\right)=\dfrac{-5}{9}+\dfrac{-4}{9}=-1\)

Bài 1:

a) \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)\(=\left(\dfrac{5}{7}+\dfrac{-5}{7}\right)+\dfrac{3}{5}\)\(=0+\dfrac{3}{5}=\dfrac{3}{5}\)

b) \(\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2\)\(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+1\)\(=\dfrac{-3}{4}-\dfrac{-3}{2}+1\)

                                            \(=\dfrac{3}{4}+1\)\(=\dfrac{7}{4}\)

a)=2

b)=1/3

a,2

b,1/3

Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{138}{30}< X< \dfrac{200}{3}\)

\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)

Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)

\(\Rightarrow X=\dfrac{105381}{52}\)