bài1  

a) \(\dfrac{7}{6}-\dfrac{13}{12}+\dfrac{3}{4}\) 

=\(\dfrac{14}{12}-\dfrac{13}{12}+\dfrac{9}{12}\) 

=\(\dfrac{1}{12}+\dfrac{9}{12}\) 

=\(\dfrac{10}{12}=\dfrac{5}{6}\)

bài 1 

b)\(1\dfrac{1}{2}.(\dfrac{-4}{5})\) + \(\dfrac{3}{10}\) 

= \(\dfrac{3}{2}.\left(-\dfrac{4}{5}\right)+\dfrac{3}{10}\) 

= \(-\dfrac{6}{5}+\dfrac{3}{10}\) 

=\(-\dfrac{12}{10}+\dfrac{3}{10}\) 

=\(-\dfrac{9}{10}\) 

Bài 4: 

a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)

\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)

\(1,25-x=\dfrac{11}{12}\)

\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)

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

b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)

\(x-\dfrac{7}{6}=\dfrac{13}{12}\)

\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)

\(x=\dfrac{27}{12}=\dfrac{9}{4}\)

c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)

\(4-\left(2x+1\right)=\dfrac{8}{3}\)

\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)

\(2x+1=\dfrac{20}{3}\)

\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)

\(2x=\dfrac{17}{3}\)

\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)

Bài 15:

a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)

\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)

\(=>x=\left(\dfrac{-2}{3}\right)^8\)

b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)

\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)

\(=>x=\left(\dfrac{4}{9}\right)^9\)

c) \(\left(x+4\right)^3=-125\)

\(\left(x+4\right)^3=\left(-5\right)^3\)

\(=>x+4=-5\)

\(x=-5-4\)

\(=>x=-9\)

d) \(\left(10-5x\right)^3=64\)

\(\left(10-5x\right)^3=4^3\)

\(=>10-5x=4\)

\(5x=10-4\)

\(5x=6\)

\(=>x=\dfrac{6}{5}\)

e) \(\left(4x+5\right)^2=81\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)

Bài 16:

a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)

\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)

\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)

b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)

\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)

\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)

c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)

\(=\dfrac{7}{4}.\dfrac{-12}{5}\)

\(=\dfrac{-21}{5}\)

\(#Wendy.Dang\)

Uh, chừa sau k dám học muộn nx

3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

bấm máy tính là ra mak

Bạn tính hai vế à.!? Hay tính vế thứ nhất rồi với vế thứ 2.!???

a: \(=6+\dfrac{4}{5}-1-\dfrac{2}{3}-3-\dfrac{4}{5}\)

\(=2-\dfrac{2}{3}=\dfrac{4}{3}\)

b: \(=7+\dfrac{5}{9}-2-\dfrac{3}{4}-3-\dfrac{5}{9}=2-\dfrac{3}{4}=\dfrac{5}{4}\)

c: =6+7/7-1-3/4-2-5/7

=3+2/7-3/4

=84/28+8/28-21/28

=84/28-13/28

=71/28

\(\dfrac{3}{16}\) - (\(x\) - \(\dfrac{5}{4}\)) - ( \(\dfrac{3}{4}\)  - \(\dfrac{7}{8}\) - 1) = 2\(\dfrac{1}{2}\)

\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{7}{8}\) + 1 = \(\dfrac{5}{2}\)

\(\dfrac{3}{16}\) - \(x\) + ( \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\)) + (\(\dfrac{7}{8}\) + 1) = \(\dfrac{5}{2}\) 

\(\dfrac{3}{16}\)  - \(x\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\) = \(\dfrac{5}{2}\)

 ( \(\dfrac{3}{16}\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\)) - \(x\) = \(\dfrac{5}{2}\) 

    \(\dfrac{41}{16}\)              - \(x\)    = \(\dfrac{5}{2}\)

                         \(x\)    = \(\dfrac{41}{16}\)  - \(\dfrac{5}{2}\)

                         \(x\)     = \(\dfrac{1}{16}\)

2, \(\dfrac{1}{2}\).( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\)) = \(\dfrac{1}{5}\) - \(x\) + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\)) 

    \(\dfrac{1}{2}\).(-\(\dfrac{11}{15}\)) = \(\dfrac{1}{5}\) - \(x\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{5}\)

   - \(\dfrac{11}{30}\) = ( \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{15}\)) - \(x\)

    - \(\dfrac{11}{30}\) = \(\dfrac{7}{15}\) - \(x\)

       \(x\)   = \(\dfrac{7}{15}\) + \(\dfrac{11}{30}\)

       \(x\)    = \(\dfrac{5}{6}\)

1.Tính

a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)

b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)

c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)

d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)

e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)

Bài 2

a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)

\(x=\dfrac{13}{49}\)

b.\(\left|x-1,5\right|=2\)

Xảy ra 2 trường hợp

TH1

\(x-1,5=2\)

\(x=3,5\)

TH2

\(x-1,5=-2\)

\(x=-0,5\)

Vậy \(x=3,5\) hoặc \(x=-0,5\) .

Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.

Ths bn nhé

`@` `\text {Ans}`

`\downarrow`

`1,`

`3/16 - (x - 5/4) - (3/4 + (-7)/8 - 1) = 2 1/2`

`=> 3/16 - x + 5/4 - (-1/8 - 1) = 2 1/2`

`=> 3/16 - x + 5/4 - (-9/8) = 2 1/2`

`=> 3/16 - x + 19/8 = 2 1/2`

`=> 3/16 - x = 2 1/2 - 19/8`

`=> 3/16 - x =1/8`

`=> x = 3/16 - 1/8`

`=> x = 1/16`

Vậy, `x = 1/16`

`2,`

`1/2* (1/6 - 9/10) = 1/5 - x + (1/15 - (-1)/5)`

`=> 1/2 * (-11/15) = 1/5 - x + 4/15`

`=> -11/30 = x + 1/5 - 4/15`

`=> x + (-1/15) = -11/30`

`=> x = -11/30 + 1/15`

`=> x = -3/10`

Vậy, `x = -3/10.`

 mik cảm ơn .

a) $A=\frac{3}{5}.\frac{6}{7}+\frac{3}{7}.\frac{3}{5}-\frac{2}{7}.\frac{3}{5}$
$=\frac{3}{5}\cdot \left(\frac{6}{7}+\frac{3}{7}-\frac{2}{7}\right)=\frac{3}{5}$
b) $B=\left(-13\cdot \frac{2}{5}+\frac{-2}{9}\cdot \frac{2}{5}+\frac{2}{5}\cdot \frac{11}{9}\right)\cdot \frac{5}{2}$
$=\left(-13-\frac{2}{9}+\frac{11}{9}\right)\cdot \frac{2}{5}\cdot \frac{5}{2}=-13+\left(\frac{11}{9}-\frac{2}{9}\right)=-12.$
c) $C=\left(\frac{-4}{5}+\frac{5}{7}\right)\cdot \frac{3}{2}+\left(\frac{-1}{5}+\frac{2}{7}\right)\cdot \frac{3}{2}=\left(\frac{-4}{5}+\frac{5}{7}+\frac{-1}{5}+\frac{2}{7}\right)\cdot \frac{3}{2}=\left(\left(\frac{-4}{5}+\frac{-1}{5}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\right)\cdot \frac{3}{2}=0.$
d) $D=\frac{4}{9}:\left(\frac{1}{15}-\frac{10}{15}\right)+\frac{4}{9}:\left(\frac{2}{22}-\frac{5}{22}\right)$
$=\frac{4}{9}:\frac{-3}{5}+\frac{4}{9}:\frac{-3}{22}=\frac{4}{9}\cdot \frac{-5}{3}+\frac{4}{9}.\frac{-22}{3}$
$=\frac{4}{9}\cdot \left(\frac{-5}{3}+\frac{-22}{3}\right)=\frac{4}{9}.\frac{-27}{3}=-4.$

a) $\mathrm{A}=\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{6}{7}+\genfrac{}{}{0ex}{}{3}{7}.\genfrac{}{}{0ex}{}{3}{5}-\genfrac{}{}{0ex}{}{2}{7}.\genfrac{}{}{0ex}{}{3}{5}$

b) $\mathrm{B}=\left(-13\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{-2}{9}\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{5}{2}$
$=\left(-13-\genfrac{}{}{0ex}{}{2}{9}+\genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{5}{2}=-13+\left(\genfrac{}{}{0ex}{}{11}{9}-\genfrac{}{}{0ex}{}{2}{9}\right)=-12.$
c) $\mathrm{C}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}+\left(\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{-1}{5}\right)+\left(\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{2}{7}\right)\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=0.$
d) $\mathrm{D}=\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{1}{15}-\genfrac{}{}{0ex}{}{10}{15}\right)+\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{2}{22}-\genfrac{}{}{0ex}{}{5}{22}\right)$