\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):\left(\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}\right)+\left(\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}\right):\left(\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}\right)\)

\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right)\)

\(D=\dfrac{1}{3}+\dfrac{2}{3}\)

\(D=1\)

D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}}\) + \(\dfrac{\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}}{\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}}\)

D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{3.\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right)}\) + \(\dfrac{2.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}{3.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}\)

D = \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\)

D = \(\dfrac{3}{3}\)

D = 1

c

C

c

1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)

\(=1-1+\dfrac{1}{2}=\dfrac{1}{2}\)

2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)

\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)

3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)

\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)

\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)

\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)

\(=12\cdot\dfrac{5}{3}=20\)

5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)

6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)

\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)

\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)

1, 54 : x - 1 = 5

54 : x = 5+1 = 6

x = 54 : 6 = 9

2, 42 : x + 0 = 8

x = 42 : 8 = 21/4

3, 24 : x - 8 = 0

24 : x = 0 + 8 = 8

x = 24 : 8 = 3

Tk mk nha

1) 54:x-x:x=3x2-1

    54:x-  1 =6-1

    54:x-   1=5

    54:x      =6

         x=54:6=9

kết quả bằng 1 đó bạn

\(\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{4}{12}+\dfrac{9}{12}+\dfrac{6}{12}=\dfrac{19}{12}\)
\(\dfrac{6}{8}+\dfrac{2}{4}+\dfrac{6}{24}=\dfrac{3}{4}+\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{6}{4}=\dfrac{3}{2}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{5}{6}=\dfrac{10}{6}=\dfrac{5}{3}\)
\(\dfrac{3}{5}+\dfrac{3}{2}+2=\dfrac{6}{10}+\dfrac{15}{10}+\dfrac{20}{10}=\dfrac{41}{10}\)

\(\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{13}{12}+\dfrac{1}{2}=\dfrac{19}{12}\\ \dfrac{6}{8}+\dfrac{2}{4}+\dfrac{6}{24}=\dfrac{5}{4}+\dfrac{6}{24}=\dfrac{3}{2}\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{5}{6}+\dfrac{5}{6}=\dfrac{10}{6}\\ \dfrac{3}{5}+\dfrac{3}{2}+2=\dfrac{21}{10}+\dfrac{20}{10}=\dfrac{41}{10}\)

`1/3 + 3/4 +1/2`

`= 4/12 + 9/12 + 6/12`

`= (4+9+6)/12`

`= 19/12`

__

`6/8 + 2/4 + 6/24`

`= 3/4 + 2/4 + 1/4`

`=(3+2+1)/4`

`= 6/4= 3/2`

__

`1/2 + 1/3 + 5/6`

`= 3/6 + 2/6 + 5/6`

`=(3+2+5)/6`

`=10/6`

`=5/3`

__

`3/5 + 3/2+2`

`= 6/10 + 15/10 + 20/10`

`=(6+15+20)/10`

`= 41/10`

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+25\sqrt{24}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}=\frac{4}{5}\)

\(\dfrac{24}{x}:\dfrac{8}{3}=\dfrac{3}{5}\)

\(\dfrac{24}{x}=\dfrac{3}{5}.\dfrac{8}{3}\)

\(\dfrac{24}{x}=\dfrac{8}{5}\)

\(\dfrac{24}{x}=\dfrac{24}{15}\)

=>x=5

Vậy x=5

\(x+3\dfrac{1}{2}+x=24\dfrac{1}{4}\)

\(\left(x+x\right)+3\dfrac{1}{2}=24\dfrac{1}{4}\)

\(x.2+\dfrac{7}{2}=\dfrac{97}{4}\)

\(x.2=\dfrac{97}{4}-\dfrac{7}{2}\)

\(x.2=\dfrac{97}{4}-\dfrac{14}{4}\)

\(x.2=\dfrac{83}{4}\)

\(x=\dfrac{83}{4}:2\)

\(x=\dfrac{83}{4}.\dfrac{1}{2}\)

\(x=\dfrac{83}{8}\)

\(x=10\dfrac{3}{8}\)

\(\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{\frac{3}{4}+\frac{3}{24}+\frac{3}{124}}+\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{127}}{\frac{3}{7}+\frac{3}{17}+\frac{3}{127}}=\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{3\left(\frac{1}{4}+\frac{1}{24}+\frac{1}{124}\right)}+\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{127}\right)}{3\left(\frac{1}{7}+\frac{1}{17}+127\right)}=\frac{1}{3}+\frac{2}{3}=\) \(1\)