Các câu dễ tự làm nha:

\(D=\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(D=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)\(D=-\dfrac{1}{100}-1\)

các bạn ơi giúp mìh với mìh đag cần gấp ai nhanh và đúng thì mih tick cho

Lời giải:

a) \(\dfrac{1}{6};\dfrac{1}{3};\dfrac{1}{2};...\)

\(\Rightarrow\dfrac{1}{6};\dfrac{2}{6};\dfrac{3}{6};...\)

Dãy có quy luật tăng dần lên 1 đơn vị ở tử số

\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{4}{6}\)

b) \(\dfrac{1}{8};\dfrac{5}{24};\dfrac{7}{24};...\)

\(\Rightarrow\dfrac{3}{24};\dfrac{5}{24};\dfrac{7}{24};...\)

Dãy có quy luật tăng dần lên 2 đơn vị ở tử số

\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{9}{24}\)

c) \(\dfrac{1}{5};\dfrac{1}{4};\dfrac{1}{3};...\)

\(\dfrac{4}{20};\dfrac{5}{20};\dfrac{6}{20};...\)

Dãy có quy luật tăng dần lên 1 đơn vị ở tử số

\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{7}{20}\)

d) \(\dfrac{4}{15};\dfrac{3}{10};\dfrac{1}{3};...\)

\(\Rightarrow\dfrac{8}{30};\dfrac{9}{30};\dfrac{11}{30};...\)

Dãy có quy luật tăng dần lên 1 đơn vị ở tử số

\(\Rightarrow\) Số tiếp theo của dãy là: \(\dfrac{12}{30}\)

\(a)\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)

\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{6}{12}\right)\)

\(=\left(\dfrac{153}{54}+\dfrac{78}{54}\right):\left(1\dfrac{-5}{12}\right)\)

\(=\dfrac{231}{54}:\dfrac{7}{12}\)

\(=\dfrac{198}{27}\)

\(b)\dfrac{0,8\left(\dfrac{4}{5}:1,25\right)}{0,64-\dfrac{1}{25}}\)

\(=\dfrac{0,8\left(0,8:1,25\right)}{0,64-0,04}\)

\(=\dfrac{0,8.0,64}{0,6}\)

\(=\dfrac{0,512}{0,6}\)\(=\dfrac{64}{75}\)

a: \(=\left(\dfrac{19}{6}-\dfrac{2}{5}\right):\left(\dfrac{29}{6}+\dfrac{7}{10}\right)\)

\(=\dfrac{19\cdot5-2\cdot6}{30}:\dfrac{290+42}{30}=\dfrac{83}{332}=\dfrac{1}{4}\)

b: \(=\dfrac{\left(\dfrac{102}{25}-\dfrac{2}{25}\right)\cdot\dfrac{17}{4}}{\left(6+\dfrac{5}{9}-3-\dfrac{1}{4}\right)\cdot\dfrac{16}{7}}\)

\(=\dfrac{4\cdot\dfrac{17}{4}}{\dfrac{16}{7}\cdot\dfrac{119}{36}}=\dfrac{17}{\dfrac{68}{9}}=17\cdot\dfrac{9}{68}=\dfrac{9}{4}\)

c: \(=\left(\dfrac{120}{60}-\dfrac{15}{60}+\dfrac{20}{60}-\dfrac{36}{60}\right):\left(\dfrac{45}{15}-\dfrac{3}{15}-\dfrac{25}{15}\right)\)

\(=\dfrac{89}{60}:\dfrac{17}{15}=\dfrac{89}{60}\cdot\dfrac{15}{17}=\dfrac{89}{68}\)

c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)

\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)

\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)

\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)

\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)

\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)

\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)

\(\Rightarrow x=-36\)

Vậy ...

a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)

\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)

\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)

\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)

\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)

Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)

Chúc bạn học tốt!!!

B=10 7/41-(2 7/41+5 3/4)

Giup minh nha!

vì mình ko thể đăng bài lên nên bạn thông cảm nha!

a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)

=>2/5x=8/5

=>x=4

b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)

\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)

=>1/3x=-6

=>x=-18

c: =>2|x-1/3|=0,24-4/5=-0,56<0

a: \(=\dfrac{32}{9}+\dfrac{13}{6}=\dfrac{32\cdot2+13\cdot3}{18}=\dfrac{64+39}{18}=\dfrac{103}{18}\)

b: \(=\dfrac{43}{8}-\dfrac{43}{6}=\dfrac{-43}{24}\)

c:\(=4-2-\dfrac{1}{6}=2-\dfrac{1}{6}=\dfrac{11}{6}\)

d: \(=5+\dfrac{2}{3}+7+\dfrac{1}{2}-3-\dfrac{1}{2}+1+\dfrac{2}{3}\)

\(=10+\dfrac{4}{3}=\dfrac{34}{3}\)

bài 1:

a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)

\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)

\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)

\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)

c) Cho mình hỏi x ở đâu vậy ???

d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)

\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)

\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)

\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)

\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)

f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)

\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)

\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)

\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)

\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)

\(x=\dfrac{-5}{7}\)

bài 2:

Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)

\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)

\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)

Vậy \(a=-15;b=81\)