Ta có: \(\dfrac{1}{19}+\dfrac{2}{18}+...+\dfrac{19}{1}=\left(\dfrac{1}{19}+1\right)+\left(\dfrac{2}{18}+1\right)+...+1\)

\(=\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}+\dfrac{20}{20}=20\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}\right)\)

Thế lại bài toán ta được

\(\dfrac{\dfrac{1}{19}+\dfrac{2}{18}+...+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}=\dfrac{20\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}=20\)

Ta có

\(\dfrac{1}{19}+\dfrac{2}{18}+\dfrac{3}{17}+...+\dfrac{19}{1}\\ =\dfrac{1}{19}+1+\dfrac{2}{18}+1+\dfrac{3}{17}+1+...+\dfrac{19}{1}+1-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+\dfrac{20}{17}+...+\dfrac{20}{1}-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}+20-19\\ =\dfrac{20}{19}+\dfrac{20}{18}+\dfrac{20}{17}+...+\dfrac{20}{2}+1+19-19\\ =\dfrac{20}{20}+\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}\\ =20\cdot\left(\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}\right)\)

Thế vào ta có:

\(\dfrac{\dfrac{1}{19}+\dfrac{2}{18}+\dfrac{3}{17}+...+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\\ =\dfrac{20\cdot\left(\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}\right)}{\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}}\\ =20\)

Bài 2:

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2016}{2017}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2016}{2017}\)

\(\Leftrightarrow\dfrac{1}{x+1}=1-\dfrac{2016}{2017}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)

\(\Leftrightarrow x+1=2017\Leftrightarrow x=2016\)

Vậy \(x=2016\)

2.x=2016

1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)

\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{19}{18}+\dfrac{5}{18}\)

\(=\dfrac{24}{18}\)

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

2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\dfrac{1}{15}+\dfrac{7}{15}\)

\(=\dfrac{8}{15}\)

3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)

\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

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

\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}\)

\(=\dfrac{7}{30}\)

4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)

\(=\dfrac{5}{7}.-\dfrac{7}{11}\)

\(=-\dfrac{35}{77}\)

\(=-\dfrac{5}{11}\)

Giải:

a) A=17¹⁸+1/17¹⁹+1

17A=17¹⁹+17/17¹⁹+1

17A=17¹⁹+1+16/17¹⁹+1

17A=1+16/17¹⁹+1

Tương tự:

B=17¹⁷+1/17¹⁸+1

17B=17¹⁸+17/17¹⁸+1

17B=17¹⁸+1+16/17¹⁸+1

17B=1+16/17¹⁸+1

Vì 16/17¹⁹+1<16/17¹⁸+1 nên 17A<17B

⇒A<B

b) A=10⁸-2/10⁸+2

    A=10⁸+2-4/10⁸+2

    A=1+-4/10⁸+2

Tương tự:

B=10⁸/10⁸+4

B=10⁸+4-4/10⁸+1

B=1+-4/10⁸+1

Vì -4/10⁸+2>-4/10⁸+1 nên A>B

c)A=20¹⁰+1/20¹⁰-1

   A=20¹⁰-1+2/20¹⁰-1

   A=1+2/20¹⁰-1

Tương tự:

B=20¹⁰-1/20¹⁰-3

B=20¹⁰-3+2/20¹⁰-3

B=1+2/20¹⁰-3

Vì 2/20¹⁰-3>2/20¹⁰-1 nên B>A

⇒A<B

Chúc bạn học tốt!

a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)

\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)

\(=-1+1=0\)

b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)

\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)

=1-1+1=1

\(=\dfrac{\left(\dfrac{1}{19}+1\right)+\left(\dfrac{2}{18}+1\right)+...+\left(\dfrac{18}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}\)

\(=\dfrac{\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}+\dfrac{20}{20}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}=20\)

Xét: \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\)

\(=\dfrac{3-2-1}{6}\)

\(=0\)

\(\rightarrow C=0\)

a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)

\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)

\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)

\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)

mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)

nên x-17=0

hay x=17

Vậy: x=17

b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)

\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)

\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)

mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)

nên x+20=0

hay x=-20

Vậy: x=-20

a: \(A=\dfrac{19}{9}+\dfrac{4}{11}+\dfrac{2}{3}=\dfrac{209}{99}+\dfrac{44}{99}+\dfrac{66}{99}=\dfrac{319}{99}\)

b: \(B=\dfrac{-50}{60}+\dfrac{-35}{60}+\dfrac{12}{60}=\dfrac{-73}{60}\)

c: \(C=\dfrac{-27}{36}+\dfrac{132}{36}+\dfrac{10}{36}=\dfrac{115}{36}\)

d: \(D=\dfrac{-19}{3}+\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{-17}{3}-\dfrac{4}{5}=\dfrac{-85-12}{15}=-\dfrac{97}{15}\)