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Xét: \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\)
\(=\dfrac{3-2-1}{6}\)
\(=0\)
\(\rightarrow C=0\)
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 \(x=2016\)
\(=\dfrac{\left(\dfrac{1}{19}+1\right)+\left(\dfrac{2}{18}+1\right)+...+\left(\dfrac{18}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}\)
\(=\dfrac{\dfrac{20}{19}+\dfrac{20}{18}+...+\dfrac{20}{2}+\dfrac{20}{20}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}=20\)
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!
17A=1719+1+16/1719+1
17A=1+16/1719+1
phần in nghiêng mình không hiểu lắm, bn giải thích cho mình được ko?
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
`S=1/19+1/19^2+1/19^3+........+1/19^20`
`=>19S=1+1/19+1/19^2+.....+1/19^19`
`=>19S-S=18S=1-1/19^20<1`
`=>S<1/18(đpcm)`
Giải:
S=\(\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}\)
19S=\(1+\dfrac{1}{19}+\dfrac{1}{19^2}+...+\dfrac{1}{19^9}\)
19S-S=\(\left(1+\dfrac{1}{19}+\dfrac{1}{19^2}+...+\dfrac{1}{19^9}\right)-\left(\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}\right)\)
18S=1-\(\dfrac{1}{19^{10}}\)
S=(1-\(\dfrac{1}{19^{10}}\) ):18
S=\(1:18-\dfrac{1}{19^{10}}:18\)
S=\(\dfrac{1}{18}-\dfrac{1}{19^{10}.18}\)
⇒S<\(\dfrac{1}{18}\) (đpcm)
Chúc bạn học tốt!
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
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:
\(\dfrac{\dfrac{1}{19}+\dfrac{2}{18}+\dfrac{3}{17}+...+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\\ =\dfrac{20\cdot\left(\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}\right)}{\dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{2}}\\ =20\)