Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Có \(A=\dfrac{1}{1+3}+\dfrac{1}{1+3+5}+...+\dfrac{1}{1+3+5+...+2017}\)
\(\Rightarrow A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{1+3+...+2017}\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2017^2}\)
Ta thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{4}\)
\(\dfrac{1}{3^2}< \dfrac{1}{3.2}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
.................
\(\dfrac{1}{2017^2}< \dfrac{1}{2016.2017}\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2016.2017}\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2016}-\dfrac{1}{2017}\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{2017}\)
\(\Rightarrow A< \dfrac{3}{4}-\dfrac{1}{2017}\)
\(\Rightarrow A< \dfrac{3}{4}\)
Vậy \(A< \dfrac{3}{4}\).
Có \(\dfrac{1}{1+3}\) + \(\dfrac{1}{1+3+5}\) +...+ \(\dfrac{1}{1+3+...+2017}\)
= \(\dfrac{1}{2^2 }\)+\(\dfrac{1}{3^2}\) + ... +\(\dfrac{1}{2017^2}\)
Lại có :
\(\dfrac{1}{2^2}\) = \(\dfrac{1}{4} \)
\(\dfrac{1}{3^2}\) <\(\dfrac{1}{2.3}\)
...
\(\dfrac{1}{2017^2}\) <\(\dfrac{1}{2016.2017}\)
\(\Rightarrow \) A< \(\dfrac{1}{4} \) +\(\dfrac{1}{2.3}\)+... +\(\dfrac{1}{2016.2017}\)
A<\(\dfrac{1}{4} \)+\(\dfrac{1}{2}\)- \(\dfrac{1}{3}\) +...+\(\dfrac{1}{2016}- \dfrac{1}{2017}\)
A< \(\dfrac{1}{4} \)+\(\dfrac{1}{2}\) -\(\dfrac{1}{2017}\)
A<\(\dfrac{3}{4}\) -\(\dfrac{1}{2017}\)
\(\Rightarrow\)A<\(\dfrac{3}{4}\) (đpcm)
chúc bạn học tốt !!!
A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)
A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)
A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)
A=\(\dfrac{7}{24}\)
B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)
B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)
B=\(1+\left(-1\right)+\left(-1\right)=-1\)
C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)
D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)
D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)
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: \(\Leftrightarrow70+18< x< 120+126+70\)
=>88<x<316
hay \(x\in\left\{89;90;...;315\right\}\)
b: \(\Leftrightarrow-\dfrac{9}{3}< x< \dfrac{8}{5}+\dfrac{9}{5}=\dfrac{17}{5}\)
=>-3<x<3,4
hay \(x\in\left\{-2;-1;0;1;2;3\right\}\)
1) \(19\dfrac{5}{8}:\dfrac{7}{12}-15\dfrac{1}{4}:\dfrac{7}{12}\)
\(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\\ =\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)\\ =\dfrac{12}{7}\cdot\dfrac{35}{8}\\ =\dfrac{15}{2}\)
2) \(\dfrac{2}{5}\cdot\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}\cdot\dfrac{1}{3}\)
\(=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}\cdot5\\ =\dfrac{1}{3}\cdot1-\dfrac{2}{3}\\ =\dfrac{1}{3}-\dfrac{2}{3}\\ =-\dfrac{1}{3}\)
3) \(\dfrac{4}{9}\cdot19\dfrac{1}{3}-\dfrac{4}{9}\cdot39\dfrac{1}{3}\)
\(=\dfrac{4}{9}\left(19\dfrac{1}{3}-39\dfrac{1}{3}\right)\\ =\dfrac{4}{9}\cdot\left(\dfrac{58}{3}-\dfrac{118}{3}\right)\\ =\dfrac{4}{9}\cdot\left(-20\right)\\ =-\dfrac{80}{9}\)
a: \(A=\left(\dfrac{-3}{4}+\dfrac{-2}{9}-\dfrac{1}{36}\right)+\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{3}{5}\right)+\dfrac{1}{57}\)
\(=\dfrac{-27-8-1}{36}+\dfrac{5+1+9}{15}+\dfrac{1}{57}\)
=1/57
b: \(B=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}-\dfrac{5}{7}-\dfrac{3}{35}\right)+\dfrac{1}{41}\)
\(=\dfrac{3+1+2}{6}+\dfrac{-7-25-3}{35}+\dfrac{1}{41}\)
=1/41
c: \(C=\left(\dfrac{-1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{107}\)
=1-1+1/107
=1/107
2) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)
1. Tìm \(x\):
a) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)
\(\dfrac{x}{5}=\dfrac{1}{5}\)
\(\Rightarrow x=1\)
b) \(\dfrac{-5}{6}-x=\dfrac{7}{12}-\dfrac{1}{3}.x\)
\(\dfrac{-5}{6}-\dfrac{7}{12}=x-\dfrac{1}{3}.x\)
\(x-\dfrac{1}{3}.x=\dfrac{-17}{12}\)
\(\dfrac{2}{3}.x=\dfrac{-17}{12}\)
\(x=\dfrac{-17}{12}:\dfrac{2}{3}\)
\(x=\dfrac{-17}{8}\)
c) \(2016^3.2016^x=2016^8\)
\(2016^x=2016^8:2016^3\)
\(2016^x=2016^{8-3}\)
\(2016^x=2016^5\)
\(\Rightarrow x=5\)
d) \(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=3\dfrac{1}{2}\)
\(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=\dfrac{7}{2}\)
\(\left(x+\dfrac{3}{4}\right)=\dfrac{7}{2}.\dfrac{5}{2}\)
\(x+\dfrac{3}{4}=\dfrac{35}{4}\)
\(x=\dfrac{35}{4}-\dfrac{3}{4}\)
\(x=\dfrac{32}{4}=8\)
e) \(\left(2,8.x-2^5\right):\dfrac{2}{3}=3^2\)
\(\left(2,8.x-2^5\right)=9.\dfrac{2}{3}\)
\(2,8.x-2^5=6\)
\(2,8.x=6+32\)
\(2,8.x=38\)
\(x=38:2,8\)
\(x=\dfrac{95}{7}\)
f) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{2}{5}\)
\(\dfrac{4}{7}.x=\dfrac{2}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}:\dfrac{4}{7}\)
\(x=\dfrac{28}{15}\)
g) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)
\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
\(\Rightarrow3x=-6\)
\(x=\left(-6\right):3\)
\(x=-2\)
2. Thực hiện phép tính:
a) \(\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}+1\dfrac{4}{5}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{3}+1\right)-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{1}{2}.\dfrac{5}{3}-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{5}{6}-\dfrac{4}{9}+\dfrac{9}{5}\)
\(=\dfrac{7}{18}+\dfrac{9}{5}\)
\(=\dfrac{197}{90}\)
b) \(\dfrac{7.5^2-7^2}{7.24+21}\)
\(=\dfrac{7.25-7.7}{7.24+7.3}\)
\(=\dfrac{7.\left(25-7\right)}{7.\left(24+3\right)}\)
\(=\dfrac{7.18}{7.27}\)
\(=\dfrac{2}{3}\)
c) \(\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{-4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)