c/m
a/\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}\) bé hơn 1
b/\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\)bé hơn 2
Ai giải đúng cho 2 tick
Tìm x:
\(\frac{x}{7}=\frac{x+1}{14}\) \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\) Bé hơn bằng x bé hơn bằng\(\frac{15}{4}+\frac{18}{8}\)
\(\frac{1}{2}+\frac{-3}{5}+\frac{1}{10}\)bé hơn bằng x bé hơn bằng \(\frac{8}{3}+\frac{14}{6}\)
\(\frac{x}{7}=\frac{x+1}{14}\Leftrightarrow14x=7x+7\Leftrightarrow7x=7\Leftrightarrow x=1\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{15}{4}+\frac{18}{8}\)
\(\Leftrightarrow1\le x\le6\Leftrightarrow x=1;2;3;4;5;6\)
\(\frac{1}{2}+\frac{-3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)
\(\Leftrightarrow\frac{1}{2}-\frac{3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)
\(\Leftrightarrow0\le x\le5\Leftrightarrow x=0;1;2;3;4;5\)
\(\frac{x}{7}=\frac{x+1}{14}\)
=> \(\frac{x\cdot2}{7\cdot2}=\frac{x+1}{14}\)
=> \(2x=x+1\)
=> \(2x-x-1=0\)
=> \(1x-1=0\)
=> \(x=1\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{15}{4}+\frac{18}{8}\)
=> \(1\le x\le6\)
=> \(x=\left\{1;2;3;4;5;6\right\}\)
\(\frac{1}{2}+\frac{-3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)
=> \(0\le x\le5\)
=> \(x=\left\{0;1;2;3;4;5\right\}\)
Cảm ơn các bạn nhé !!!
1) Tính:
a) \(\frac{\left(1+\frac{17}{1}\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right).....\left(1+\frac{17}{19}\right)}{\left(1+\frac{19}{1}\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right).....\left(1+\frac{19}{17}\right)}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
c) \(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}\)
e) \(\frac{\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}\)
2) CMR: \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{800}}< \frac{1}{3}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
Tính:
a) \(\frac{\left(1+\frac{17}{1}\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right).....\left(1+\frac{17}{19}\right)}{\left(1+\frac{19}{1}\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right).....\left(1+\frac{19}{17}\right)}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
c) \(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}\)
e) \(\frac{\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}\)
2) CMR: \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{800}}< \frac{1}{3}\)
Làm tiếp:
\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)
\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)
Bài 2:
Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)
\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)
\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)
\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)
Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)
Bài 1:Tính
a, Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)
Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)
\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)
\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)
\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)
\(=\frac{\left(n+1\right)\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=1\)
Áp dụng vào bài toán ta có đáp số là:1
b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)
c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)
d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)
e,Xét mẫu số ta có:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)
Cho A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{127}\)
CMR 3 bé hơn A bé hơn 6
Làm nhanh giùm mình!!!!!!!
CMR là gì vậy chị nếu em biết được thì có thể giải giùm chị em có công thức đây(lớp 5)
Tính A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}}{\frac{19}{1}+\frac{18}{2}+\frac{17}{3}+...+\frac{3}{17}+\frac{2}{18}+\frac{1}{19}}\)
* Cách làm : Tử giữ nguyên,còn mẫu ta biến đổi như sau:
Mẫu : ( \(\frac{19}{1}\)+ 1 ) + ( \(\frac{18}{2}\)+ 1 ) + ( \(\frac{17}{3}\)+ 1 ) +...+ ( \(\frac{3}{17}\)+ 1 ) + ( \(\frac{2}{18}\)+ 1 ) + ( \(\frac{1}{19}\)+ 1 ) - 19 ( vì ta cộng với 19 số 1 nên phải trừ 19 )
= \(\frac{20}{1}\)+ \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)- 19
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ ( \(\frac{20}{1}\)- 19)
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+ ...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ \(\frac{20}{20}\)
= 20.( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+...+ \(\frac{1}{17}\)+ \(\frac{1}{18}\)+ \(\frac{1}{19}\)+ \(\frac{1}{20}\))
=> \(\frac{Tử}{Mâu}\)= \(\frac{1}{20}\)
Phùng Quang Thịnh biến đổi sai 1 chỗ kìa
-19 = \(\frac{20}{20}-20\)chứ mà bạn
Chứng minh rằng: Nếu C= \(\frac{1}{5^2}\)+ \(\frac{1}{6^2}\)+ \(\frac{1}{7^2}\)+...+ \(\frac{1}{100^2}\)thì \(\frac{1}{6}\)bé hơn A và bé hơn \(\frac{1}{4}\)
\(A=0,2113727891\)
\(\frac{1}{6}=0,166666667\)
\(\frac{1}{4}=0,25\)
\(\Leftrightarrow\frac{1}{6}< A< \frac{1}{4}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}}{\frac{19}{1}+\frac{18}{2}+\frac{17}{3}+...+\frac{3}{17}+\frac{2}{18}+\frac{1}{19}}\)
Minh còn kém .Ai thông minh giải rõ cho minh bài tính tổng này \(1\frac{2}{3}+4\frac{3}{5};2\frac{5}{12}+3\frac{4}{9}\)\(\frac{7}{31}+\frac{14}{29}+\frac{27}{31};8\frac{3}{19}+7\frac{12}{19}+1\frac{4}{19};12\frac{3}{5}+\frac{17}{24}+\frac{7}{5};4\frac{2}{3}+2\frac{1}{5}+2\frac{2}{3}+3\frac{3}{5}\)
Các bạn giải rõ từng câu ra nhé trả lời được hết thì mới cho đúng