1/5^2+1/6^2+1/7^2+1/8^2+......+1/2017^2<1/4
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làm bừa thui,ai tích mình mình tích lại
Số số hạng là :
Có số cặp là :
50 : 2 = 25 ( cặp )
Mỗi cặp có giá trị là :
99 - 97 = 2
Tổng dãy trên là :
25 x 2 = 50
Đáp số : 50
\(\frac{x-2017}{5}-\frac{x-2017}{6}=\frac{x-2017}{7}-\frac{x-2017}{8}\)
\(\Leftrightarrow\frac{x-2017}{5}-\frac{x-2017}{6}-\frac{x-2017}{7}+\frac{x-2017}{8}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\right)=0\)
\(\Rightarrow x-2017=0\Rightarrow x=2017\)
Vậy x=2017
1.
a, => 21-x+3 < 0
=> 24-x < 0
=> x < 24
b, => 7+x > 0
=> x > -7
c, => x-1 < 0 ; x+2 > 0 ( vì x-1 < x+2 )
=> x < 1 ; x > -2
=> -2 < x < 1
Tk mk nha
Bài 2:
Ta thấy: 52 > 4.5
62 > 5.6
72 > 6.7
....
20172 > 2016.2017
\(\Rightarrow\)\(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
\(\frac{1}{7^2}< \frac{1}{6.7}\)
....
\(\frac{1}{2017^2}< \frac{1}{2016.2017}\)
Cộng vế với nhau, ta có:
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2017^2}\) < \(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2016.2017}\)
\(\Rightarrow\)A < \(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow\)A < \(\frac{1}{4}-\frac{1}{2017}\)
\(\Rightarrow\)A < \(\frac{1}{4}\)( vì \(\frac{1}{2017}>0\))
k giúp mik ✅
a) \(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}\)
⇒ \(x=\dfrac{21}{52}\)
b) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)
\(\dfrac{x}{3}=\dfrac{14}{21}+\dfrac{-3}{21}\)
\(\dfrac{x}{3}=\dfrac{11}{21}\)
⇒ \(x=\dfrac{11.3}{21}=\dfrac{33}{21}\)
⇒ \(x=\dfrac{11}{7}\)
c) \(\dfrac{-8}{3}+\dfrac{1}{3}< x< \dfrac{-2}{7}+\dfrac{-5}{7}\)
\(\dfrac{-17}{7}< x< -1\)
⇒ \(-17< x< -7\)
⇒ \(x\in\left\{-16;-15,....;-6\right\}\)
d) \(\dfrac{1}{6}+\dfrac{2}{5}\)
\(=\dfrac{5}{30}+\dfrac{12}{30}\)
\(=\dfrac{17}{30}\)
e) \(\dfrac{3}{5}+\dfrac{-7}{4}\)
\(=\dfrac{12}{20}+\dfrac{-35}{20}\)
\(=\dfrac{-23}{20}\)
f) \(\dfrac{4}{13}+\dfrac{-12}{30}\)
\(=\dfrac{4}{13}+\dfrac{-2}{5}\)
\(=\dfrac{20}{65}+\dfrac{-26}{65}\)
\(=\dfrac{-6}{65}\)
g) \(\dfrac{-3}{29}+\dfrac{16}{58}\)
\(=\dfrac{-6}{58}+\dfrac{16}{58}\)
\(=\dfrac{10}{58}\)
h) \(\dfrac{8}{40}+\dfrac{-36}{45}\)
\(=\dfrac{1}{5}+\dfrac{-4}{5}\)
\(=\dfrac{-3}{5}\)
j) \(\dfrac{-8}{18}+\dfrac{15}{27}\)
\(=\dfrac{-2}{9}+\dfrac{5}{9}\)
\(=\dfrac{3}{9}\)
\(=\dfrac{1}{3}\)
Đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2017^2}\) ta có :
\(A< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2016.2017}\)
\(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A< \frac{1}{4}-\frac{1}{2017}< \frac{1}{4}\)
\(\Rightarrow\)\(A< \frac{1}{4}\) ( đpcm )
Vậy \(A< \frac{1}{4}\)
Chúc bạn học tốt ~
\(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2017^2}\)
=> \(A< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2016.2017}\)
=> \(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2016}-\frac{1}{2017}=\frac{1}{4}-\frac{1}{2017}< \frac{1}{4}\)
=> \(A< \frac{1}{4}\)