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Ta có:\(\dfrac{1}{4^2}< \dfrac{1}{2.4}\)
\(\dfrac{1}{6^2}< \dfrac{1}{4.6}\)
\(\dfrac{1}{8^2}< \dfrac{1}{6.8}\)
...
\(\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right).2n}\)
=>\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2n-2\right)2n}=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\left(đpcm\right)\)
Đặt A = \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}\)
\(A=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\right)\)
Đặt \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)
Ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\) ( vì 1 > 0 ; 0 < 1.2 < 22 )
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\) ( vì 1 > 0 ; 0 < 2.3 < 32 )
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\) ( vì 1 > 0 ; 0 < 3.4 < 42 )
...
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\) ( vì 1 > 0 ; 0 < ( n - 1 ) n < n2 )
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow B< 1-\dfrac{1}{n}< 1\Rightarrow A< 1.\dfrac{1}{4}\Rightarrow A< \dfrac{1}{4}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
Từ gt ta có:
\(\dfrac{13}{3}.\left(-\dfrac{1}{3}\right)\le x\le\dfrac{2}{3}.\left(-\dfrac{11}{12}\right)\)
\(\Leftrightarrow\dfrac{-13}{9}\le x\le-\dfrac{11}{18}\)
\(\Leftrightarrow\dfrac{-26}{18}\le x\le-\dfrac{11}{18}\)
Suy ra \(26\ge x\ge11\)
Vậy \(11\le x\le26\) ( x thuộc Z ) là các giá trị cần tìm
\(4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
\(\dfrac{13}{3}.\dfrac{-1}{3}\le x\le\dfrac{2}{3}.\dfrac{-11}{12}\)
\(\dfrac{-13}{9}\)\(\le x\le\)\(\dfrac{-11}{18}\)
\(\dfrac{-26}{18}\)\(\le x\le\dfrac{-11}{18}\)
\(\Rightarrow x\in\left\{\dfrac{-12}{18};\dfrac{-13}{18};\dfrac{-14}{18};\dfrac{-15}{18};...;\dfrac{-24}{18};\dfrac{-25}{18}\right\}\)Tick hộ mình nha bạn
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!
2155-(174+2155)+(-68+174)=2155-174-2155-68+174
= -68
( 1 - \(\dfrac{1}{2}\) ) ( 1- \(\dfrac{1}{3}\)) ( 1 - \(\dfrac{1}{4}\)) ( 1 - \(\dfrac{1}{5}\)) = \(\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}.\dfrac{1}{5}\)
= \(\dfrac{1}{120}\)
Mình ps có 2 câu à ^.^!
Lời giải:
Ta có:
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{(2n)^2}< \frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}(*)\)
Mà:
\(\frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n-1)(2n+1)}\)
\(=\frac{1}{2}\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2n+1}\right)\)
\(< \frac{1}{6}< \frac{1}{4}(**)\)
Từ \((*);(**)\Rightarrow N< \frac{1}{4}\) (đpcm)