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\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+........+\dfrac{1}{100^2}\)
Ta có :
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
...................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+.......+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+......+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}=\dfrac{6}{25}\)
Mà \(\dfrac{1}{6}< \dfrac{5}{26}< \dfrac{1}{4}\)
Mà \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+.........+\dfrac{1}{100^2}< \dfrac{6}{25}\)
\(\Leftrightarrow\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+.......+\dfrac{1}{100^2}< \dfrac{1}{4}\left(đpcm\right)\) \(\left(1\right)\)
Bài 1:
a: \(=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{3}+\dfrac{4}{3}=\dfrac{4}{3}-1+\dfrac{1}{2}=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\)
b: \(=\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{1}{9}-1-\dfrac{2}{5}+\dfrac{5}{4}=2-1+\dfrac{1}{9}=\dfrac{10}{9}\)
c: \(=\left(\dfrac{-3}{2}\cdot\dfrac{4}{3}\right)\cdot\dfrac{-9}{2}-\dfrac{1}{2}=9-\dfrac{1}{2}=8.5\)
\(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\Rightarrow4S=1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}+......-\dfrac{1}{2^{2002}}\Rightarrow4S+S=5S=1-\dfrac{1}{2^{2004}}< 1\Rightarrow S< 0,2\left(\text{đpcm}\right)\)
4/ \(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{y}{20}\\\dfrac{y}{20}=\dfrac{z}{24}\end{matrix}\right.\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=k\) (đặt k)
Suy ra \(x=15k;y=20k;z=24k\)
Thay vào,ta có:
\(M=\dfrac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
\(a)3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right):2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)\)
\(=\dfrac{7}{2}.\dfrac{4}{49}-\dfrac{88}{27}:\left(\dfrac{-42}{7}\right)\)
\(=\dfrac{2}{7}-\dfrac{-220}{567}\)
\(=\dfrac{382}{567}\)
các phần con lại dễ nên bn tự lm đi nhé mk bn lắm
Chúc bạn học tốt!
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
a/ \(\left(x+1\right)\left(x-2\right)< 0\)
TH1:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\) (vô lý)
TH2:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow-1< x< 2\)
Vậy.........
b/ \(\left(x-3\right)\left(x-4\right)>0\)
TH1:\(\left\{{}\begin{matrix}x-3>0\\x-4>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>3\\x>4\end{matrix}\right.\)\(\Rightarrow x>4\)
TH2:\(\left\{{}\begin{matrix}x-3< 0\\x-4< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 3\\x< 4\end{matrix}\right.\)\(\Rightarrow x< 3\)
Vậy...............
c/ \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}-\dfrac{1}{8}\)
\(\Rightarrow\dfrac{-1}{12}< x< -\dfrac{5}{48}\)
Vậy...............
Để ( x + 1 ) ( x - 2 ) < 0
=> x + 1 và x - 2 phải khác dấu mà x + 1 > x + 2
=> x + 1 dương x + 2 âm
Tức là x + 1 > 0 => x > - 1 và x - 2 < 0 => x < 2
a, 1/3-3/4+3/5+1/4-2/9-1/36+1/15
=(1/3+3/5+1/15)-(3/4-1/4+2/9+1/36)
=1 - 3/4
=1/4
b, 3-1/4+2/3-5-1/3+6/5-6+7/4-3/2
=(3-5-6)-(1/4-7/4)+(2/3-1/3)+(6/5-3/2)
=-8 +3/2 +1/3 -3/10
=-97/15
\(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{100}}\)
\(2^2A=2^2\left(\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{100}}\right)\)
\(4A=1+\dfrac{1}{2^2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}\)
\(4A-A=\left(1+\dfrac{1}{2^2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{100}}\right)\)\(3A=1-\dfrac{1}{2^{100}}\)
\(A=\dfrac{1-\dfrac{1}{2^{100}}}{3}\)
\(A=\dfrac{1}{3}-\dfrac{\dfrac{1}{2^{100}}}{3}< \dfrac{1}{3}\)
Vậy \(A< \dfrac{1}{3}\)