2N
3
K
Khách
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JK
26 tháng 7 2019
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\); \(\frac{1}{3^2}< \frac{1}{2\cdot3}\); \(\frac{1}{4^2}< \frac{1}{3\cdot4}\); ....; \(\frac{1}{9^2}< \frac{1}{8\cdot9}\)
\(\Rightarrow S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)
\(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow S< 1-\frac{1}{9}\)
\(\Rightarrow S< \frac{8}{9}\) (1)
\(\frac{1}{2^2}>\frac{1}{2\cdot3};\frac{1}{3^2}>\frac{1}{3\cdot4};\frac{1}{4^2}>\frac{1}{4\cdot5};...;\frac{1}{9^2}>\frac{1}{9\cdot10}\)
\(\Rightarrow S>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}\)
\(\Rightarrow S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow S>\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow S>\frac{2}{5}\) (2)
(1)(2) => 2/5 < S < 8/9
26 tháng 7 2019
\(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}< \frac{1}{a^2}\)
\(\frac{1}{a}-1-\frac{1}{a}=-1< \frac{1}{a^2}\) Vì \(\frac{1}{a^2}>0;-1< 0\)
Khi đó thì ĐỀ SAI
23 tháng 6 2019
\(\left(a-\frac{2}{3}\right)^2=\frac{5}{6}\)
\(\Rightarrow a\in\varnothing\)
1 tháng 8 2019
a) \(\frac{3x-6}{x+4}=\frac{2\left(x+5\right)+\left(x-3\right)}{x-2}\)
\(\frac{3\left(x-2\right)}{x+4}=\frac{2\left(x+5\right)+x-3}{x-2}\)
\(\frac{3\left(x-4\right)}{x+4}=\frac{3x+7}{x-2}\)
\(3\left(x-2\right)\left(x-2\right)=\left(3x+7\right)\left(x+4\right)\)
\(3\left(x-2\right)^2=\left(3x+7\right)\left(x+4\right)\)
\(3x^2-12x+12=3x^2+12x+7x+28\)
\(3x^2-12x+12=3x^2+19x+28\)
\(-12x+12=19x+28\)
\(12=19x+28+12x\)
\(19x+28+12x=12\) (chuyển vế)
\(31x+28=12\)
\(31x=12-28\)
\(31x=-16\)
\(x=-\frac{16}{31}\)
\(\Rightarrow x=-\frac{16}{31}\)
NB
4
29 tháng 4 2020
\(\frac{2x+4}{-10}=\frac{2}{5}\)
\(\frac{2x+4}{-10}=\frac{-4}{-10}\)
\(\Leftrightarrow2x+4=-4\Leftrightarrow2x=-8\Leftrightarrow x=-4\)
Cách khác :
\(\frac{2x+4}{-10}=\frac{2}{5}\)
\(\Leftrightarrow5\left(2x+4\right)=-20\)
\(\Leftrightarrow10x+20=-20\Leftrightarrow10x=-40\Leftrightarrow x=-4\)
29 tháng 4 2020
Lớp 6 :\(\frac{2x+4}{-10}=\frac{2}{5}\)
\(\Rightarrow\frac{\left(2x+4\right):\left(-2\right)}{\left(-10\right):\left(-2\right)}=\frac{2}{5}\)
\(\Rightarrow\left(2x+4\right):\left(-2\right)=2\)
\(\Rightarrow2x+4=-4\)
\(\Rightarrow2x=-8\)
\(\Rightarrow x=-4\)
Lớp 7 : \(\frac{2x+4}{-10}=\frac{2}{5}\)
\(\Rightarrow\left(2x+4\right)\cdot5=-10\cdot2\)
\(\Rightarrow10x+20=-20\)
\(\Rightarrow10x=-40\)
\(\Rightarrow x=-4\)
5 tháng 12 2018
B=-4/5+4/52-4/53+...+4/5200
5B=-4+4/5-4/52+...+4/5201
5B+B=-4+4/5200
6B=-4x5200/5200+4/5200
6B=-4+4x5200/5200
Còn lại bạn tính nốt nha
BT
3
MV
5 tháng 5 2017
a)
\(2x-\dfrac{1}{4}=\dfrac{1}{2}\\ 2x=\dfrac{1}{2}+\dfrac{1}{4}\\ 2x=\dfrac{3}{4}\\ x=\dfrac{3}{4}:2\\ x=\dfrac{3}{8}\)
b)
\(\dfrac{-2}{3}\cdot x+\dfrac{1}{5}=\dfrac{3}{10}\\ \dfrac{-2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\\ \dfrac{-2}{3}x=\dfrac{1}{10}\\ x=\dfrac{1}{10}:\dfrac{-2}{3}\\ x=\dfrac{-3}{20}\)
AT
5 tháng 5 2017
a) \(2x-\dfrac{1}{4}=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}+\dfrac{1}{4}\)
\(2x=\dfrac{4}{8}+\dfrac{2}{8}=\dfrac{6}{8}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}:2=\dfrac{3}{4}.\dfrac{1}{2}=\dfrac{3}{8}\)
Vậy x = \(\dfrac{3}{8}\)
b) \(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(\dfrac{-2}{3}.x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(\dfrac{-2}{3}.x=\dfrac{3}{10}-\dfrac{2}{10}\)
\(\dfrac{-2}{3}.x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:\dfrac{-2}{3}=\dfrac{1}{10}.\dfrac{3}{-2}\)
\(x=\dfrac{3}{-20}\)
Vậy x = \(\dfrac{3}{-20}\)
\((\frac23)^5-2x=\frac23\)
\(=\frac{32}{243}-2x=\frac{162}{243}\)
\(=2x=\frac{32}{243}-\frac{162}{243}\)
\(=-2x=\frac{130}{243}\)
\(x=-\frac{65}{243}\)
(\(\frac23\))\(^{5-2x}\) = \(\frac23\)
(\(\frac23\))\(^{5-2x}\) = (\(\frac23\))\(^1\)
5 - 2\(x\) = 1
2\(x\) = 5 - 1
2\(x\) = 4
\(x\) = 4 : 2
\(x\) = 2
Vậy \(x\) = 2