(1+1/2).(1.1/3)x ...x(1+1/2019)
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Những câu hỏi liên quan
CW
1 tháng 8 2016
a) \(1-\left(\frac{2x}{3}+2\right)=-1\cdot\frac{1}{3}\)
\(1-\frac{2}{3}x-2=-\frac{1}{3}\)
\(-\frac{2}{3}x-1=-\frac{1}{3}\)
\(-\frac{2}{3}x=\frac{2}{3}\)
\(x=-1\)
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b) \(\frac{2}{5}x-1\cdot\frac{1}{2}x+x=\frac{1}{3}\)
\(\left(\frac{2}{5}-\frac{1}{2}+1\right)x=\frac{1}{3}\)
\(\frac{9}{10}x=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{9}{10}\)
\(x=\frac{10}{27}\)
HP
3 tháng 9 2021
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{499}{500}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{499}{500}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{499}{500}\)
\(\Leftrightarrow x=499\)
MM
1
H9
HT.Phong (9A5)
CTVHS
8 tháng 10 2023
\(A=\left(2020\times2019+2019\times2018\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)
\(A=\left[2019\times\left(2020+2018\right)\right]\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times0\)
\(A=0\)
NM
20 tháng 9 2023
\(x\times2+x\times\dfrac{1}{5}=1.\dfrac{1}{3}\\ \Rightarrow x\times\left(2+\dfrac{1}{5}\right)=\dfrac{1}{3}\\ \Rightarrow x\times\dfrac{11}{5}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{5}{33}\)
BL
0
BL
1
GD
1
XO
9 tháng 3 2020
Ta có A = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}\)(1)
=> 3A = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)(2)
Lấy (2) trừ (1) theo vế ta có :
3A - A = \(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}\right)\)
2A = \(1-\frac{1}{3^{2019}}\)
Khi đó : \(\left(2A+\frac{1}{3^{2019}}\right).x=2\)
\(\Leftrightarrow\left(1-\frac{1}{3^{2019}}+\frac{1}{3^{2019}}\right).x=2\)
\(\Rightarrow x=2\)
C
2
29 tháng 8 2023
1: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)
\(=3+2\sqrt{2}+\sqrt{5}-2=2\sqrt{2}+\sqrt{5}+1\)
2: \(\dfrac{1}{\sqrt{3}+\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}+\dfrac{2\left(1+\sqrt{7}\right)}{-6}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}-\dfrac{1+\sqrt{7}}{3}\)
\(=\dfrac{3\left(\sqrt{7}-\sqrt{3}\right)-4\left(\sqrt{7}+1\right)}{12}=\dfrac{-\sqrt{7}-3\sqrt{3}-4}{12}\)
3:
\(=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{2-\sqrt{a}}=-\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=-\sqrt{a}\)
4:
\(=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{xy}\)
H9
HT.Phong (9A5)
CTVHS
29 tháng 8 2023
1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=\dfrac{3+2\sqrt{2}}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(=\dfrac{3+2\sqrt{2}}{3^2-\left(2\sqrt{2}\right)^2}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}\right)^2-2^2}\)
\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)
\(=3+2\sqrt{2}+\sqrt{5}-2\)
\(=2\sqrt{2}+\sqrt{5}+1\)
2) \(\dfrac{1}{\sqrt{3}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{\left(1-\sqrt{7}\right)\left(1+\sqrt{7}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}\right)^2-\left(\sqrt{7}\right)^2}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{1^2-\left(\sqrt{7}\right)^2}\)
\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{2\cdot\left(1+\sqrt{7}\right)}{6}\)
\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{1+\sqrt{7}}{3}\)
\(=\dfrac{-3\sqrt{3}-3\sqrt{7}}{12}-\dfrac{4+4\sqrt{7}}{12}\)
\(=\dfrac{-3\sqrt{3}-3\sqrt{7}-4-4\sqrt{7}}{12}\)
\(=\dfrac{-3\sqrt{3}-7\sqrt{7}-4}{12}\)
3) \(\dfrac{a-2\sqrt{a}}{2-\sqrt{a}}\)
\(=-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\)
\(=-\dfrac{\sqrt{a}\cdot\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\)
\(=-\sqrt{a}\)
4) \(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{xy}+\sqrt{y}\cdot\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{\sqrt{xy}\cdot\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{xy}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)..\left(1+\frac{1}{2019}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{2020}{2019}\)
\(=\frac{2020}{2}=1010\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{2019}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{2020}{2019}\)
\(=\frac{2020}{2}\)
\(=1010\)