Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot3}+...+\frac{1}{x\cdot\left(x+1\right)}-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(=1-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=1-\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2019}{2019}-\frac{2018}{2019}=\frac{1}{2019}\)
Đến đây bn tự tính nhé !!
a) \(\dfrac{1}{3}x-\dfrac{1}{2}=\dfrac{3}{4}x+\dfrac{1}{15}\)
\(\Rightarrow\dfrac{1}{3}x-\dfrac{3}{4}x=\dfrac{1}{2}+\dfrac{1}{15}\)
\(\Rightarrow\dfrac{4}{12}x-\dfrac{9}{12}x=\dfrac{15}{30}+\dfrac{2}{30}\)
\(\Rightarrow\dfrac{-5}{12}x=\dfrac{17}{30}\)
\(\Rightarrow x=\dfrac{-102}{75}\)
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\Rightarrow\left(x-\dfrac{2}{9}\right)^3=\dfrac{64}{729}\)
\(\Rightarrow x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(\Rightarrow x=\dfrac{2}{3}\)
\(\dfrac{x-1}{2019}+\dfrac{x-2}{2018}=\dfrac{x-3}{2017}+\dfrac{x-4}{2016}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2019}-1\right)+\left(\dfrac{x-2}{2018}-1\right)=\left(\dfrac{x-3}{2017}-1\right)+\left(\dfrac{x-4}{2016}-1\right)\)
\(\Leftrightarrow\dfrac{x-2020}{2019}+\dfrac{x-2020}{2018}-\dfrac{x-2020}{2017}-\dfrac{x-2010}{2016}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)
\(\Rightarrow x-2020=0\Leftrightarrow x=2020\)
vậy.......
Ta có:
\(C=\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2017}=1-\frac{1}{2018}+1-\frac{1}{2019}+1+\frac{2}{2017}=3+\left(\frac{2}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)\)Mà ta có:
\(\frac{2}{2017}=\frac{1}{2017}+\frac{1}{2017}>\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow\frac{2}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)
\(\Rightarrow C>3\)
Ta có :
\(A=\frac{2018^{2017}+1}{2018^{2017}-1}=\frac{2018^{2017}-1+2}{2018^{2017}-1}=\frac{2018^{2017}-1}{2018^{2017}-1}+\frac{2}{2018^{2017}-1}=1+\frac{2}{2018^{2017}-1}\)
\(B=\frac{2018^{2017}-1}{2018^{2017}-3}=\frac{2018^{2017}-3+2}{2018^{2017}-3}=\frac{2018^{2017}-3}{2018^{2017}-3}+\frac{2}{2018^{2017}-3}=1+\frac{2}{2018^{2017}-3}\)
Vì \(2018^{2017}-1>2018^{2017}-3\) nên \(\frac{2}{2018^{2017}-1}< \frac{2}{2018^{2017}-3}\)
\(\Rightarrow\)\(1+\frac{2}{2018^{2017}-1}< 1+\frac{2}{2018^{2017}-3}\)
\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
ta có nếu \(\frac{a}{b}\)>1 thì \(\frac{a}{b}\)>\(\frac{a+m}{b+m}\)
mà B> nên B=\(\frac{2018^{2017}-1}{2018^{2017}-3}\)>\(\frac{2018^{2017}-1+2}{2018^{2017}-3+2}\)=\(\frac{2018^{2017}+1}{2018^{2017}-1}\)=A
vậy B>A
a) 70-5(x-3)=45
=> 5(x-3)=70-45
=> 5(x-3)=25
=> x-3=5
=> x=8
\(2^{x+1}.3^y=12^x\)
\(\Rightarrow2^{x+1}.3^y=3^x.4^x\)
\(\Rightarrow2^{x+1}.3^y=3^x.2^{2x}\)
\(\Rightarrow\orbr{\begin{cases}2^{x+1}=2^{2x}\\3^y=3^x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+1=2x\\y=x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\\text{Vì y = x}\Rightarrow y=1\end{cases}}\)
\(\dfrac{x+5}{2017}+\dfrac{x+4}{2018}+\dfrac{x+3}{2019}=-3\\ \dfrac{x+5}{2017}+1+\dfrac{x+4}{2018}+1+\dfrac{x+3}{2019}=-3+3\\ \dfrac{x+5}{2017}+\dfrac{2017}{2017}+\dfrac{x+4}{2018}+\dfrac{2018}{2018}+\dfrac{x+3}{2019}+\dfrac{2019}{2019}=0\\ \dfrac{x+2022}{2017}+\dfrac{x+2022}{2018}+\dfrac{x+2022}{2019}=0\\ x+2022.\left(\dfrac{1}{2017}+\dfrac{1}{2018}+\dfrac{1}{2019}\right)=0\)
⇒x+2022=0 (vì \(\dfrac{1}{2017}+\dfrac{1}{2018}+\dfrac{1}{2019}\)\(\ne0\))
⇒x=0-2022
⇒x=-2022
Cảm ơn nka