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\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{20}{41}\div\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Leftrightarrow x+2=41\)
\(\Leftrightarrow x=41-2\)
\(\Leftrightarrow x=39\)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
Cop thì ghi cái nguồn ra không thì đưa cái link cho người ta.
Nguồn: Câu hỏi của Tran Thi Minh Thu - Toán lớp 7 | Học trực tuyến
\(\frac{x+18}{2018}+\frac{x+17}{2017}+\frac{x+16}{2016}=3\)
\(\Rightarrow\frac{x+18}{2018}-1+\frac{x+17}{2017}-1+\frac{x+16}{2016}-1=3-3\)
\(\Rightarrow\frac{x+18-2018}{2018}+\frac{x+17-2017}{2017}+\frac{x+16-2016}{2016}=0\)
\(\Rightarrow\frac{x-2000}{2018}+\frac{x-2000}{2017}+\frac{x-2000}{2016}=0\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\ne0\)
=> x - 2000 = 0
=> x = 2000
Ta có :
\(\frac{x+18}{2018}+\frac{x+17}{2017}+\frac{x+16}{2016}=3\)
\(\Leftrightarrow\)\(\left(\frac{x+18}{2018}-1\right)+\left(\frac{x+17}{2017}-1\right)+\left(\frac{x+16}{2016}-1\right)=3-3\) ( trừ hai vế cho 3 )
\(\Leftrightarrow\)\(\frac{x-2000}{2018}+\frac{x-2000}{2017}+\frac{x-2000}{2016}=0\)
\(\Leftrightarrow\)\(\left(x-2000\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\ne0\)
Nên \(x-2000=0\)
\(\Rightarrow\)\(x=2000\)
Vậy \(x=2000\)
Chúc bạn học tốt ~
\(\frac{x-2017}{2018}-\frac{x-2018}{2017}=\frac{2017}{x-2018}-\frac{2018}{x-2017}\)
\(\Leftrightarrow\)\(\frac{2017\left(x-2017\right)-2018\left(x-2018\right)}{2017.2018}=\frac{2017\left(x-2017\right)-2018\left(x-2018\right)}{\left(x-2017\right)\left(x-2018\right)}\)
Do \(2017\left(x-2017\right)-2018\left(x-2018\right)\ne0\) nên \(\left(x-2017\right)\left(x-2018\right)=2017.2018\)
\(\Leftrightarrow\)\(x^2-4035x+2017.2018=2017.2018\)
\(\Leftrightarrow\)\(x\left(x-4035\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\left(l\right)\\x=4035\left(n\right)\end{cases}}\)
Vậy x = 4035