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TP
1
23 tháng 11 2018
Em chuyển sang cùng một vế rồi ghép cái đầu vói cái thứ 3 cái thứ 2 với cái cuối. :)Dùng quy đồng :)
HK
1
FM
13 tháng 10 2018
ta có:
\(\frac{x-2016}{2015}+\frac{x-2017}{2016}-\frac{2018-x}{2017}=-3\)
\(\Leftrightarrow\left(\frac{x-2016}{2015}+1\right)+\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2018}{2017}+1\right)=0\)
\(\Leftrightarrow\frac{x-1}{2015}+\frac{x-1}{2016}+\frac{x-1}{2017}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
DT
3
22 tháng 7 2018
\(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)
\(\Rightarrow\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2}{2018}+1+\frac{x+1}{2019}+1\)
\(\Rightarrow\frac{x+4+2016}{2016}+\frac{x+3+2017}{2017}=\frac{x+2+2018}{2018}+\frac{x+1+2019}{2019}\)
\(\Rightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}=\frac{x+2020}{2018}+\frac{x+2020}{2019}\)
\(\Rightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{2018}-\frac{x+2020}{2019}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)
\(\Rightarrow x+2020=0\) vì \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)
\(\Rightarrow x=-2020\)
7 tháng 8 2018
\(a)\) 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 ~
9 tháng 4 2018
\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(100=2x+4\)
\(\Leftrightarrow\)\(2x=96\)
\(\Leftrightarrow\)\(48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
9 tháng 4 2018
\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(49=x+1\)
\(\Leftrightarrow\)\(x=48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
HX
3
22 tháng 3 2018
Ta có :
\(\frac{x-1}{2017}+\frac{x-2}{2016}+\frac{x-3}{2015}+\frac{x-4}{2014}=2^2\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2017}-1\right)+\left(\frac{x-2}{2016}-1\right)+\left(\frac{x-3}{2015}-1\right)+\left(\frac{x-4}{2014}-1\right)=2^2-4\)
\(\Leftrightarrow\)\(\frac{x-2018}{2017}+\frac{x-2018}{2016}+\frac{x-2018}{2015}+\frac{x-2018}{2014}=4-4\)
\(\Leftrightarrow\)\(\left(x-2018\right)\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\right)=0\)
Vì \(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\)
Nên \(x-2018=0\)
\(\Rightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
22 tháng 3 2018
\(\frac{x-1}{2017}+\frac{x-2}{2016}+\frac{x-3}{2015}+\frac{x-4}{2014}=2^2\)
\(\left(\frac{x-1}{2017}-1\right)+\left(\frac{x-2}{2016}-1\right)+\left(\frac{x-3}{2015}-1\right)+\left(\frac{x-4}{2014}-1\right)=0\)
\(\frac{x-2018}{2017}+\frac{x-2018}{2016}+\frac{x-2018}{2015}+\frac{x-2018}{2014}=0\)
\(\left(x-2018\right).(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014})=0\)
\(x-2018=0\left(Vì\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\right)\)
\(\Rightarrow x=2018\)
30 tháng 3 2018
Bài 1 : dễ bạn tự làm được :)
Bài 2 :
Ta có :
\(B=\frac{2015+2016+2017}{2016+2017+2018}=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì :
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
30 tháng 3 2018
Ta có : B = 2016 + 2017 + 2018 2015 + 2016 + 2017 = 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 Vì : 2016 2015 > 2016 + 2017 + 2018 2015 2017 2016 > 2016 + 2017 + 2018 2016 2018 2017 > 2016 + 2017 + 2018 2017 Nên 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 ⇔ 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 ⇔A > B Vậy A > B Chúc bạn học tốt ~
12 tháng 5 2018
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{20}{41}\div\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Leftrightarrow x+2=41\)
\(\Leftrightarrow x=41-2\)
\(\Leftrightarrow x=39\)
\(\frac{x+4}{2015}+\frac{x+3}{2016}=\frac{x+2}{2017}+\frac{x+1}{2018}\)
\(\Rightarrow\frac{x+4}{2015}+1+\frac{x+3}{2016}+1=\frac{x+2}{2017}+1+\frac{x+1}{2018}+1\)
\(\Rightarrow\frac{x+4+2015}{2015}+\frac{x+3+2016}{2016}=\frac{x+2+2017}{2017}+\frac{x+1+2018}{2018}\)
\(\Rightarrow\frac{x+2019}{2015}+\frac{x+2019}{2016}-\frac{x+2019}{2017}-\frac{x+2019}{2018}=0\)
\(\Rightarrow\left(x+2019\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
Vì \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\)
=> x + 2019 = 0
=> x = -2019
Vậy x = -2019
x=-2019