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mk làm câu c cho nó dễ
c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010
=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010
=1-1/x+1=2009/2010
=1/x+1=1-2009/2010
=1/x+1=1/2010
=) x+1=2010
x =2010-1
x =2009
\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}\right)=\frac{1}{2}.\frac{2015}{2017}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{2015}{4034}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2015}{4034}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}\)
\(\frac{1}{x+1}=\frac{1}{2017}\)
\(\Rightarrow\)x+1=2017
\(\Rightarrow\)x=2017-1
x=2016
Vậy x=2016
Chúc bạn học tốt+-*/
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Rightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Rightarrow2\cdot\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}\div2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4032}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4032}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4032}\)
\(\Rightarrow x+1=4032\Rightarrow x=4031\)
Vậy \(x=4031\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.x+1}\right)=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}:2\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2032}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{2032}\)
=> \(\frac{1}{x+1}=\frac{1}{2032}\)
Vì 1 = 1
=> x + 1 = 2032
=> x = 2032 - 1
=> x = 2031
b,\(\Rightarrow\)\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right):2=\frac{2013}{2015}:2\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{4030}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)
\(\Rightarrow\)\(x+1=2015\)
\(\Rightarrow x=2014\)
a, 2/3x -3/2.x-1/2x=5/12
x.(2/3-3/2-1/2)=5/12
x. -4/3=5/12
x=5/12:-4/3
x=-5/16
b,2/6+2/12+2/20+...+2/x.(x+1)=2013/2015
2/2.3+2/3.4+2/4.5+...+2/x.(x+1)=2013/2015
1/2(1-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)=2013/2015
1/2(1-1/x+1)=2013/2015
1-1/x+1=2013/2015 : 1/2
1-1/x+1=4206/2015
suy ra đề sai
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2015}{2017}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\left(x+1\right)}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2017}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{40341}=\frac{1}{2017}\)
\(\Rightarrow x+1=2017\Rightarrow x=2016\)
2/6+2/12+2/20+...+2/x.(x+1)=2013/2015
2.[1/6+1/12+1/20+...+1/x.(x+1)]=2013/2015
1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=2013/4030
1/2-1/3+1/3-1/4+...+1/x-1/x+1=2013/4030
1/2-1/x+1=2013/4030
1/x+1=1/2015
=> x+1=2015
x=2014
Vậy x=2014
Đặt A=Vế trái
Ta có :
\(A \over 2 \)\(= \)\({1\over 6 } +{1\over 12 }+{1\over 20 }+...+{1\over x(x+1)}\)
=\({1\over 2}-{1\over 3}+{1\over 3}-{1\over 4}+{1\over4}-{1\over 5}+...+{1\over x-1}-{1\over x}+{1\over x}-{1\over x+1}\)
=\({1\over2}-{1\over x+1}\)
Từ đó suy ra: \({1\over2}-{1\over x+1}={2013\over4030}\)
=> x=2014
a. 60%x + 0,4x + x : 3 = 2
0.6x + 0,4x + x : 3 = 2
x(0,6 + 0,4 : 3 ) = 2
\(x.\frac{1}{3}=2=>x=2:\frac{1}{3}=\frac{1}{6}\)
câu B tự làm nha .
what ?
Bạn trả lời kiểu j vậy
ko biết làm thì đừng trả lời