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\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{2015}:2\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2013}{4030}\)
\(\frac{1}{x+1}=\frac{1}{2015}\)
=>x+1=2015
=>x=2014
$1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}$1+13 +16 +110 +...+2x(x+1) =120132015
Toán lớp 6\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{6.2}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
\(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=1\frac{2013}{2015}\)
\(2\left(\frac{1}{x}-\frac{1}{x+1}\right)=1\frac{2013}{2015}\)
tự làm tiếp nhé
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
nhân cả 2 vế của đẳng thức với 1/2 ta được
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-......+\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=>\frac{1}{x+1}=\frac{1}{2}-\frac{2014}{2015}\)
\(\frac{1}{x+1}=-\frac{2013}{4030}\)
hay \(1:\left(x+1\right)=-\frac{2013}{4030}\)
\(x+1=-\frac{4030}{2013}\)
\(=>x=-\frac{6043}{2013}\)
có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1
=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)
vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015
x=2015
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
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