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Ta có :
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\) ( đpcm )
Vậy với mọi \(x\inℕ^∗\) thì ta có công thức \(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
Chúc bạn học tốt ~
Có \(\frac{1}{x\left(x+1\right)}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
Vậy với mọi x \(\inℕ^∗\)ta luôn có \(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)( đpcm )
Chúc bạn học tốt!
Để mình đưa công thức tổng quát luôn khỏi mất công bạn đăng câu hỏi cho mệt =.=
Với mọi \(a,n\inℕ^∗\)
Cần chứng minh :
\(\frac{n}{a\left(a+n\right)}=\frac{1}{a}-\frac{1}{a+n}\)
Ta có :
\(\frac{1}{a}-\frac{1}{a+n}=\frac{a+n}{a\left(a+n\right)}-\frac{a}{a\left(a+n\right)}=\frac{a+n-a}{a\left(a+n\right)}=\frac{n}{a\left(a+n\right)}\) ( đpcm )
Vậy với mọi \(a,n\inℕ^∗\) thì \(\frac{n}{a\left(a+n\right)}=\frac{1}{a}-\frac{1}{a+n}\)
Chúc bạn học tốt ~
Ta có :
\(\frac{1}{x}-\frac{1}{x+2}=\frac{x+2}{x\left(x+2\right)}-\frac{x}{x\left(x+2\right)}=\frac{x+2-x}{x\left(x+2\right)}=\frac{2}{x\left(x+2\right)}\) ( đpcm )
Vậy với mọi \(x\inℕ^∗\) ta luôn có \(\frac{2}{x\left(x+2\right)}=\frac{1}{x}-\frac{1}{x+2}\)
Chúc bạn học tốt ~
Ta có: \(\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}=\frac{a+2}{a\left(a+1\right)\left(a+2\right)}-\frac{a}{a\left(a+1\right)\left(a+2\right)}\)
\(=\frac{a+2-a}{a\left(a+1\right)\left(a+2\right)}=\frac{2}{a\left(a+1\right)\left(a+2\right)}\left(đpcm\right)\)
Có: \(\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{\left(a+2\right)-a}{a\left(a+1\right)\left(a+2\right)}=\frac{a+2}{a\left(a+1\right)\left(a+2\right)}-\frac{a}{a\left(a+1\right)\left(a+2\right)}\)\(=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)
Bài 1 :
\(\left(-2\right)\left(x+1\right)-3\left(1-x\right)=4\)
\(\Leftrightarrow-2x-2-3+3x=4\)
\(\Leftrightarrow x=4+2+3=9\)
Bài 2 :
Cho \(S=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\)
\(\Leftrightarrow S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)
\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\Rightarrow S< \left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)\)
\(+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)
\(\Leftrightarrow S< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)(1)
Lại có :
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)
\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\Leftrightarrow S>\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)
\(+\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)
\(\Leftrightarrow S>\frac{10}{40}+\frac{10}{50}+\frac{10}{60}=\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\)(2)
Từ (1) và (2) , ta có :
\(\frac{3}{5}< S< \frac{4}{5}hay\frac{3}{5}< \frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}< \frac{4}{5}\)
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
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
1/x-1/x+1=a/x.(x+a)
ta có : 1/x-1/x+1=(x+1)-x/x.(x+1)=1/x.(x+1)
Vay ....