CD
1
K
Khách
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PV
24 tháng 4 2016
* ĐK: \(x\ne0\)
Đề ra ...<=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
<=> \(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{1}{9}\)
<=> \(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
<=>\(\frac{1}{6}-\frac{1}{x+1}+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
<=>\(\frac{1}{x+1}\left(1-\frac{1}{x}\right)=\frac{1}{6}-\frac{1}{9}\)
<=> \(\frac{x-1}{x\left(x+1\right)}=\frac{1}{36}\)
<=> \(\frac{x-1}{x\left(x-1\right)}=\frac{x-1}{36.\left(x-1\right)}\)
=> x(x-1) = 36. (x-1) => x =36
TN
24 tháng 4 2016
\(\frac{2}{2}.\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x+\left(x+1\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)
\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
x+1=18
x=18-1
x=17
24 tháng 3 2016
ban viet sai de bai roi khong phai1/27 ma la1/21
ta co: 1/21+1/28+1/36+...+2/x(x+1)=2/9
1/21.2+1/28.2+1/36.2+...+2/x(x+1).2=2/9.1/2
1/42+1/56+1/72+...+1/x(x+1)=1/9
1/6.7+1/7.8+1/8.9+...+1/x(x+1)=1/9
1/6-1/7+1/7-1/8+1/8-1/9+...+1/x-1/x+1=1/9
1/6-1/x+1=1/9(khu lien tiep)
1/x+1=1/6-1/9
1/x+1=1/18
=>x+1=18
=>x=17
24 tháng 3 2016
khu lien tiep là khu liên tiếp đúng ko bạn, tại bạn đánh ko dấu nên mình ko hiểu
BV
1
BI
18 tháng 3 2016
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}\div2\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=18-1\Rightarrow x=17\)
24 tháng 4 2016
1/21 + 1/28 + 1/36 + ...+ 1/x(x+1)
=> 2/42 + 2/56 + 2/72 +....+ 2/x(x+1)
=> 2.(1/42 + 1/56 + 1/72 + ... + 1/x.(x+1))
=> 2 .(1/6.7 + 1/7.8 + 1/8.9 + ..+ 1/x.(x+1))
=> 2. ( 1/6 - 1/7 + 1/7-1/8 + ...+ 1/x - 1/x+1
=> 2 . (1/6 - 1/x+1)
=>1/3 - 2/x+1
DX
5
27 tháng 5 2015
1/21 +1/28 +1/36 ...2/x (x+1)=2/9
=>1/3.7 +1/4.7+1/4.9 ....2/x.(x+1)=2/9
=>2/6.7+2/7.8+2/8.9....+2/x.(x+1)=2/9
=>2.(1/6-1/x+1)=2/9
=>1/6 - 1/x +1 =1/9
=>1/x + 1 =1/18
=>x+1=18
=>x=17
DS:17
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2022
\(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(\Leftrightarrow x=17\)
<=> \(\frac{1}{3.7}+\frac{1}{4.7}+\frac{1}{4.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
<=> \(\frac{2}{2.3.7}+\frac{2}{2.4.7}+\frac{2}{2.4.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
<=> \(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
<=> \(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{\frac{x-1}{x+1}}=\frac{2}{9}\right)\)
<=> \(2.\left(\frac{1}{6}-\frac{\frac{1}{x-1}}{x+1}\right)=\frac{2}{9}\)
<=> \(\frac{1}{6}+\frac{1}{x+1}=\frac{1}{9}\)
<=> \(\frac{1}{x-1}=\frac{1}{6}-\frac{1}{9}\)
<=> \(\frac{1}{x+1}=\frac{1}{18}\)
<=> x + 1 = 18
=> x = 18 - 1
x = 17