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Tìm x biết : \(\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+...+\frac{2}{x.\left(x+1\right)}=\frac{1}{9}\)
\(\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+...+\frac{2}{x.\left(x+1\right)}=\frac{1}{9}\)
\(\frac{2}{90}+\frac{2}{110}+\frac{2}{132}+...+\frac{2}{x.\left(x+1\right)}=\frac{1}{9}\)
\(2\left(\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{1}{9}\)
\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{1}{9}\)
\(2\left(\frac{1}{9}-\frac{1}{\left(x+1\right)}\right)=\frac{1}{9}\)
\(\frac{1}{9}-\frac{1}{\left(x+1\right)}=\frac{1}{18}\)
\(\frac{1}{\left(x+1\right)}=\frac{1}{18}\)
\(x=17\)
khi ko mún tích thì tích 1 tích
khi mún tích thì tích 50 tích
\(\left(1\frac{1}{4}-\frac{3}{5}\right):\frac{17}{20}< \frac{x}{17}< \left(5\frac{1}{3}-3\frac{1}{2}\right).\frac{12}{17}\)
= \(\left(\frac{5-3}{4}\right):\frac{17}{20}< \frac{x}{17}< \left(\frac{16}{3}-\frac{7}{2}\right).\frac{12}{17}\)
= \(\frac{1}{2}:\frac{17}{20}< \frac{x}{17}< \left(\frac{32-21}{6}\right).\frac{12}{17}\)
= \(\frac{10}{17}< \frac{x}{17}< \frac{3}{2}.\frac{12}{17}\)
= \(\frac{10}{17}< \frac{x}{17}< \frac{18}{17}\)
( Mik thấy mẫu giống nhau mik sẽ bỏ mẫu đi mik sẽ tìm tử )
=> 10 < 11 ; 12 ; 13 ; 14 ; 15 ; 16 ; 17 < 18
=> x = { 11 ; 12 ; 13 ; 14 ; 15 ; 16 ; 17 }
k mik nha làm ơn đó
Ta có :
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\) ( cái đề hình như có 1 phân số \(\frac{2}{9}\) đúng không bạn )
\(\Leftrightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\)\(x+1=1:\frac{1}{18}\)
\(\Leftrightarrow\)\(x+1=18\)
\(\Leftrightarrow\)\(x=18-1\)
\(\Leftrightarrow\)\(x=17\)
Vậy \(x=17\)
Chúc bạn học tốt ~
#)Giải :
Đặt \(A=\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+...+\frac{2}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{9}-\frac{1}{x+1}=\frac{1}{9}\)
Đến đây thì ez rùi nhé ^^