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\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{3.7}+\frac{1}{4.7}+\frac{1}{4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{2.3.7}+\frac{2}{2.4.7}+\frac{2}{2.4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6}-\frac{2}{7}+\frac{2}{7}-\frac{2}{8}+....+\frac{2}{x}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{6}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{3}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{3}{9}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)
câu a khó quá.Để nghĩ.
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{21\cdot2}+\frac{2}{28\cdot2}+\frac{2}{36\cdot2}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x-1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{x-5}{6x+6}=\frac{1}{9}\)
\(\Rightarrow9\left(x-5\right)=6x+6\)
\(\Rightarrow9x-45=6x+6\)
\(\Rightarrow9x-6x=51\)
\(\Rightarrow3x=51\)
Tới đây bí:v
\(A=3.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{101}-\frac{1}{105}\right)\)
\(A=3.\left(1-\frac{1}{105}\right)\)
\(A=3.\frac{104}{105}\)
\(A=\frac{104}{35}\)
Em yêu cầu bác nhìn xuống dưới và bác sẽ biết cách làm
Bác thấy rồi mà còn đăng
Thay số mà làm nhé
:))
\(4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{x.\left(x+4\right)}\)
\(4A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+4}\)
\(4A=1-\frac{1}{x+4}\)
\(4A=\frac{x+4-1}{x+4}\)
\(A=\frac{x+3}{\text{4(x+4)}}\)
Bạn tự thay rồi tính nhé
\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+........+\frac{1}{x\cdot\left(x+4\right)}\)
\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+........+\frac{4}{x\cdot\left(x+4\right)}\)
\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.......+\frac{1}{x}-\frac{1}{x+4}\)
\(4A=1-\frac{1}{x+4}\)
\(A=\left(1-\frac{1}{x+4}\right):4\)
Khi x = 12 => \(A=\left(1-\frac{1}{12+4}\right):4\)
A = \(\left(1-\frac{1}{16}:4\right)\)
A = \(\frac{15}{16}:4=\frac{15}{64}\)
Khi x = 2 => \(A=\left(1-\frac{1}{2+4}\right):4\)
A = \(\left(1-\frac{1}{6}\right):4\)
A \(=\frac{5}{6}:4=\frac{5}{24}\)
Khi x = \(\frac{5}{6}\)=> \(A=\left(1-\frac{1}{\frac{5}{6}+4}\right):4\)
A = \(\left(1-\frac{1}{\frac{29}{6}}\right):4\)
A = \(\frac{23}{29}:4=\frac{23}{116}\)
\(D=12\cdot\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{22}+...+\frac{1}{97}-\frac{1}{202}\right)\)
\(D=12\cdot\left(\frac{1}{6}-\frac{1}{202}\right)\)
\(D=12\cdot\frac{49}{303}\)
\(D=\frac{588}{303}\)
\(F=\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\)
\(\Rightarrow\)\(\frac{1}{2}F=\frac{1}{2}.\left(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\right)\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{380}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{19.20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{19}-\frac{1}{20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{5}-\frac{1}{20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{4}{20}-\frac{1}{20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{3}{20}\)
\(\Rightarrow\)\(F=\frac{3}{20}\div\frac{1}{2}\)
\(\Rightarrow\) \(F=\frac{3}{20}.2\)
\(\Rightarrow\)\(F=\frac{3}{10}\)
\(F=\frac{1}{15}+\frac{ 1}{21}+...+\frac{1}{190}\)
\(F=\frac{2}{30}+\frac{2}{21}+...+\frac{2}{380}\)
\(F=\frac{2}{5.6}+...+\frac{2}{19.20}\)
\(F=2.\left(\frac{1}{5.6}+...+\frac{1}{19.20}\right)\)
\(F=2.\left(\frac{1}{5}-\frac{1}{6}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(F=2\left[\frac{1}{5}-\left(\frac{1}{6}-\frac{1}{6}\right)-...-\left(\frac{1}{19}-\frac{1}{19}\right)-\frac{1}{20}\right]\)
\(F=2.\left(\frac{1}{5}-\frac{1}{20}\right)\)
\(F=2.\frac{3}{20}\)
\(F=\frac{6}{20}=\frac{3}{10}\)
\(G=\frac{12}{84}+\frac{12}{210}+...+\frac{12}{2100}\)
\(G=\frac{4}{28}+\frac{4}{70}+...+\frac{4}{700}\)
\(G=\frac{4}{4.7}+\frac{4}{7.10}+...+\frac{4}{25.28}\)
\(G=\frac{4}{3}.\left(\frac{3}{4.7}+...+\frac{3}{25.28}\right)\)
\(G=\frac{4}{3}.\left(\frac{1}{4}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(G=\frac{4}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(G=\frac{4}{3}.\frac{6}{28}\)
\(G=\frac{2}{7}\)
Tổng của G và F là : \(\frac{3}{10}+\frac{2}{7}=\frac{21}{70}+\frac{20}{70}=\frac{41}{70}\)
\(A=\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\)
\(A=\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{380}\) ( nhân cả tử và mẫu với 2 )
\(A=\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{19.20}=2\left(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{19.20}\right)\)
A = \(2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\right)=2\left(\frac{1}{5}-\frac{1}{20}\right)=2.\frac{3}{20}=\frac{3}{10}\)
B = \(\frac{12}{84}+\frac{12}{210}+\frac{12}{390}+...+\frac{12}{2100}\)
\(B=\frac{4}{28}+\frac{4}{70}+\frac{4}{130}+...+\frac{4}{700}\) ( chia cả tử và mẫu của mỗi phân số cho 3 )
B = \(\frac{4}{4.7}+\frac{4}{7.10}+\frac{4}{10.13}+...+\frac{4}{25.28}=\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)
B = \(\frac{4}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{4}{3}.\frac{6}{28}=\frac{2}{3}\)
\(A=\frac{12}{1.5}+\frac{12}{5.9}+\frac{12}{9.13}+.............+\frac{12}{101.105}\)
\(=3.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+............+\frac{4}{101.105}\right)\)
\(=3\left(1-\frac{1}{105}\right)\)
\(=3.\frac{104}{105}=\frac{312}{105}\)