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\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+....+\frac{3}{418}+\frac{3}{550}\)
\(\Leftrightarrow\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{19.22}+\frac{3}{22.25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)
Nhớ k cho m nhé!
Lướt qua rồi! không phải bạn k mà ấn tượng "đừng lướt qua"
\(A=\frac{3a}{4.1}+\frac{3a}{7.4}+\frac{3a}{10.7}+\frac{3a}{13.10}+..+\frac{3a}{22.19}+\frac{3a}{25.22}=\frac{48}{25}\)
\(a.\left(\frac{3}{4.1}+\frac{3}{7.4}+\frac{3}{10.7}+\frac{3}{13.10}+..+\frac{3}{22.19}+\frac{3}{25.22}\right)=\frac{48}{25}\)
\(B=\left(\frac{3}{4.1}+\frac{3}{7.4}+\frac{3}{10.7}+\frac{3}{13.10}+..+\frac{3}{22.19}+\frac{3}{25.22}\right)\)
\(B=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{22}-\frac{1}{25}\)
\(B=\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)
\(A=a.B=\frac{24a}{25}=\frac{48}{25}\Rightarrow a=2\)
\(\frac{3a}{4}+\frac{3a}{28}+\frac{3a}{70}+...+\frac{3a}{418}+\frac{3a}{550}=\frac{48}{25}\)
\(\Rightarrow a\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{19.22}+\frac{3}{22.25}\right)=\frac{48}{25}\)
\(\Rightarrow a\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\right)=\frac{48}{25}\)
\(\Rightarrow a\left(1-\frac{1}{25}\right)=\frac{48}{25}\)
\(\Rightarrow a.\frac{24}{25}=\frac{48}{25}\)
\(\Rightarrow a=2\)
\(\frac{2}{9}\times\frac{5}{17}+\frac{7}{9}+\frac{2}{9}\times\frac{12}{17}\)
\(=\frac{2}{9}\times\left(\frac{5}{17}+\frac{12}{17}\right)+\frac{7}{9}\)
\(=\frac{2}{9}\times1+\frac{7}{9}\)
\(=\frac{2}{9}+\frac{7}{9}\)
\(=1\)
\(\frac{2}{9}\)× \(\frac{5}{17}\)+ \(\frac{7}{9}\)+\(\frac{2}{9}\)×\(\frac{12}{17}\)
= \(\frac{2}{9}\)×( \(\frac{5}{17}\)+\(\frac{12}{17}\)) + \(\frac{7}{9}\)
= \(\frac{2}{9}\)× \(\frac{17}{17}\)+\(\frac{7}{9}\)
= \(\frac{2}{9}\)+ \(\frac{7}{9}\)
= 1
\(=\frac{\frac{28}{31}.\frac{31}{7}}{\frac{8}{9}.\frac{9}{4}}=\frac{4}{2}=2\)
= 14 : ( 14/3 - 14/9 ) + 14 : ( 2/3 + 8/9 )
= 14 : 28/9 + 14 : 14/9
= 14 * 9/28 + 14 * 9/14
= 14 * ( 9/28 + 9/14 )
= 14 * 27/28
= 27/2
Ta có:
A=\(14:\left(\frac{14}{3}-\frac{14}{5}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)
\(=14:\left(\frac{14}{3}-\frac{14}{5}+\frac{2}{3}+\frac{8}{9}\right)\)
\(=14:\left(\frac{42+6+8}{3}-\frac{14}{5}\right)\)
\(=14:\left(\frac{56}{3}-\frac{14}{5}\right)\)
\(=14:\left(\frac{280-42}{15}\right)=14:\left(\frac{238}{15}\right)=14\cdot\frac{15}{238}=\frac{210}{238}\)
ko biết có tính sai chỗ nào ko
Đặt A = 9/4 + 9/28 +.. + 9/550
A = 9/1.4 + 9/4.7 +... + 9/22.25
A = 3( 3/1.4 + 3/4.7 + .. + 3/22.25)
A = 3 . (1/1 - 1/4 + 1/4 - 1/7 + ... +1/22 - 1/25)
A = 3 (1 - 1/25)
A = 3. 24 / 25
A = 72/25