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gọi tổng này là một số A
ta có
A=7/4x(1/1x5+1/5x9+1/9x13+1/13x17+1/17x21)
A= 7/4x(1-1/5+1/5-1/9+1/9-1/13+1/13-1/17+1/17-1/21)
A= 7/4x(1-1/21)=7/4x20/21
suy ra A=5/3
7/1×5 + 7/5x9 + 7/9x13 + 7/13x17 + 7/17x21
= 7/4x(4/1x5 + 4/5x9 + 4/9x13 + 4/13x17 + 4/17x21)
= 7/4x(1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + 1/17 - 1/21)
= 7/4x(1-1/20)
= 7/4x19/20
= 133/80
Bài 2:
\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(3A=\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}\)
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(3A=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)
\(A=\frac{15}{34}\times\frac{1}{3}=\frac{5}{34}\)
Bài 2:
\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(3A=\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(3A=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)
\(A=\frac{15}{34}\times\frac{1}{3}=\frac{5}{34}\)
Ta có : \(\frac{5}{7}=\frac{x}{63}\)
\(\Rightarrow\) \(\frac{45}{63}=\frac{x}{63}\)
\(\Rightarrow\) \(x=45\)
Vậy x = 45
Cbht !!! ❤️❤️❤️
\(\frac{5}{7}=\frac{x}{63}\\ \Leftrightarrow5\times63=7x.\)
\(\Leftrightarrow315=7x\)
\(\Leftrightarrow x=45\)
\(=\dfrac{195}{81}-\dfrac{12}{7}+\dfrac{44}{5}=\dfrac{8971}{945}\)
\(\frac{7}{4}\times\frac{15}{7}+\frac{13}{7}\times\frac{9}{4}-\frac{2}{7}\)
\(=\left(\frac{7}{4}+\frac{9}{4}\right)\times\left(\frac{15}{7}+\frac{13}{7}\right)-\frac{2}{7}\)
\(=4\times4-\frac{2}{7}\)
\(=16-\frac{2}{7}\)
\(=\frac{112}{7}-\frac{2}{7}\)
\(=\frac{110}{7}\)
Đề\(\Rightarrow\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{15}-\frac{1}{21}+\frac{3}{x}=\frac{39}{25}\)
\(\Rightarrow1-\frac{1}{21}+\frac{3}{x}=\frac{39}{25}\Rightarrow\frac{20}{21}+\frac{3}{x}=\frac{39}{25}\)
\(\Rightarrow\frac{3}{x}=\frac{39}{25}-\frac{20}{21}\)
đến đây bạn tự tính ra rồi tìm x là được nha, như thế này là rút gọn về đến đơn giản rồi nên bạn có thể làm được
Thiengl2015
7/4 x 15/7 + 13/7 x 9/4 - 7/4 x2/7
=7/4 x 15/7- 7/4 x2/7+ 13/7 x 9/4
=7/4 x (15/7 - 2/7) + 13/7 x 9/4
=7/4 x 13 /7 + 13/7 x 9/4
=13/7 x (7/4+9/4)
=13/7 x 16/4
=13/7 x 4
=52/7
k mk nha
sorry câu trả lời vừa nãy mk ấn nhầm
7/4 x 15/7 + 13/7 x 9/4 - 7/4 x2/7
=7/4 x 15/7 - 7/4 x 2/7 + 13/7 x 9/4
\(A=\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+...+\frac{7}{17\cdot21}=\)
\(\frac{4}{7}A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{17\cdot21}=\)
\(\frac{4}{7}A=\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+...+\left(\frac{1}{17}-\frac{1}{21}\right)=\)
\(\frac{4}{7}A=1-\frac{1}{21}=\)
\(\frac{4}{7}A=\frac{20}{21}\)
\(A=\frac{20}{21}\div\frac{4}{7}\)
\(A=\frac{20}{21}\times\frac{7}{4}=\frac{140}{84}=\frac{5}{3}\)
\(\frac{7}{1.5}+\frac{7}{5.9}+\frac{7}{9.13}+...+\frac{7}{17.21}\)
\(=\frac{7}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{17.21}\right)\)
\(=\frac{7}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{17}-\frac{1}{21}\right)\)
\(=\frac{7}{4}.\left(1-\frac{1}{21}\right)\)
\(=\frac{7}{4}.\frac{20}{21}=\frac{5}{3}\)