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\(A=\frac{7}{1\cdot5}+\frac{7}{5\cdot9}+...+\frac{7}{17\cdot21}=\)
\(\frac{4}{7}A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{17\cdot21}=\)
\(\frac{4}{7}A=\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+...+\left(\frac{1}{17}-\frac{1}{21}\right)=\)
\(\frac{4}{7}A=1-\frac{1}{21}=\)
\(\frac{4}{7}A=\frac{20}{21}\)
\(A=\frac{20}{21}\div\frac{4}{7}\)
\(A=\frac{20}{21}\times\frac{7}{4}=\frac{140}{84}=\frac{5}{3}\)
\(\frac{7}{1.5}+\frac{7}{5.9}+\frac{7}{9.13}+...+\frac{7}{17.21}\)
\(=\frac{7}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{17.21}\right)\)
\(=\frac{7}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{17}-\frac{1}{21}\right)\)
\(=\frac{7}{4}.\left(1-\frac{1}{21}\right)\)
\(=\frac{7}{4}.\frac{20}{21}=\frac{5}{3}\)
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:
\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(3A=\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(3A=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)
\(A=\frac{15}{34}\times\frac{1}{3}=\frac{5}{34}\)
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
a.(4276+5724)+(2357+7643)
=10000 + 10000
=20000
b.câu này không biết:)))
c.câu này cũng không biết :)))
a)4276+2357+5724+7643
= (4276+5724)+(2357+7643)
= 1000+1000
= 2000
b)3145-246+2347-145+4246-347
= (3145-145)+(4246-246)+(2347-347)
= 3000+4000+2000
= 9000
c)(45-5x9)x2x3x4x5x6x7x8x9
= (45-45)x2x3x4x5x6x7x8x9
=0x2x3x4x5x6x7x8x9
=0
4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 x 2 : 2
= 32 + 4 x 2 : 2
= 32 + 8 : 2
= 32 + 4
= 36
1 - \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) = \(\dfrac{4}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) = \(\dfrac{2}{4}\) = \(\dfrac{1}{2}\)
Vậy a, S còn b, Đ
\(\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+\dfrac{4}{13\times17}+\dfrac{4}{17\times21}+\dfrac{4}{21\times25}\)
\(=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{21}-\dfrac{1}{25}\)
\(=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)