Tính A = 1/2x5 + 1/3x5 + 1/3x7 +1/4x7+...+1/9x19+1/10x19.
Giúp mình với!!!!!!!!!!!
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\(A=\frac{1}{2.5}+\frac{1}{3.5}+\frac{1}{3.7}+\frac{1}{4.7}+...+\frac{1}{9.19}+\frac{1}{10.19}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{4.5}+\frac{1}{6.5}+\frac{1}{6.7}+\frac{1}{8.7}+...+\frac{1}{18.19}+\frac{1}{20.19}\)
\(\Rightarrow\frac{1}{2}A=\frac{5-4}{4.5}+\frac{6-5}{6.5}+\frac{7-6}{6.7}+...+\frac{20-19}{20.19}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{19}-\frac{1}{20}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{4}-\frac{1}{20}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{5}\)
\(\Rightarrow A=\frac{2}{5}\)
Mình có cách giải khác:
A= \(\frac{1}{2.5}+\frac{1}{3.5}+\frac{1}{3.7}+\frac{1}{4.7}+...+\frac{1}{9.19}+\frac{1}{10.19}\)
A= \(\frac{2.1}{2.2.5}+\frac{2.1}{2.3.5}+\frac{2.1}{2.3.7}+\frac{2.1}{2.4.7}+...+\frac{2.1}{2.9.19}+\frac{2.1}{2.10.19}\)
A= \(\frac{2.1}{4.5}+\frac{2.1}{5.6}+\frac{2.1}{6.7}+\frac{2.1}{7.8}+...+\frac{2.1}{18.19}+\frac{2.1}{19.20}\)
A= \(2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
A=\(2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
A= \(2.\left(\frac{1}{4}+0+0+0+...+0+0-\frac{1}{20}\right)\)
A=\(2.\left(\frac{1}{4}-\frac{1}{20}\right)\)
A=\(2.\left(\frac{5}{20}-\frac{1}{20}\right)\)
A= \(2.\frac{1}{5}\)
A=\(\frac{2}{5}\)
Xong rùi đó!!!!! :))
\(\dfrac{1}{1.3}+\dfrac{1}{2.3}+\dfrac{1}{2.5}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.9}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{3.2}+\dfrac{1}{2.5}+\dfrac{1}{5.3}+\dfrac{1}{3.7}+\dfrac{1}{7.4}+\dfrac{1}{4.9}\)
\(=\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right):\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right):\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{9}\right):\dfrac{1}{2}\)
\(=\dfrac{7}{18}:\dfrac{1}{2}\)
\(=\dfrac{7}{9}\)
a=511/256
b=647/20
c=mình đang suy nghĩ,nhưng nếu bạn k cho mình thì bạn sẽ có câu trả lời
a. 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
= 1 + ( 1 - 1/2) + ( 1/2 - 1/4) + ( 1/4 - 1/8) + ( 1/8 - 1/16) + ( 1/16 - 1/32) + (1/32 - 1/64) + ( 1/64 - 1/128) + (1/128 - 1/256)
= 1 + 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256
= 2 - 1/256
= 511/256
Câu b bạn có viết sai đề không vậy?
\(\frac{3}{2\times5}+\frac{2}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{602\times605}\)
\(=\frac{5-2}{2\times5}+\frac{8-5}{5\times8}+\frac{11-8}{8\times11}+...+\frac{605-602}{602\times605}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\)
\(=\frac{1}{2}-\frac{1}{605}=\frac{603}{1210}\)
\(\frac{4}{3\times7}+\frac{5}{7\times12}+\frac{1}{12\times13}+\frac{2}{13\times15}\)
\(=\frac{7-4}{3\times7}+\frac{12-7}{7\times12}+\frac{13-12}{12\times13}+\frac{15-13}{13\times15}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(\frac{1}{1.3}+\frac{1}{3.2}+\frac{1}{2.5}+...+\frac{1}{99.100}\)
= \(2.\left(\frac{1}{1.3.2}+\frac{1}{3.2.2}+\frac{1}{2.5.2}+...+\frac{1}{99.50.2}\right)\)
= \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{100}\right)\)
= \(2.\frac{49}{100}\)
= \(\frac{49}{50}\)
\(A=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{37\times40}\\ =\dfrac{1}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{37\times40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\dfrac{39}{40}\\ =\dfrac{13}{40}\)
\(A=\dfrac{1}{2.5}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+...+\dfrac{1}{9.19}+\dfrac{1}{10.19}\)
\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{18.19}+\dfrac{2}{19.20}\)
\(A=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\)
\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{20}\right)\)
\(A=2.\dfrac{1}{5}\)
\(A=\dfrac{2}{5}\)