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Tìm y:
-y:1/2-5/2=4+1/2
-y:1/2 = 4+1/2+5/2
-y:1/2 = 7
-y = 7.2
y = -14
Vậy y = -14
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\\ A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\\ A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot29\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot31\right)}{\left(2\cdot3\cdot4\cdot...\cdot30\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot30\right)}\\ A=\dfrac{1\cdot2\cdot3\cdot...\cdot29}{2\cdot3\cdot4\cdot...\cdot30}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot31}{2\cdot3\cdot4\cdot...\cdot30}\\ A=\dfrac{1}{30}\cdot\dfrac{31}{2}\\ A=\dfrac{31}{60}\)
A=\(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.....\dfrac{899}{900}\)
A=\(\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)
A=\(\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)
A=\(\dfrac{\left(1.2.3.....28.29\right).\left(3.4.5....29.31\right)}{\left(2.3.4....29.30\right).\left(2.3.4....29.30\right)}\)
A=\(\dfrac{1.31}{30.2}\)
A=\(\dfrac{31}{60}\)
Vậy A = \(\dfrac{31}{60}\)
\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}\)
\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)
\(A=\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)
\(A=\dfrac{1.2.3.....29}{2.3.4....30}.\dfrac{3.4.5.....31}{2.3.4.....30}\)
\(A=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}\)
\(B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.....\dfrac{49.51}{50.50}\)
\(B=\dfrac{2.4.3.5.4.6.....49.51}{3.3.4.4.5.5....50.50}\)
\(B=\dfrac{2.3.4......49}{3.4.5....50}.\dfrac{4.5.6.....51}{3.4.5....50}\)
\(B=\dfrac{2}{50}.\dfrac{51}{3}=\dfrac{17}{25}\)
Giải:
\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}.\)
\(A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{29.31}{30^2}.\)
\(A=\dfrac{1.2.3.....29}{2.3.4.....30}.\dfrac{2.3.4.....31}{2.3.4.....30}.\)
\(A=\dfrac{1}{30}.31=\dfrac{30}{31}.\)
Vậy \(A=\dfrac{30}{31}.\)
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\)
\(A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot...\cdot30\right)^2}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot30}\)
\(A=\dfrac{1\cdot31}{30}=\dfrac{31}{30}\)
Ta có : \(\dfrac{1}{101}>\dfrac{1}{300}\)
...
\(\dfrac{1}{299}>\dfrac{1}{300}\)
Do đó :
\(\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{1}{300}+\dfrac{1}{300}..+\dfrac{1}{300}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{200}{300}=\dfrac{2}{3}\)
Vậy...
\(\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}\)
= \(\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)
= \(\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)
=\(\dfrac{\left(1.2.3.....29\right).\left(3.4.5......31\right)}{\left(2.3.4......30\right).\left(2.3.4.....30\right)}\)
= \(\dfrac{1.31}{2.30}\)
= \(\dfrac{31}{60}\)
Ta có: \(\dfrac{3}{2^2}\cdot\dfrac{8}{3^2}\cdot\dfrac{15}{4^2}\cdot...\cdot\dfrac{899}{30^2}\)
\(=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot2^2}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\)
\(=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot29\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot31\right)}{\left(2\cdot3\cdot4\cdot...\cdot30\right)\left(2\cdot3\cdot4\cdot...\cdot30\right)}\)
\(=\dfrac{1\cdot31}{2\cdot30}=\dfrac{31}{60}\)
a: \(=\dfrac{8}{9}\cdot\dfrac{9}{4}\cdot\dfrac{12}{19}\cdot\dfrac{19}{24}=\dfrac{1}{2}\cdot2=1\)
b: \(=\dfrac{5}{16}\cdot\dfrac{17}{15}\cdot\dfrac{8}{17}=\dfrac{5}{16}\cdot\dfrac{8}{15}=\dfrac{40}{240}=\dfrac{1}{6}\)
c: \(=\dfrac{4}{13}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)-\dfrac{3}{26}=\dfrac{4}{13}-\dfrac{3}{26}=\dfrac{5}{26}\)
c: \(=\dfrac{3}{4}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{1}{5}=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}\)
Ta có: \(S=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)
\(=\dfrac{3}{2^2}\cdot\dfrac{2^3}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{99}{10^2}\)
\(=\dfrac{11}{20}\)
bạn có thể giải thích rõ tại sao S=\(\dfrac{11}{20}\) đc ko
\(a.\)
\(\dfrac{3}{16}:\dfrac{?}{8}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3}{16}\cdot\dfrac{8}{?}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3}{2?}=\dfrac{3}{4}\)
\(\Leftrightarrow?=2\)
\(b.\)
\(\dfrac{1}{25}:-\dfrac{3}{?}=-\dfrac{1}{15}\)
\(\Leftrightarrow\dfrac{1}{25}\cdot\dfrac{-?}{3}=-\dfrac{1}{15}\)
\(\Leftrightarrow\dfrac{-?}{75}=-\dfrac{1}{15}\)
\(\Leftrightarrow?=\dfrac{75}{15}=5\)
\(c.\)
\(\dfrac{?}{12}:-\dfrac{4}{9}=-\dfrac{3}{16}\)
\(\Leftrightarrow\dfrac{?}{12}\cdot\dfrac{-9}{4}=-\dfrac{3}{16}\)
\(\Leftrightarrow\dfrac{-3?}{16}=-\dfrac{3}{16}\)
\(\Leftrightarrow?=1\)
Mk gọi ? = x nha
a) \(\dfrac{3}{16}:\dfrac{x}{8}=\dfrac{3}{4}\)
\(\dfrac{x}{8}=\dfrac{3}{16}:\dfrac{3}{4}\)
\(\dfrac{x}{8}=\dfrac{1}{4}\)
⇒\(x=\dfrac{1.8}{4}=2\)
b) \(\dfrac{1}{25}:\dfrac{-3}{x}=\dfrac{-1}{15}\)
\(\dfrac{-3}{x}=\dfrac{1}{25}:\dfrac{-1}{15}\)
\(\dfrac{-3}{x}=\dfrac{-3}{5}\)
⇒x=5
c) \(\dfrac{x}{12}:\dfrac{-4}{9}=\dfrac{-3}{16}\)
\(\dfrac{x}{12}=\dfrac{-3}{16}.\dfrac{-4}{9}\)
\(\dfrac{x}{12}=\dfrac{1}{12}\)
⇒x=1
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\\ =\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\\ =\dfrac{1\cdot2\cdot3\cdot...\cdot29}{2\cdot3\cdot4\cdot...\cdot30}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot31}{2\cdot3\cdot4\cdot...\cdot30}\\ =\dfrac{1}{30}\cdot\dfrac{31}{2}\\ =\dfrac{31}{60}\)
Vậy ...