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\(y=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1275}\)\(=2\cdot\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\right)=2\cdot\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{50.51}\right)\)\(=2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)=2\cdot\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)
\(y:\frac{5}{2}=\frac{7}{4}:\frac{7}{3}\)
\(y:\frac{5}{2}=\frac{3}{4}\)
\(y=\frac{3}{4}.\frac{5}{2}\)
\(y=\frac{15}{8}\)
Vậy \(y=\frac{15}{8}\)
Chúc bạn zui ~^^
\(y:\frac{5}{2}=\frac{7}{4}:\frac{7}{3}\)
\(y:\frac{5}{2}=\frac{3}{4}\)
\(y=\frac{3}{4}\cdot\frac{5}{2}\)
\(y=1.875\)
Vậy y = 1.875
y:\(\frac{7}{3}=\frac{7}{4}:\frac{7}{3}\)
y.\(\frac{3}{7}=\frac{7}{4}.\frac{3}{7}\)
y.3/7=3/4
y=3/4:3/7
y=7/3
Vậy y=7/3
\(1\frac{1}{3}+1\frac{1}{5}.y-\frac{4}{5}=2\frac{4}{5}\)
\(\Rightarrow\frac{4}{3}+\frac{6}{5}.y=2\frac{4}{5}+\frac{4}{5}\)
\(\Rightarrow\frac{4}{3}+\frac{6}{5}.y=2\frac{8}{5}=\frac{18}{5}\)
\(\Rightarrow\frac{6}{5}y=\frac{18}{5}-\frac{4}{3}\)
\(\Rightarrow\frac{6}{5}y=\frac{34}{15}\)
\(\Rightarrow y=\frac{34}{15}:\frac{6}{5}\)
\(\Rightarrow y=\frac{34}{15}.\frac{5}{6}=\frac{17}{9}\)
Trả lời:
\(y\times\frac{15}{2}-\frac{1}{3}\times\left(\frac{1}{4}+y\right)=96\frac{2}{3}\)
\(\Leftrightarrow y\times\frac{15}{2}-\frac{1}{12}-\frac{1}{3}\times y=\frac{290}{3}\)
\(\Leftrightarrow y\times\left(\frac{15}{2}-\frac{1}{3}\right)=\frac{387}{4}\)
\(\Leftrightarrow y\times\frac{43}{6}=\frac{387}{4}\)
\(\Leftrightarrow y=\frac{27}{2}\)
Vậy \(y=\frac{27}{2}\)
1 / 7/4 - y . 5/6 = 1/2 + 1/3
7/4 - 5/6y = 5/6
5/6y = 7/4 - 5/6
5/6y = 11/12
y = 11/12 : 5/6
y = 11/10
2 . 136 là số chia 9 dư 1
=> y = 136 - 1 = 135
3 . Dựa theo quy luật của dãy thì số tiếp theo là :
1/16 : 2 = 1/32
1: y = \(\frac{11}{10}\)
2: Y = 135
3: Số hạng tiếp theo của dãy: \(\frac{1}{32}\)
$=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}$
$1+\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{1}{2008}\right)$
$\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}$
$2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)$
A=$\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}}$
A=2009
\(y\div2\frac{1}{3}=1\frac{3}{4}\div2\frac{1}{3}\)
\(1\frac{3}{4}=\frac{7}{4};2\frac{1}{3}=\frac{7}{3}\)
\(y\div2\frac{1}{3}=1\frac{3}{4}\div2\frac{1}{3}\)
\(\Rightarrow y\div\frac{7}{3}=\frac{3}{4}\)
\(\Rightarrow y=\frac{3}{4}\times\frac{7}{3}\)
\(\Rightarrow y=\frac{7}{4}\)
~ Ủng hộ nhé ~
\(y:2\frac{1}{3}=1\frac{3}{4}:2\frac{1}{3}\)
\(y:\frac{7}{3}=\frac{7}{4}:\frac{7}{3}\)
\(y:\frac{7}{3}=\frac{3}{4}\)
\(y=\frac{3}{4}\times\frac{7}{3}\)
\(y=\frac{7}{4}\)