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\(\frac{Xx0,75+2,4}{3,8-0,8}\)= \(\frac{Xx0,75+2,4}{3}\)= 6,4
=> X x 0,75 + 2,4= 6,4 x 3
=> X x 0,75 + 2,4 = 19,2
=> X x 0,75 = 19,2 - 2,4
=> X x 0,75 = 16,8
=> X = 16,8 : 0,75
=> x= 22,4
vậy x= 22,4
\(\left(\frac{3}{4}\cdot x\right):\frac{1}{2}=\frac{4}{5}\)
\(\frac{3}{4}\cdot x=\frac{4}{5}\cdot\frac{1}{2}\)
\(\frac{3}{4}\cdot x=\frac{2}{5}\)
\(x=\frac{2}{5}:\frac{3}{4}\)
\(x=\frac{8}{15}\)
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
Ta có:
\(\frac{1}{3.4}.x+\frac{1}{4.5}.x+...+\frac{1}{49.50}.x=1\)
\(\Rightarrow x.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)=1\)
\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)
\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{50}\right)=1\Leftrightarrow x.\frac{47}{150}=1\)
\(\Rightarrow x=1:\frac{47}{150}\Leftrightarrow x=\frac{150}{47}\)
`x xx0,75+x xx1/4=123,5`
`=>x xx 3/4 + x xx 1/4 = 123,5`
`=>x xx(3/4+1/4)=123,5`
`=>x xx 1 = 123,5`
`=> x = 123,5`
dỗi anh V chx bé :))