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y:0,04+yx3/2-yx6,5=2014
yx25+yx1,5-yx6,5=2014
yx(25+1,5-6,5)=2014
yx20=2014
y=2014:20
y=100,7
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{y\times\left(y+1\right)}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{y}-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow\frac{1}{y+1}=1-\frac{996}{997}=\frac{1}{997}\)
\(\Leftrightarrow y+1=997\Leftrightarrow y=996\)
Vậy y = 996
a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006
không nói linh tinh nha thánh troll trả lời thì trả lời đi bị trừ điểm đó
\(\hept{\begin{cases}\frac{x}{3}+\frac{y}{5}=12\\\frac{x}{5}+\frac{y}{7}=8\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{15}+\frac{y}{25}=\frac{12}{5}\\\frac{x}{15}+\frac{y}{21}=\frac{8}{3}\end{cases}}}\)
\(\Rightarrow\left(\frac{x}{15}+\frac{y}{21}\right)-\left(\frac{x}{15}+\frac{y}{25}\right)=\frac{12}{5}-\frac{8}{3}\)
\(\Rightarrow\frac{y}{21}-\frac{y}{25}=\frac{-4}{15}\)
\(\Rightarrow\frac{4y}{525}=-\frac{4}{15}\Rightarrow\frac{4y}{525}=\frac{-140}{525}\)
\(\Rightarrow y=-35\Rightarrow x=65\)
`y xx 1/8 + y xx 3/8 + y : 2 =2021`
`y xx 1/8 + y xx 3/8 + y xx 1/2 = 2021`
`y xx ( 1/8 + 3/8 + 1/2) =2021`
`y xx ( 1/8 + 3/8 +4/8) = 2021`
`y xx ((1+3+4)/8)=2021`
`y xx 8/8=2021`
`y xx 1=2021`
`y=2021:1`
`y=2021`
Vậy `y=2021`