Tìm số tự nhiên x thỏa mãn: 5/1.6 + 5/6.11 +...+ 5/(5x+1).(5x+6) =2010/2011. b) 1/3.5+1/5.7+.,.+ 1/(2x+1).(2x+3) =15/93.
Tìm số tự nhiên x thỏa mãn:
a) 5/1.6 + 5/6.11 +...+ 5/(5x + 1).(5x + 6) = 2010/2011
b) 1/3.5 + 1/5.7 + 1/7.9 +...+ 1/(2x + 1).(2x + 3) = 15/93
tìm các số nguyên x thoả mãn:
5/1.6+5/6.11+...+5/(5x+1)(5x+6)=2010/2011
=> 1 - 1/6 + 1/6 - 1/11 + ......+ 1/5x+1 - 1/5x + 6 = 2010/2011
=> 1 - 1/5x+6 = 2010/2011
=> 1/5x+6 = 1 - 2010/2011
=> 1/5x + 6 = 2009/2011
=> ...........................Còn lại bạn tự làm nha!
Ai k mk mk k lại !!!
tìm x,y thỏa mãn
a.\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2010}{2011}\)
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2011}\)
\(\Rightarrow5x+6=2011\)
\(\Rightarrow5x=2011-6\)
\(\Rightarrow5x=2005\)
\(\Rightarrow x=401\)
Tìm x biết 5/1.6+5/6.11+...+5/(5x+1)(5x+6)=2010/2011
Ta có:
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}\)
\(=1-\frac{1}{5x+6}\)
\(=\frac{5x+5}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow5x+5=2010\)
\(\Rightarrow5x=2010-5=2005\)
\(\Rightarrow x=2005:5=401\)
Vậy x=401
Tìm x:
Bài 3:
a) 1/3 + 2/3 : x = -7
b) 3 1/2 - 1/2 x = 2/3
c) [(x + 1/3) * 3/4 +5] : 2=3
d) (1 2/3 - x). 0,75 - 2 = 1/2
e) x + 75% x = -1,6
f) (x - 2/3) (2x+1) = 0
g) 1 - (2x + 1/2) 2 = 3/4
h) 5/1.6 + 5/6.11 + ... + 5/(5x + 1).(5x + 6) = 2020/2021
Giúp mình bài 3 với, ai làm được đúng hết mình sẽ tik hết nhé! Hurry up!
Tìm số nguyên x thỏa mãn :
1. -1/4 : -3/4 + 1/2 < x < 7/8 - 1/2 : -5/6
2. 5/1.6 + 5/6.11 +...+ 5/(5x+1).(5x+6)
Bạn nào trả lời nhanh và đúng nhất, trc 8h15 thì có thưởng 1 đúng nhé. Nhưng làm ơn nghiêm túc. Xin cảm ơn !
tìm stn x thỏa mãn
a. 7/x+4/5.9+4/9.13+4/13.17+...+4/41.45=29/45
b. 1/3.5+1/5.7+1/7.9+...+1/(2x+1)(2x+3)=15/93
Ta có \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)(đk : \(x\ne0\))
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
=> \(\frac{7}{x}=\frac{7}{15}\)
=> x = 15 (tm)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
=> \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)
=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
=> \(\frac{1}{3}-\frac{1}{n+3}=\frac{10}{31}\)
=> \(\frac{1}{2x+3}=\frac{1}{93}\)
=> 2x + 3 = 93
=> 2x = 90
=> x = 45
tìm x biết
5/1.6 + 5/6.11 + 5/11.16 + .... + 5/5x+1.5x+6 = 2010/2011
giải cụ thể nha
=> 1 - 1/6 + 1/6 - 1/11 +.......+1/5x+1 - 1/5x+6=2010/2011
=> 1 - 1/5x+6 = 2010/2011
=> 1/5x+6 = 1/2011
=> 5x + 6 = 2011
=> 5x = 2005
=> x = 401(tm)
Tìm số tự nhiên x thỏa mãn :
a) (2x+3).(5x-15)=0 b) (3x+1).(3x-9)=0
a) \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-15=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\3x-9=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=3\end{matrix}\right.\)
a. \(\left[{}\begin{matrix}2x+3=0\\5x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=3\end{matrix}\right.\)
b. \(\left[{}\begin{matrix}3x+1=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=3\end{matrix}\right.\)
a)\(\left(2x+3\right).\left(5x-15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-3\\5x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1,5\\x=3\end{matrix}\right.\)
b)\(\left(3x+1\right).\left(3x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\3x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=3\end{matrix}\right.\)