Tìm x:
a)4/1.5+4/5.9+...+4/9.13+...97/101=2.x+4/101
b)1/5.8+1/8.11+....+1/x.(x+3)=101/1840
c)1+1/3+1/6+1/10+...+2/x.(x+1)=1 2015/2017
câu c 1 2015/2017 là hỗn số nhé
Tìm x biết
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
b) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
c) \(\frac{x-1}{2017}+\frac{x-2}{2016}=\frac{x-3}{2015}+\frac{x-4}{2014}\)
d) \(\frac{x+1}{2017}+\frac{x+2}{2016}=\frac{x+3}{2015}+\frac{x+4}{2014}\)
\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(100=2x+4\)
\(\Leftrightarrow\)\(2x=96\)
\(\Leftrightarrow\)\(48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(49=x+1\)
\(\Leftrightarrow\)\(x=48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
Tìm x,biết rằng:
a)1/5.8+1/8.11+1/11.14+...1/x(x+3)=101/1540
b)1+1/3+1/6+1/10+...+1/x(x+1):2=1 1991/1993
*1 1991/1993 là hỗn số nha
Giúp mik vs, cần gấp:
Tìm x ( "." là dấu nhân nhé)
a. x . (x + 1) - x - 1 = 0
b.(6 - x)2014 = ( 6 - x)2015
c.34x+4 = 9x + 2
d. 3 . ( 1/5.8 + 1/8.11 + 1/11.14 + ...+ 1/97.100 + x) = 319/100
e. 1/21 + 1/28 + 1/36 + ... + 2/x(x+1) = 2/9
Tìm x biết: a) 34x+4=9x+2
b) 3.(1/5.8 + 1/8.11 + 1/11.14 +........+ 1/97.100+x) =319/100
c) (6-x)2014=(6-x)2015
d) 2+4+6+...+2.x=210
e) 1/21+1/28+1/36+...+2/x.(x+1) = 2/9
Ta có : (6 - x)2014 = (6 - x)2015
=> (6 - x)2014 - (6 - x)2015 = 0
<=> (6 - x)2014(1 - 6 - x) = 0
<=> \(\orbr{\begin{cases}\left(6-x\right)^{2014}=0\\1-6-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}6-x=0\\-5-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=6\\x=-5\end{cases}}\)
sory bạn trừng hợp hai mk nhầm :
1 - (6 - x) = 0
=> 1 - 6 + x = 0
=> -5 + x = 0
=> x = 5
Ta có : 2 + 4 + 6 + ...... + 2x = 210
Nên : số số hạng là ; (2x - 2) : 2 + 1 = x - 1 + 1 = x (số)
Ta có : tổng là : (2x + 2) * x : 2 = 210
=> (2x2 + 2x) : 2 = 210
=> x2 + x = 210
=> x(x + 1) = 20.21
=> x = 20
tìm x biết
1/5.8+1/8.11+1/11.14+...+1/x(x+3)=101/15401+1/3+1/6+1/10+...+1/x(x+1):2=1 1991/1993tìm x :
a, 1/5.8+1/8.11+11/11.14+...+1/x.(x+3)=101/1540
b, 1+1/3+1/6+1/10+...+1/x.(x+1):2=1 và 1991/1993
Các bạn giải giùm mik cái nha mik gấp lắm rồi
1/Tính tổng
a,1+(-2)+(-3)+4+5+(-6)+(-7)+8+...+2013-2014-2016+2017
b.Tính
1+(-3)+(-7)+....+97+(-99)+101)
c.Tìm x thuộc Z biết
2017=2017+2016+2015+....+x
d.Tìm x thuộc Z biết
x+(x+1)+(x+2)+....+2016+2017=2017
D. Tìm x thuộc Z biết
x+(x+1)+(x+2)+....+2016+2017=2017
=> ( x + x + x + ..+ x ) + ( 1 + 2 + 3+...+2016 + 2017 ) = 2017
<=> 2017x + 2035153 = 2017
=> 2017x = -2033136
=> x = -1008
Vậy ...
cảm ơn bạn nhưng bạn có biết những câu hỏi còn lại ko
Tìm x biết
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
b) \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2017}{2019}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Leftrightarrow x=308-3\)
\(\Leftrightarrow x=305\)
Vậy \(x=305\)
Tìm x biết :
a) \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)(1)
Gọi B = \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\)
\(\Rightarrow3A=3\left(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\right)\)
\(\Rightarrow3A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}\)
\(\Rightarrow3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(\Rightarrow3A=\frac{1}{5}-\frac{1}{x+3}\)
\(\Rightarrow A=\left(\frac{1}{5}-\frac{1}{x+3}\right):3\)(2)
Thay (2) vào (1) , ta có : \(\left(\frac{1}{5}-\frac{1}{x+3}\right):3=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
Vậy x=305
Tương tự như câu b) nha
Chúc bạn học tốt!! nhớ mik nha
Tim x
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
2/ tim x
\(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7} +\frac{x+2018}{8}\)
3/ tim x
\(\frac{1}{3}+\frac{1}{6}+\frac{99}{101}+\frac{1}{15}+... +\frac{1}{x\left(2x+1\right)}=\frac{1}{10}\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
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