Tìm x nguyên biết: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{16}{34}\)
tìm số nguyên x thỏa mãn: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.........+\frac{1}{x\left(x+2\right)}=\frac{16}{34}\)
Trả lời: x=
=> \(2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{16}{34}\)
=>\(2.\left(1-\frac{1}{x+2}\right)=\frac{16}{34}\)
=>\(1-\frac{1}{x+2}=\frac{4}{17}\)
=> \(\frac{1}{x+2}=\frac{13}{17}\)
=>\(x=-\frac{9}{13}\)
Tìm số tự nhiên x thoả mãn \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{16}{34}\)
Số tự nhiên x thỏa mãn \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..........+\frac{1}{x\left(x+2\right)}=\frac{16}{34}\) là ....................
tìm x nguyên biết
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{\left(2x-1\right)}-\frac{1}{\left(2x+1\right)}\)
\(2A=1-\frac{1}{2x+1}=\frac{2x}{2x+1}\)
\(A=\frac{x}{2x+1}\)
Mà \(A=\frac{49}{99}\) \(\Leftrightarrow\frac{x}{2x+1}=\frac{49}{99}\Leftrightarrow x=49\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right)\left(2x-1\right)}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}+\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2x+1}=\frac{49}{99}\)
\(\Rightarrow99x=49\left(2x+1\right)\)
\(\Rightarrow99x=98x+49\)
\(\Rightarrow x=49\)
Vậy : \(x=49\)
Tìm x thuộc N, biết:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{8}{17}\)
A\(A=\frac{1}{1.3}+..+\frac{1}{x\left(x+1\right)}\)
\(2A=\frac{1}{1}-\frac{1}{\left(x+1\right)}\)
\(A=\frac{x}{2.\left(x+1\right)}=\frac{8}{17}=\frac{16}{2.17}\)
X=16
Tìm x, biết:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
bạn nào giải giúp mình với
nếu đúng thì mình sẽ ***
=1/2*(1-1/3+1/3-1/5+....+1/x+1/x+2)
=1/2*(1-1/x+2)
=>1/2*x+1/x+2=20/21
Đến đó đưa về giống tìm x nha
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}\right)=\frac{20}{41}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{40}{41}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)
\(=1-\frac{1}{x+2}=\frac{40}{41}\)
\(\frac{1}{x+2}\)\(=1-\frac{40}{41}\)
\(\frac{1}{x+2}=\frac{1}{41}\)
=> x+2=41
x=41-2
x=39
Số tự nhiên thỏa mãn \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{16}{34}\) là
Kèm theo cách giải
ta nhân vế trái vs 2:
\(\frac{2}{1.3}+\frac{2}{3.5}+......+\frac{2}{x\left(x+2\right)}=\frac{8}{17}\)
\(\frac{1}{ }-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{8}{17}\)
\(1-\frac{1}{x+2}=\frac{8}{17}\)
\(\Rightarrow17\left(x+1\right)=8\left(x+2\right)\)
\(\Rightarrow17x+17=8x+16\)
\(\Rightarrow17x-8x=-17+16\)
\(\Rightarrow9x=-1\)
\(\Rightarrow x=\frac{-1}{9}\)
2(1/1.3+1/3.5+1/5.7+...+1/x(x+2) )=16/34 *2
2/1.3+2/3.5+2/5.7+...+2/x(x+2)=32/34=16/17
1/1-1/3+1/3-1/5+1/5-1/7+...+1/x-1/x+2=16/17
1/1-1/x+2=16/17
1/x+2=1/1-16/17
1/x+2=1/17
suy ra x+2=17
x=17=2=15
Tìm x
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow20\left(x+2\right)=41\)
\(\Leftrightarrow x-2=\frac{41}{20}\)
\(\Leftrightarrow x=\frac{41}{20}+2\)
\(\Leftrightarrow x=\frac{81}{20}\)
\(\frac{1}{1.3}+...+\frac{1}{a\left(a+2\right)}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{a\left(a+2\right)}\right)=\frac{1}{2}\left(1-\frac{1}{3}+....-\frac{1}{a+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{a+2}\right)=\frac{20}{41}\Rightarrow a+2=41\Leftrightarrow a=39\)
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Rightarrow x+2=41\)
\(\Rightarrow x=41-2\)
\(\Rightarrow x=39\)
Vậy x = 39
BÀI 1: tìm x biết : \(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)
BÀI 2: tìm số tự nhiên x thỏa mãn: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{16}{34}\)
BÀI 3: Cho x;y thỏa mãn : \(\left(x-2014\right)^{2010}+\left(y-2010\right)^{2014}\le0\)
bài 1
[(x+2)/1010]+ [(x+2)/1111]= [(x+2)/1212]+[(x+2)/1313]
=>[(x+2)/1010]+[(x+2)/1111] - [(x+2)/1212]-[(x+2)/1313] = 0
=>(x+2).[(1/1010)+(1/1111)-(1/1212)-(1/1313)=0
Vì [(1/1010)+(1/1111)-(1/1212)-(1/1313)] khác 0
=>x+2=0
=>x=-2
Bài 1 : -2
Bài 2 : 15
Bải 3 : x =2014 ; y = 2010