Tìm x, biết:
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{11}{48}\) (x ϵ N , x ≥ 2)
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{11}{48}\left(x\in N;x\ge12\right)\)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right)2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right)2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Rightarrow x=12\) (nh)
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{11}{48}\left(x\in N;x\ge12\right)\)
\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2x-2\right)2x}=\dfrac{11}{48}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2x-2\right)2x}\right)=\dfrac{11}{48}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}\right)=\dfrac{11}{48}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Leftrightarrow2x=24\Leftrightarrow x=12\) (thỏa mãn)
Tìm x
\(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) +...+ \(\dfrac{1}{\left(2x-2\right).2x}\) = \(\dfrac{1}{8}\) ( x ∈ N , x ≥ 2 )
có lời giải chi tiết
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)
\(\Leftrightarrow\dfrac{1}{4}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.......+\dfrac{1}{\left(x-1\right)x}\right)=\dfrac{1}{8}\) ( đk x khác 0 , x khác 1)
\(\Leftrightarrow\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{x-1}-\dfrac{1}{x}\right)=\dfrac{1}{8}\)
\(\Leftrightarrow1-\dfrac{1}{x}=\dfrac{1}{2}\)
=> x =2 ( tm)
Tìm x biết :
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right)2x}=\dfrac{1}{8}\left(x\in N,x\ge2\right)\)
mình ko biết mình làm đúng hay sai bạn nhé, mong mọi người góp ý
= 1/2.( 1/2.4+1/4.6+....+1/(2x-2)2x)=1/8
= 1/2.(1/2-1/4+1/4-1/6+....+1/(2x-2)-1/2x)=1/8
= 1/2.( 1/2-1/2x)=1/8
( 1/2-1/2x)=1/8:1/2
1/2-1/2x=1/4
1/2x =1/2-1/4
1/2x =1/4
2x = 4
x =4:2
x =2
tim x ϵ N* biết \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left[1+\dfrac{1}{x\left(x+2\right)}\right]=\dfrac{31}{16}\)
\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).........\left[1+\dfrac{1}{x.\left(x+2\right)}\right]=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}........\dfrac{\left(x+1\right)^2}{x.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left[2.3.4.............\left(x+1\right)\right].\left[2.3.4.............\left(x+1\right)\right]}{\left(1.2.3...................x\right).\left(3.4.5..........................\left(x+2\right)\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left(x+1\right).2}{1.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Leftrightarrow16.2\left(x+1\right)=31.\left(x+2\right)\)
\(\Rightarrow32x+32=31x+62\)
\(\Rightarrow x=30\)
Vậy x=30
Chúc bn học tốt
ĐKXĐ: \(x\notin\left\{0;-2\right\}\)
Ta có: \(\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{x\left(x+2\right)}\right)=\dfrac{31}{16}\)
\(\Leftrightarrow\dfrac{1\cdot3+1}{1\cdot3}+\dfrac{1+2\cdot4}{2\cdot4}+\dfrac{1+3\cdot5}{3\cdot5}\cdot...\cdot\dfrac{1+x\left(x+2\right)}{x\left(x+2\right)}=\dfrac{31}{16}\)
\(\Leftrightarrow\dfrac{2\cdot2}{1\cdot3}+\dfrac{3\cdot3}{2\cdot4}+\dfrac{4\cdot4}{3\cdot5}+...+\dfrac{\left(x+1\right)\left(x+1\right)}{x\left(x+2\right)}=\dfrac{31}{16}\)
\(\Leftrightarrow\dfrac{1\cdot2\cdot3\cdot...\cdot\left(x+1\right)}{1\cdot2\cdot3\cdot...\cdot x}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot\left(x+1\right)}{3\cdot4\cdot5\cdot...\cdot\left(x+2\right)}=\dfrac{31}{16}\)
\(\Leftrightarrow\left(x+1\right)\cdot\dfrac{2}{x+2}=\dfrac{31}{16}\)
\(\Leftrightarrow\dfrac{2x+2}{x+2}=\dfrac{31}{16}\)
\(\Leftrightarrow\dfrac{32x+32}{16\left(x+2\right)}=\dfrac{31\left(x+2\right)}{16\left(x+2\right)}\)
Suy ra: \(32x+32=31x+62\)
\(\Leftrightarrow x=30\)(thỏa ĐK)
Vậy: S={30}
\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+ ..........+ \(\dfrac{1}{\left(2x-2\right).2x}\)= \(\dfrac{1}{8}\)( x thuộc N ; x lớn hơn hoặc bằng 2)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right)2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right)2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-1}{2x}=\dfrac{-1}{4}\)
\(\Rightarrow x=2\)
Ta có: \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}\right)=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2x}\right)=\dfrac{1}{8}\Rightarrow\dfrac{1}{2}.\dfrac{x-1}{2x}=\dfrac{1}{8}\Rightarrow\dfrac{x-1}{4x}=\dfrac{1}{8}\)
\(\Rightarrow8\left(x-1\right)=4x\Rightarrow8x-8=4x\Rightarrow4x=8\Rightarrow x=2\)
\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+...+\(\dfrac{1}{x.\left(x+2\right)}\)=\(\dfrac{1}{10}\)
Gọi biểu thức trên là A
Ta có:
2A = (\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+...+\(\dfrac{1}{x.\left(x+2\right)}\)).2
2A = \(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+...+\(\dfrac{2}{x\left(x+2\right)}\)
2A = \(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+2}\)
2A = \(\dfrac{1}{2}\)-\(\dfrac{1}{x+2}\)
mà A = \(\dfrac{1}{10}\)(đề bài)
nên 2A = \(\dfrac{2}{10}\) hay \(\dfrac{1}{2}\) - \(\dfrac{1}{x+2}\) = \(\dfrac{2}{10}\)
suy ra \(\dfrac{1}{x+2}\) = \(\dfrac{1}{2}\)-\(\dfrac{2}{10}\)=\(\dfrac{3}{10}\)
tìm x:
\(x\left(x+2\right)=\dfrac{45}{x+4}\)
\(\dfrac{1}{x^2+2x-3}=\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{48}\)
tìm x
\(\dfrac{3-x}{5-x}=\dfrac{6}{11}\) \(\left(1\dfrac{1}{3}-25\%.x-\dfrac{5}{12}\right)-2x=1,6:\dfrac{3}{5}\)
\(\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)-\dfrac{1}{3}.\left(2x-3\right)=x\)
\(2.\left(\dfrac{1}{2}-x\right)-3\left(x-\dfrac{1}{3}\right)=\dfrac{7}{2}\)
a: =>11(x-3)=6(x-5)
=>11x-33=6x-30
=>5x=3
=>x=3/5
b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3
=>-9/4x+11/12=8/3
=>-9/4x=32/12-11/12=21/12=7/4
=>x=-7/9
c: =>1/2x-1/3-2/3x-1=x
=>-1/6x-4/3=x
=>-7/6x=4/3
=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7
d: =>1-2x-3x+1=7/2
=>-5x=3/2
=>x=-3/10