Tìm x:
2/2.4+2/4.6+......+2/x.(×+2)=4/9
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Giải
1/(2.4) + 1/(4.6) + … + 1/[(2x – 2).2x] = 1/8
=> 2/(2.4) + 2/(4.6) + ...+ 2/[(2x - 2).2x] = 2/8
=>1-1/4+1/4-1/6+...+1/(2x-2) - 1/2x = 2/8
=>1 - 1/2x = 2/8
=>1/2x = 1 - 2/8
=>1/2x = 6/8 = 3/4
=>1.4 = 2.x.3
=>4 = 6x
=> x thuộc rỗng
Vậy x thuộc rỗng
bạn tk mình một lần cho mình biết đi mình chưa được ai tk lần nào
\(\text{Sửa đề:}\)
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)
\(\text{Đặt biểu thức là A:}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}=\frac{1}{8}\times2=\frac{1}{4}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{4}\)
\(2A=\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(A=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{2x}\right)=\frac{1}{8}\)
\(A=\frac{1}{2}\times\frac{1}{2}-\frac{1}{2}\times\frac{1}{2x}=\frac{1}{8}\)
\(A=\frac{1}{4}-\frac{1}{4x}=\frac{1}{8}\)
\(\Rightarrow\frac{1}{4x}=\frac{1}{4}-\frac{1}{8}=\frac{2}{8}-\frac{1}{8}=\frac{1}{8}\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=8\div4=2\)
1/(2.4) + 1/(4.6) + … + 1/[(2x – 2).2x] = 1/8
suy ra 2/(2.4) + 2/(4.6) + ...+ 2/[(2x - 2).2x] = 2/8
suy ra 1-1/4+1/4-1/6+...+1/(2x-2) - 1/2x = 2/8
suy ra 1 - 1/2x = 2/8
suy ra 1/2x = 1 - 2/8
suy ra 1/2x = 6/8 = 3/4
suy ra 1.4 = 2.x.3
suy ra 4 = 6x
suy ra x thuộc rỗng
Vậy x thuộc rỗng
k cho mình nha. Chúc bạn học tốt!
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x.\left(x+2\right)}=\dfrac{4}{9}\)
=\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x-2}\)
=\(\dfrac{1}{2}-\dfrac{1}{x+2}\)=\(\dfrac{4}{9}\)
=>\(\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}=\dfrac{1}{18}\)
=>\(x+2=18\)
=>x=16
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{x+2-2}{2\left(x+2\right)}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{x}{2x+4}=\dfrac{4}{9}\)
\(\Rightarrow9x=8x+16\)
\(\Rightarrow x=16\)
Vậy x = 16
\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{x\left(x+2\right)}=\dfrac{2011}{2013}\)
\(\Leftrightarrow2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{2011}{2013}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x-2}=\dfrac{2011}{4026}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x-2}=\dfrac{2011}{4026}\)\(\Leftrightarrow\dfrac{1}{x-2}=\dfrac{1}{2013}\)
\(\Rightarrow x-2=2013\Rightarrow x=2015\)
\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)\cdot2x}=\frac{1}{8}\)
\(\Rightarrow\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{\left(2x-2\right)\cdot2x}=\frac{2}{8}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(S=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
\(S=\frac{4-2}{2.4}+\frac{6-2}{4.6}+...+\frac{\left(x+2\right)-x}{x\left(x+2\right)}=\frac{4}{9}\)
\(S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
\(S=\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2}-\frac{4}{9}=\frac{1}{18}\)
\(\Rightarrow x+2=18\Rightarrow x=18-2=16\)
Vậy x=16
\(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
\(\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
\(\frac{1}{x+2}=\frac{1}{2}-\frac{4}{9}\)
\(\frac{1}{x+2}=\frac{1}{18}\)
\(\Leftrightarrow x+2=18\)
=> x = 18 - 2
x = 16
Vậy x =16