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\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x.\left(x+2\right)}=\dfrac{4}{9}\)
=\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x-2}\)
=\(\dfrac{1}{2}-\dfrac{1}{x+2}\)=\(\dfrac{4}{9}\)
=>\(\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}=\dfrac{1}{18}\)
=>\(x+2=18\)
=>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
\(S=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
\(S=\frac{4-2}{2.4}+\frac{6-2}{4.6}+...+\frac{\left(x+2\right)-x}{x\left(x+2\right)}=\frac{4}{9}\)
\(S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
\(S=\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2}-\frac{4}{9}=\frac{1}{18}\)
\(\Rightarrow x+2=18\Rightarrow x=18-2=16\)
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
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{x+2-2}{2\left(x+2\right)}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{x}{2x+4}=\dfrac{4}{9}\)
\(\Rightarrow9x=8x+16\)
\(\Rightarrow x=16\)
Vậy x = 16
2/2.4+2/4.6+......+2/x.(x+2)=4/9
1/2-1/4+1/4-1/6+....+1/x-1/x+2=4/9
1/2-1/x+2=4/9
1/x+2=1/2-4/9
1/x+2=1/18
=>x+2=18
x=18-2=16
\(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
<=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
<=> \(\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
<=> \(\frac{1}{x+2}=\frac{1}{18}\)
=> \(x+2=18\)
<=> \(x=16\)
Vậy...
bạn tk mình một lần cho mình biết đi mình chưa được ai tk lần nào
\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{x\left(x+2\right)}=\dfrac{2011}{2013}\)
\(\Leftrightarrow2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{2011}{2013}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x-2}=\dfrac{2011}{4026}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x-2}=\dfrac{2011}{4026}\)\(\Leftrightarrow\dfrac{1}{x-2}=\dfrac{1}{2013}\)
\(\Rightarrow x-2=2013\Rightarrow x=2015\)
\(a,\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\frac{11}{15}x=\frac{2}{5}\)
\(x=\frac{6}{11}\)
b,\(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
Vậy