Question

\(tìmxbiết\dfrac{2}{2.4}+\dfrac{2}{4.6}+.....+\dfrac{2}{x.\left[x+2\right]}=\dfrac{4}{9}\)

Answer 1

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{4}{9}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)

\(\Rightarrow\dfrac{x+2-2}{2\left(x+2\right)}=\dfrac{4}{9}\)

\(\Rightarrow\dfrac{x}{2x+4}=\dfrac{4}{9}\)

\(\Rightarrow9x=8x+16\)

\(\Rightarrow x=16\)

Vậy x = 16

Answer 2

bài này mà ko biết á????

Answer 3

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

Answer 4

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x.\left(x+2\right)}=\dfrac{4}{9}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)

\(=\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)

\(=\dfrac{\left(x+2\right)-2}{2.\left(x+2\right)}=\dfrac{4}{9}\)

\(=\dfrac{x}{2x+4}=\dfrac{4}{9}\Rightarrow\left\{{}\begin{matrix}9x=8x+16\\\Rightarrow x=16\end{matrix}\right.\)

Vậy \(x=16\)

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