Question

\(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+..............+\frac{2}{x\cdot\left(x+2\right)}=\frac{4}{9}\)

Answer 1

ĐKXĐ:  \(x\ne0;x\ne-2\)

\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)

\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)

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

\(\Leftrightarrow\)\(\frac{1}{x+2}=\frac{1}{18}\)

\(\Rightarrow\)\(x+2=18\)

\(\Leftrightarrow\)\(x=16\) (t/m ĐKXĐ)

Vậy...

Answer 2

1/2(1-1/4+1/4-1/6+1/6-1/8+...+1/x-1/x+2)=4/9

1/2(1-1/x+2)=4/9

1-  1/x+2=4/9:1/2

1  -  1  /x+2=8/9

1/x+2=1-8/9

1/x+2=1/9

suy ra x+2=9

        x=9-2

        x=7

Answer 3

\(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+................+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\Rightarrow\frac{x+2-2}{2x+4}=\frac{4}{9}\Rightarrow9x=8x+16\)

\(\Rightarrow x=16\)