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

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

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

=>\(\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}=\dfrac{1}{18}\)

=>\(x+2=18\)

=>x=16

\(S=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
\(S=\frac{4-2}{2.4}+\frac{6-2}{4.6}+...+\frac{\left(x+2\right)-x}{x\left(x+2\right)}=\frac{4}{9}\)
\(S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
\(S=\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2}-\frac{4}{9}=\frac{1}{18}\)
\(\Rightarrow x+2=18\Rightarrow x=18-2=16\)
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

bạn tk mình một lần cho mình biết đi mình chưa được ai tk lần nào

Nhưng bn phải trả lời chứ

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

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

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

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

=>x+2=18

x=18-2

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

bài này mà ko biết á????

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

\(a,\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\frac{11}{15}x=\frac{2}{5}\)

\(x=\frac{6}{11}\)

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

\(\left(3x-1\right).\left(-\frac{1}{2}x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\-\frac{1}{2}x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}}\)\

Vậy

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

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

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

<=> \(\frac{1}{x+2}=\frac{1}{18}\)

=>  \(x+2=18\)

<=>  \(x=16\)

Vậy...

a) \(\left(x+1\right)^2=64\Leftrightarrow\left|x+1\right|=8\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=8\\x+1=-8\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=7\\x=-9\end{array}\right.\)

b) \(\frac{9}{2.4}+\frac{9}{4.6}+...+\frac{9}{96.98}+\frac{9}{98.100}=\frac{9}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{48.49}+\frac{1}{49.50}\right)\)

\(=\frac{9}{4}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{9}{4}\left(1-\frac{1}{50}\right)=\frac{441}{200}\)

a)(x+1)²=64

x + 1      = 8

x            = 8 - 1

x            = 7

a)(x+1)²=64

=>(x+1)²=8²

=>x+1=8

=>x=7

b)\(\frac{9}{2.4}+\frac{9}{4.6}+...+\frac{9}{98.100}\)

\(=\frac{9}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

\(=\frac{9}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{9}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{9}{2}\cdot\frac{49}{100}\)

\(=\frac{441}{200}\)