Sửa đề: \(C=\dfrac{17^{99}+1}{17^{99}-1}\)

\(C=\dfrac{17^{99}-1+2}{17^{99}-1}=1+\dfrac{2}{17^{99}-1}\)

\(D=\dfrac{17^{98}-1+2}{17^{98}-1}=1+\dfrac{2}{17^{98}-1}\)

17^99>17^98

=>17^99-1>17^98-1

=>C<D

\(\left|x+1\right|+\left|x-5\right|=3x+1\left(đk:x\ge-\dfrac{1}{3}\right)\)

\(\Leftrightarrow x+1+\left|x-5\right|=3x+1\)

\(\Leftrightarrow\left|x-5\right|=2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=2x\left(x\ge5\right)\\x-5=-2x\left(-\dfrac{1}{3}\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\left(ktm\right)\\x=\dfrac{5}{3}\left(tm\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-x-1+5-x=3x+1\left(x< -1\right)\\x+1+5-x=3x+1\left(-1\le x< 5\right)\\x+1+x-5=3x+1\left(x\ge5\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(ktm\right)\\x=\dfrac{5}{3}\left(tm\right)\\x=-5\left(ktm\right)\end{matrix}\right.\Rightarrow x=\dfrac{5}{3}\)

0,44 x (\(x+x\times\) 5 - \(\dfrac{23}{55}\)) + \(\dfrac{3}{14}\) x 2,24 = 1

 0,44 x (\(x+x\times\)5 - \(\dfrac{23}{55}\)) + 0,48 = 1

 0,44 x (\(x+x\times\) 5 - \(\dfrac{23}{55}\)) = 1 - 0,48

0,44 x (\(x+x\times5\) - \(\dfrac{23}{55}\)) = 0,52

           \(x+x\times5\) - \(\dfrac{23}{55}\) = 0,52 : 0,44

          \(x\) x (1 + 5) -\(\dfrac{23}{55}\)  = \(\dfrac{13}{11}\)

         \(x\) x 6 - \(\dfrac{66}{10}\) = \(\dfrac{13}{11}\) + \(\dfrac{23}{55}\)

         \(x\) x 6 = \(\dfrac{65}{55}\) + \(\dfrac{23}{55}\)

          \(x\times\) 6 = \(\dfrac{8}{5}\)

         \(x\) x 6 = \(\dfrac{8}{5}\)  : 6

          \(x\)      = \(\dfrac{4}{15}\)

Vậy \(x=\dfrac{4}{15}\)

Lời giải:
$2\times A=\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...+\frac{2}{19\times 21}$
$2\times A=\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+...+\frac{21-19}{19\times 21}$

$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{19}-\frac{1}{21}$

$=1-\frac{1}{21}=\frac{20}{21}$

$\Rightarrow A=\frac{20}{21}: 2= \frac{10}{21}$

=2

=4

kb nha

1+1 =2 

2+2=4

1 + 1 = 2

 2 + 2 = 4