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 \(x=\dfrac{4}{15}\)

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

\(=\lim\limits\dfrac{n^2+an+2020-n^2}{\sqrt{n^2+an+2020}+n}+\lim\limits\dfrac{n^3-bn^3-6n^2-3n-2021}{n^2+\sqrt[3]{\left(bn^3+6n^2+3n+2021\right)^2}+n\sqrt[3]{bn^3+6n^2+3n+2021}}\)

\(=\lim\limits\dfrac{\dfrac{an}{n}+\dfrac{2020}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{an}{n^2}+\dfrac{2020}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{\left(1-b\right)n^3}{n^2}-\dfrac{6n^2}{n^2}-\dfrac{3n}{n^2}-\dfrac{2021}{n^2}}{\dfrac{n^2}{n^2}+\dfrac{\sqrt[3]{\left(bn^3+6n^2+3n+2021\right)^2}}{n^2}+\dfrac{n\sqrt[3]{bn^3+6n^2+3n+2021}}{n^2}}\)

\(=\dfrac{1}{2}a+\lim\limits\dfrac{\left(1-b\right)n-6}{1+\sqrt[3]{b^2}+\sqrt[3]{b}}\)

De gioi han bang 0 thi \(\left(1-b\right)=0\Leftrightarrow b=1\Rightarrow\lim\limits\dfrac{\left(1-b\right)n-6}{1+\sqrt[3]{b^2}+\sqrt[3]{b}}=-\dfrac{6}{3}=-2\)

\(\Rightarrow\dfrac{1}{2}a-2=0\Leftrightarrow a=4\)

\(\Rightarrow P=4^{2020}+2^{2021}-1\)

P/s: Tổng này hỏi có bao nhiêu chữ số thì tui còn tìm được, chứ viết hẳn ra thì..chắc nhờ siêu máy tính của nasa :v

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

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

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

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

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

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

205 : 5 + 1=

=41 + 1

=42

205:5+1=42

 tự hỏi tự trả lời à

A=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}.\frac{2015}{2016}\)

A=\(\frac{1.2.3.4...2015}{2.3.4...2016}=\frac{1}{2016}\)

Hok tốt

A = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2015}\right).\left(1-\frac{1}{2016}\right)\)

= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}.\frac{2015}{2016}\)

= \(\frac{1}{2016}\)

Vậy ...

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2015}\right)\left(1-\frac{1}{2016}\right)\)

\(\Rightarrow A=\frac{1.2.3.4....2015}{2.3.4....2016}\)

\(\Rightarrow A=\frac{1}{2016}\)

#hok tốt

ket qua=5

bn k nha 

5^1=5

5^1 = 5