\(5S=5+1+\frac{1}{5^2}+...+\)\(\frac{1}{5^{2019}}\)

\(5S-S=4S=5-\frac{1}{5^{2019}}\)

\(\Rightarrow S=\frac{\frac{5^{2020}}{5^{2019}}}{4}\)

\(S=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2020}}\)

\(5S=5\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2020}}\right)\)

\(5S=5+1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}\)

\(5S-S=4S\)

\(=5+1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}-\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2020}}\right)\)

\(=5+1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}-1-\frac{1}{5}-\frac{1}{5^2}-\frac{1}{5^3}-...-\frac{1}{5^{2020}}\)

\(=5-\frac{1}{5^{2020}}\)

\(4S=5-\frac{1}{5^{2020}}\Rightarrow S=\frac{5-\frac{1}{5^{2020}}}{4}\)

4⁵ . 9⁴ - 2 . 6⁹/2¹⁰ . 3⁸ + 6⁸ . 20\(=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)

Có thể

KHÔNG THỂ

TH1:\(x\ge-2\Rightarrow\left|x+2\right|=x+2;\left|x+5\right|=x+5\)

\(\Rightarrow x+2+x+5=x\)

\(\Rightarrow x=-7\text{(loại vì ko thỏa mãn đk x\ge2 đang xét)}\)

TH2:\(-5\le x< -2\Rightarrow\hept{\begin{cases}\left|x+2\right|=-\left(x+2\right)=-x-2\\\left|x+5\right|=x+5\end{cases}}\)

\(\Rightarrow-x-2+x+5=x\)

\(\Rightarrow3=x\text{ hay }x=3\left(\text{loại}\right)\)

TH3:\(x< -5\Rightarrow\hept{\begin{cases}\left|x+2\right|=-x-2\\\left|x+5\right|=-x-5\end{cases}}\)

\(\Rightarrow-x-2-x-5=x\)

\(\Rightarrow-3x=7\)

\(\Rightarrow x=\frac{-7}{3}\left(\text{loại}\right)\)

\(\text{Vậy phương trình vô nghiệm}.\)

Bài làm ;

Ta có :

\(B=2020-|x-1|\le2020\) ( do \(-|x-1|\le0\) )

Dấu "=" xảy ra khi \(\Leftrightarrow|x-1|=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

Vậy GTLN của B là 2020 khi x = -3 .

Học tốt nhé

Ta có 

\(|x-1|\ge0\forall x\) 

\(2020-|x-1|\le2020\) 

Dấu = xảy ra 

\(\Leftrightarrow x-1=0\) 

\(x=0+1\) 

\(x=1\) 

Vậy GTLN của B là \(2020\Leftrightarrow x=1\)

vì tổng 2 góc kề bù=180 độ
tia phân giác cắt góc kề bù thành 2 nửa bằng nhau nên ta có:

180:2=90 độ

vậy tia phân giác của 2 góc kề bù vuông góc với nhau
(học tốt em nhé!)

\(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.......+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)

\(\Leftrightarrow\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+........+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)

\(\Leftrightarrow2.\left[\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+.......+\frac{1}{x\left(x+1\right)}\right]=\frac{11}{40}\)

\(\Leftrightarrow\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+......+\frac{1}{x\left(x+1\right)}=\frac{11}{80}\)

\(\Leftrightarrow\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+.......+\frac{1}{x\left(x+1\right)}=\frac{11}{80}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+.......+\frac{1}{x}-\frac{1}{x+1}=\frac{11}{80}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{11}{80}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{16}\)

\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\)

Vậy \(x=15\)

giup toi

\(\frac{3}{4}x-14\frac{2}{3}:\left(\frac{11}{15}+\frac{1111}{3535}+\frac{111111}{636363}\right)=12\)

\(\frac{3}{4}x-\frac{44}{3}:\left(\frac{11}{15}+\frac{11}{35}+\frac{11}{63}\right)=12\)

\(\frac{3}{4}x-\frac{44}{3}:\frac{11}{9}=12\)

\(\frac{3}{4}x-12=12\)

\(\frac{3}{4}x=12+12\)

\(\frac{3}{4}x=24\)

\(x=24:\frac{3}{4}\)

\(x=32\)

vậy \(x=32\)

\(3x-\frac{15}{5\cdot8}-\frac{15}{8\cdot11}-\frac{15}{11\cdot14}-...-\frac{15}{47\cdot50}=2\frac{1}{10}\)

<=> \(3x-5\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{47\cdot50}\right)=\frac{21}{10}\)

<=> \(3x-5\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{47}-\frac{1}{50}\right)=\frac{21}{10}\)

<=> \(3x-5\left(\frac{1}{5}-\frac{1}{50}\right)=\frac{21}{10}\)

<=> \(3x-5\cdot\frac{9}{50}=\frac{21}{10}\)

<=> \(3x-\frac{9}{10}=\frac{21}{10}\)

<=> \(3x=3\)

<=> \(x=1\)