\(\frac{1}{5}-\frac{1}{4}+\frac{3}{20}=\frac{4}{20}-\frac{5}{20}+\frac{3}{20}\)

                                \(=\frac{1}{10}\)

Ta có

  \(\frac{a}{b}=\frac{1}{10}\Rightarrow a.10=b\)

Thay \(b=10a\)vào \(\left(a.b\right)^2\), ta có

  \(\left(a.b\right)^2=a^2.b^2=a^2.10.a=10a^3\)

vì xOy và yOt bù nhau

\(\Rightarrow\widehat{xOy}+\widehat{yOt}=180^o\left(1\right)\)

ta có \(\widehat{xOy}=\widehat{yOt}+50^o\left(2\right)\)(giả thiết)

thay (2) vào (1) 

\(\Leftrightarrow\widehat{yOt}+50^o+\widehat{yOt}=180^o\)

\(\Leftrightarrow50^o+2\widehat{yOt}=180^o\)

\(\Leftrightarrow2\widehat{yOt}=130^o\Leftrightarrow\widehat{yOt}=130^o:2=65^o\)

vậy ^yOt = 65

Theo hình vẽ ta có

  AM + MN + BN = AB

hay MN =  4 - ( 2 + 1 )

             =  1   ( cm )

   Vậy MN = 1 cm

a>b>0

\(\Rightarrow\frac{a}{b}>\frac{b}{b}=1\)

8 hình

\(\overline{9abcd}-\overline{abcd7}=90000+\overline{abcd}-\overline{abcd0}-7=51104\)

                                          \(89993+\overline{abcd}\left(1-10\right)=51104\)

                                          \(89993-51104=-\overline{abcd}\left(-9\right)\)

                                           \(38889=9\overline{abcd}\)

                                            \(\overline{abcd}=4321\)

  Vậy \(\overline{abcd}=4321\)

Vì \(\left|x-1\right|\ge0\forall x;\left|y+2\right|\ge0\forall y\)

\(\Rightarrow\left|x-1\right|+\left|y-2\right|+2020\ge2020\)

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=0\\\left|y+2\right|=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\y=-2\end{cases}}\)

Vậy minA = 2020 <=> x = 1 ; y = - 2

\(A=\left|x-1\right|+\left|y+2\right|+2020\)

+ Ta có:

\(\hept{\begin{cases}\left|x-1\right|\ge0\text{∀}x\\\left|y+2\right|\ge0\text{∀}y\end{cases}}\)

\(\Rightarrow\left|x-1\right|+\left|y+2\right|\ge0\text{∀}x,y.\)

\(\Rightarrow\left|x-1\right|+\left|y+2\right|+2020\ge0+2020\)

\(\Rightarrow\left|x-1\right|+\left|y+2\right|+2020\ge2020\text{∀}x,y\)

\(\Leftrightarrow A\ge2020.\)

Dấu '' = '' xảy ra khi và chỉ khi:

\(\hept{\begin{cases}\left|x-1\right|=0\\\left|y+2\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0+1\\y=0-2\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

Vậy \(MIN_A=2020\) khi và chỉ khi \(x=1\) và \(y=-2.\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)

\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{50}\)

\(\Rightarrow1-\frac{1}{n+1}=\frac{49}{50}\)

\(\Rightarrow\frac{1}{n+1}=\frac{1}{50}\)

\(\Rightarrow n+1=50\)

\(\Rightarrow n=49\)

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)

\(\Rightarrow\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{50}{51}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2n+1}=\frac{50}{51}\)

\(\Rightarrow\frac{1}{2n+1}=\frac{1}{51}\)

\(\Rightarrow2n+1=51\)

\(\Rightarrow2n=50\)

\(\Rightarrow n=25\)