\(2^{2\cdot25}=2^{50}\)   

\(3^{150}=\left(3^3\right)^{50}=27^{50}\)   

\(2^{50}< 27^{50}\)   

\(2^{2\cdot25}< 3^{150}\)

Ta có: \(x=-2y\)

\(\Leftrightarrow\frac{x}{-2}=\frac{y}{1}\)

Áp dụng t/c dãy tỉ số bằng nhau ta được:

\(\frac{x}{-2}=\frac{y}{1}=\frac{x+y}{-2+1}=\frac{10}{-1}=-10\)

\(\Rightarrow\hept{\begin{cases}x=20\\y=-10\end{cases}}\)

Cách khác : 

Thay x = -2y vào x + y = 10 ta có : \(-2y+y=10\Leftrightarrow-y=10\Leftrightarrow y=-10\)

Thay y = -10 ta có : \(x-10=10\Leftrightarrow x=20\)

a, Ta có : \(\frac{x}{y}=\frac{1}{3}\Leftrightarrow\frac{x}{1}=\frac{y}{3}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có : 

\(\frac{x}{1}=\frac{y}{3}=\frac{x-3y}{1-3.3}=\frac{1}{\frac{2}{-8}}=-\frac{1}{16}\)

\(x=-\frac{1}{16};y=-\frac{3}{16}\)

Ta có :\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)

=> \(\left(\frac{x+4}{2018}+1\right)+\left(\frac{x+3}{2019}+1\right)=\left(\frac{x+2}{2020}+1\right)+\left(\frac{x+1}{2021}+1\right)\)

=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)

=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)

=> \(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)

Vì \(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\)

=> x + 2022 = 0

=> x = -2022

Vậy x = -2022

\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)  

\(\frac{x+4}{2018}+1+\frac{x+3}{2019}+1=\frac{x+2}{2020}+1+\frac{x+1}{2021}+1\) 

\(\frac{x+4}{2018}+\frac{2018}{2018}+\frac{x+3}{2019}+\frac{2019}{2019}=\frac{x+2}{2020}+\frac{2020}{2020}+\frac{x+1}{2021}+\frac{2021}{2021}\)   

\(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)   

\(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)   

\(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)   

\(x+2022=0\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)   

\(x=0-2022\) 

\(x=-2022\)

a) \(A=0,5-\left|x-3,5\right|\le0,5\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|x-3,5\right|=0\Rightarrow x=3,5\)

Vậy Max(A) = 0,5 khi x = 3,5

b) \(C=1,7+\left|3,4-x\right|\ge1,7\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|3,4-x\right|=0\Rightarrow x=3,4\)

Vậy Min(C) = 1,7 khi x = 3,4

Ta có: \(\frac{2x-2}{3}=\frac{7x+3}{2-1}\)

\(\Leftrightarrow\frac{2x-2}{3}=7x+3\)

\(\Leftrightarrow2x-2=21x+9\)

\(\Leftrightarrow19x=-11\)

\(\Rightarrow x=-\frac{11}{19}\)

\(\frac{2x-2}{3}=\frac{7x+3}{2-1}\Leftrightarrow\frac{2x-2}{3}=7x+3\)

\(\Leftrightarrow2x-2=21x+9\Leftrightarrow-19x=11\Leftrightarrow x=-\frac{11}{19}\)

Ta có: \(\frac{x+1}{2019}+\frac{x+1}{2020}=\frac{x+1}{2021}\)

\(\Leftrightarrow\frac{x+1}{2019}+\frac{x+1}{2020}-\frac{x+1}{2021}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}\right)=0\)

Mà \(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}>0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

\(\frac{x+1}{2019}+\frac{x+1}{2020}=\frac{x+1}{2021}\)   

\(\frac{x+1}{2019}+\frac{x+1}{2020}-\frac{x+1}{2021}=0\)   

\(\left(x+1\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}\right)=0\)   

\(x+1=0\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)   

\(x=0-1=-1\)

Ta có: \(\left(\frac{2}{3}-x\right)^2=\left(-2x+\frac{1}{3}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}-x=-2x+\frac{1}{3}\\\frac{2}{3}-x=2x-\frac{1}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\3x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}\)