em ko biết

2x=-7y

=>\(\frac{-2}{7}\)=\(\frac{x}{y}\)

=>\(\frac{-2}{x}\)=\(\frac{7}{y}\)

\(\frac{-2}{x}\)=\(\frac{7}{y}\)=>1.\(\frac{-2}{x}\)=\(\frac{7}{y}\)=\(\frac{2}{2}\).\(\frac{-2}{x}\)=\(\frac{7}{y}\)=\(\frac{-4}{2x}\)=\(\frac{7}{y}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có :

\(\frac{-4}{2x}\)=\(\frac{7}{y}\)=\(\frac{\left(-4\right)+7}{2x+y}\)=\(\frac{3}{3}\)=1

\(\frac{-2}{x}\)=1 => x = -2 : 1 =>x = -2

\(\frac{7}{y}\)= 1 => y = 7 : 1 => y = 7

vậy x= -2 ; y = 7

\(\frac{x}{5-x}=\frac{7+x}{3-x}ĐK:x\ne5;3\)

\(\Leftrightarrow\frac{3x-x^2}{\left(5-x\right)\left(3-x\right)}=\frac{\left(5-x\right)\left(7+x\right)}{\left(5-x\right)\left(3-x\right)}\)

Khử mẫu : \(3x-x^2=\left(5-x\right)\left(7+x\right)\)

\(\Leftrightarrow3x-x^2=35+5x-7x-x^2\)

\(\Leftrightarrow5x-35=0\Leftrightarrow x=7\)

3^400                             4^300

=3^4.100                       =4^3.100

=3^4                              =4^3

=48                                =38

=> 3^400>4^300

\(3^{400}=\left(3^4\right)^{100}=81^{100}\)   

\(4^{300}=\left(4^3\right)^{100}=64^{100}\)   

Vì \(81>64\) 

Nên \(81^{100}>64^{100}\)   

Vậy \(3^{400}>4^{300}\)