Question

Tìm x, y biết:

a) \(3^x=9^{y-1}\) với \(8^y=2^{x+8}\)

b) \(\left(2x-1\right)^2+\left(4y-1\right)^4=0\)

Answer 1

a) 3^x=9^y-1 => 3^x= 3^2(y-1) => x=2(y-1)=2y-2

8^y=2^x+8 => 2^3y=2^x+8 => 3y=x+8

3y-(2y-2)=x+8-x

y+2=8 => y=6

x+8=3y=3.6=18 => x=10

Answer 2

a) Ta có:

+) \(3^x=9^{y-1}\)

\(\Rightarrow3^x=3^{2.\left(y-1\right)}\)

\(\Rightarrow x=2.\left(y-1\right)\left(1\right)\)

+) \(8^y=2^{x+8}\)

\(\Rightarrow2^{3y}=2^{x+8}\)

\(\Rightarrow3y=x+8\left(2\right)\)

Thay (1) vào (2) ta được:

\(3y=\left[2.\left(y-1\right)\right]+8\)

\(\Rightarrow3y=2y-2+8\)

\(\Rightarrow3y=2y+6\)

\(\Rightarrow y=6\)

\(\Rightarrow x=2.\left(6-1\right)=10\)

Vậy \(x=10;y=6\)

Answer 3

b) (2x-1)²+(4y-1)⁴=0 => (2x-1)²=(4y-1)⁴=0

=> 2x-1=0 =>x=1/2

=>4y-1=0=>y=1/4