\(2^{x+1}.3^y=12^x\)

\(\Rightarrow2^{x+1}.3^y=3^x.4^x\)

\(\Rightarrow2^{x+1}.3^y=3^x.2^{2x}\)

\(\Rightarrow\orbr{\begin{cases}2^{x+1}=2^{2x}\\3^y=3^x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+1=2x\\y=x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\\text{Vì y = x}\Rightarrow y=1\end{cases}}\)

a) \(\Leftrightarrow2^{x+1}.3^y=2^{2x}.3^x\)\(\Leftrightarrow\hept{\begin{cases}x+1=2x\\y=x\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=x=1\end{cases}}}\)

b) \(\Leftrightarrow2^x.5^{x-y}=2^{2y}.5^y\Leftrightarrow\hept{\begin{cases}x=2y\\x-y=y\end{cases}\Leftrightarrow x=2y}\)

c) \(2^x=4^{y-1}\Leftrightarrow2^x=2^{2y-2}\Leftrightarrow x=2y-2\)

\(27^y=3^{x+8}\Leftrightarrow3^{3y}=3^{x+8}\Leftrightarrow3y=x+8\)

\(\Rightarrow\hept{\begin{cases}x=10\\y=6\end{cases}}\)

nhầm, sorry:

a) 10^x : 5^y = 20^y

=> 10^x  = 20^y . 5^y

=> 10^x = 100^y

=> 10^x = (10²)^y

=> x = 2y

x = 2y nha bn

a) x = y = 1

b) x = 2; y = 1

c) 

a)x=1; y=1

a ) 2^x+1 . 3^y = 12^x

=>2^x+1*3^y=(3*2²)^x

=>2^x+1*3^y=3^x*2^2x

=>2^x+1=2^2x và 3^x=3^y

=>x+1=2x và x=y

=>x=1 và x=y 

=>x=y=1

c)2^x=4^y-1 và 27^y=3^x+8

=>2^x=(2²)^y-1 và (3³)^y=3^x+8

=>2^x=2^2y-1 và 3^3y=3^x+8

=>x=2y-1 và 3y=x+8

Thay x=2y-1 vào 3y=x+8 ta có:

3y=2y-1+8 =>3y=2y+7

=>y=7 =>x=2*7-1=13

Vậy y=7 và x=13

a) 2x+1⋅3y=12x⇔2x+1⋅3y=22x⋅3x2x+1⋅3y=12x⇔2x+1⋅3y=22x⋅3x

⇒{x+1=2xy=x⇔{x=1y=1⇒{x+1=2xy=x⇔{x=1y=1

b) 10x:5y=20y⇔20y⋅5y=10x⇔(20⋅5)y=10x⇔100y=10x⇔102y=10x⇔2y=x10x:5y=20y⇔20y⋅5y=10x⇔(20⋅5)y=10x⇔100y=10x⇔102y=10x⇔2y=x

c) {2x=4y−127y=3x+8⇔{2x=22y−233y=3x+8⇔{x=2y−23y=x+8{2x=4y−127y=3x+8⇔{2x=22y−233y=3x+8⇔{x=2y−23y=x+8

⇔{x=2y−23y=2y−2+8⇔{x=10y=6

a, \(2^{x+1}.3^y=12^x\Rightarrow2^{x+1}.3^y=3^x.4^x\Rightarrow2^{x+1}.3^y=2^{2x}.3^x\)

=> x + 1 = 2x  ; y = x

=> x = 1 ; y = x = 1

b, \(10^x:5^y=20^y\Rightarrow2^x.5^x:5^y=4^y.5^y\Rightarrow2^x.5^{x-y}=2^{2y}.5^y\)

=> x = 2y ; x- y  = y => x = 2y 

VẬy mọi số tự nhiên x,y đều thỏa mãn miễn x = 2y ( thử xem)

c, \(2^x=4^{y-1}\Rightarrow2^x=2^{2\left(y-1\right)}\Rightarrow x=2\left(y-1\right)\Rightarrow x=2y-2\)

\(27^y=3^{x+8}\Rightarrow3^{3y}=3^{x+8}\Rightarrow3y=x+8\Rightarrow3y=2y-2+6\)

=> 2y + 4 = 3y => y = 4 ; 

x = 2.4 - 2 = 6 

tặng thang tran 3 **** về sự cần cù

a) \(2^{x+1}\cdot3^y=12^x\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=2x\\y=x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

b) \(10^x:5^y=20^y\Leftrightarrow20^y\cdot5^y=10^x\Leftrightarrow\left(20\cdot5\right)^y=10^x\Leftrightarrow100^y=10^x\Leftrightarrow10^{2y}=10^x\Leftrightarrow2y=x\)

c) \(\left\{{}\begin{matrix}2^x=4^{y-1}\\27^y=3^{x+8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2^x=2^{2y-2}\\3^{3y}=3^{x+8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-2\\3y=x+8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-2\\3y=2y-2+8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=6\end{matrix}\right.\)