a) Từ đề bài suy ra

2^x+1.3^y=(3.2^2)^x

2^x+1.3^y=3^x.(2^2)^x.Vì cách phân tích là duy nhất.

2^x+1=2^2x và 3^y=3^x

x+1=2x;y=x

x=y=1

b) 10^x:5^y=20^y

10^x =20^y.5^y

10^x = (20.5)^y

10^x = 100^y

10^x = 10^2y

x = 2y

Vậy x= 2y

a) Ta có: \(2^{x-1}\cdot3^{y+1}=12^{x+y}\)

\(\Leftrightarrow2^{x-1}\cdot3^{y+1}=4^{x+y}\cdot3^{x+y}\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=2x+2y\\y+1=x+y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1-1=2\cdot1+2y\\x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2+2y=0\\x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y=-2\\x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)

a) x = y = 1

b) x = 2; y = 1

c) 

a)x=1; y=1

a: \(=\left(-1\right)^{10}+\left(-1\right)^9+\left(-1\right)^8+...+\left(-1\right)^2+\left(-1\right)\)

\(=\left(1-1\right)+\left(1-1\right)+...+\left(1-1\right)\)

=0

b: \(=\left(-1\right)^{100}+\left(-1\right)^{99}+...+\left(-1\right)^2+\left(-1\right)\)

\(=\left(1-1\right)+...+\left(1-1\right)\)

=0

c: \(=1^{100}-1^{99}+1^{98}-1^{97}+...+1^2-1\)

=0

f: \(=3\cdot\sqrt{9-5}+7=3\cdot2+7=13\)

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

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

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

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

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

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