Question

Tìm x ϵ N biết:

a)\(3^x-3^{x+3}=-234.\) c)\(2^{2x+1}+4^{x+3}=264.\) d)\(9^x-3^x=702.\)

b)\(2^{x+1}.3^x-6^x=216.\)

Answer 1

a)Ta có:

\(3^x-3^{x-3}=-234\)

\(\Rightarrow3^x-3^x\cdot3^3=-234\)

\(\Rightarrow3^x\cdot\left(1-3^3\right)=-234\)

\(\Rightarrow3^x\cdot\left(-26\right)=-234\)

\(\Rightarrow3^x=9\)

\(\Rightarrow x=2\)

Vậy x=2

\(\Rightarrow3^x=3^2\)

Answer 2

b) Ta có:

\(2^{x+1}\cdot3^x-6^x=216\)

\(\Rightarrow2^x\cdot2\cdot3^x-2^x\cdot3^x=216\)

\(\Rightarrow\left(2^x\cdot3^x\right)\cdot\left(2-1\right)=216\)

\(\Rightarrow6^x\cdot1=216\)

\(\Rightarrow6^x=6^3\)

\(\Rightarrow x=3\)

Vậy x=3

Answer 3

c) Ta có:

\(2^{2x+1}+4^{x+3}=264\)

\(\Rightarrow2^{2x+1}+\left(2^2\right)^{x+3}=264\)

\(\Rightarrow2^{2x+1}+2^{2x+6}=2^{264}\)

\(\Rightarrow2^{2x+1}+2^{2x+1+5}=264\)

\(\Rightarrow2^{2x+1}+2^{2x+1}\cdot2^5=264\)

\(\Rightarrow2^{2x+1}\cdot\left(1+2^5\right)=264\)

\(\Rightarrow2^{2x+1}\cdot26=264\)

\(\Rightarrow2^{2x+1}=\frac{132}{13}\)

\(\Rightarrow2^{2x+1}\notin N\)

\(\Rightarrow2x+1\notin N\)

\(\Rightarrow x\notin N\)

\(\Rightarrow\) Không có giá trị x thỏa mãn.

Vậy không có giá trị x thỏa mãn.