Question

Tim so huu ti x, biet:

a/ \(5< 5^x< 625\)

b/ \(2^{x-1}=16\)

c/ \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)\(\left(x\in Z\right)\)

Answer 1

Giải:

a) \(5< 5^x < 625\)

\(\Leftrightarrow5< 5^x< 5^4\)

Vì \(5=5=5\)

Nên \(1< x< 4\)

\(\Leftrightarrow x\in\left\{2;3\right\}\)

Vậy ...

b) \(2^{x-1}=16\)

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

Vì \(2=2\)

Nên \(x-1=4\)

\(\Leftrightarrow x=4+1=5\)

Vậy ...

c) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)^{x+2}}{\left(x-1\right)^{x+2}}=\dfrac{\left(x-1\right)^{x+6}}{\left(x-1\right)^{x+2}}\)

\(\Leftrightarrow1=\left(x-1\right)^{x+4}\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=1+1=2\)

Vậy ...