Question

tìm x

a)\(\left(x-\dfrac{1}{7}\right)^4=\left(x-\dfrac{1}{7}\right)^2\)

b)\(3^{x+2}+3^x=7290\)

Answer 1

b) \(3^x\cdot3^2+3^x=7290\)

\(3^x\left(3^2+1\right)=720\)

\(3^x\cdot10=7290\)

\(=>3^x=729=3^6\)

=> \(x=6\)

Answer 2

ôi bạn ơi

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Answer 3

a: Ta có: \(\left(x-\dfrac{1}{7}\right)^4=\left(x-\dfrac{1}{7}\right)^2\)

\(\Leftrightarrow\left(x-\dfrac{1}{7}\right)^2\left(x-\dfrac{1}{7}-1\right)\left(x-\dfrac{1}{7}+1\right)=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{7}\right)\left(x-\dfrac{8}{7}\right)\left(x+\dfrac{6}{7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=\dfrac{8}{7}\\x=-\dfrac{6}{7}\end{matrix}\right.\)