Question

Bài 2. Tìm x, biết:

j) \(\dfrac{2}{9}\) - \(\sqrt{\left(x+1\right)^2}\) = \(\dfrac{-23}{9}\)

k) ( \(\dfrac{-3}{5}\) )^x = ( 3 . \(\sqrt{\dfrac{1}{25}}\) )2

Answer 1

\(a,\dfrac{2}{9}-\sqrt{\left(x+1\right)^2}=\dfrac{-23}{9}\)
\(\sqrt{\left(x+1\right)^2}=\dfrac{2}{9}-\dfrac{-23}{9}\)
\(\sqrt{\left(x+1\right)^2}=\dfrac{25}{9}\)
\(\left|x+1\right|=\dfrac{25}{9}\)
\(x+1=\dfrac{25}{9}\) hoặc \(x+1=\dfrac{-25}{9}\)
\(x=\dfrac{25}{9}-1\)          \(x=\dfrac{-25}{9}-1\)
\(x=\dfrac{16}{9}\)                \(x=\dfrac{-34}{9}\)
Vậy ...
\(b,\left(\dfrac{-3}{5}\right)^x=\left(3.\sqrt{\dfrac{1}{25}}\right)^2\)
\(\left(\dfrac{-3}{5}\right)^x=\left(3.\dfrac{1}{5}\right)^2\)
\(\left(\dfrac{-3}{5}\right)^x=\left(\dfrac{3}{5}\right)^2\)
\(\left(\dfrac{-3}{5}\right)^x=\left(\dfrac{-3}{5}\right)^2\)
\(x=2\)
Vậy ...