3x + 2 + 3x = 90
(x +\(\dfrac{1}{2}\))2 - \(\dfrac{1}{3}=\dfrac{1}{9}\)
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\(3^{x+2}+3^x=90\Leftrightarrow3^x.3^2+3^x=90\Leftrightarrow3^x\left(3^2+1\right)=90\Leftrightarrow3^x.10=90\Leftrightarrow3^x=9\Leftrightarrow3^x=3^2\Leftrightarrow x=2\)
Vậy ...
\(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{2}{3}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{6}\)
Vậy ...
1.
a) \(3^{x+2}+3^x=90\)
\(\Leftrightarrow3^x\left(3^2+1\right)=90\)
\(\Leftrightarrow3^x.10=90\)
\(\Leftrightarrow3^x=9=3^2\)
\(\Leftrightarrow x=2\)
vậy...
b) \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{2}{3}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{3}\\x+\dfrac{1}{2}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{-7}{6}\end{matrix}\right.\)
Vậy...
tik mik nha !!!