bài 1:tìm x a ($\frac{4}{3}$ -$\frac{1}{2}$.x )=$\frac{49}{9}$ b $x^{3}$ =$x^{2}$ c $27^{x}$ :$3^{x}$ =9 d $64^{x}$ :$4^{x^+2}$= 16 e $\frac{81}

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\dfrac{4}{3} – \dfrac{1}{2}x = \dfrac{{49}}{9}\\
 \Leftrightarrow \dfrac{1}{2}x = \dfrac{4}{3} – \dfrac{{49}}{9}\\
 \Leftrightarrow \dfrac{1}{2}x =  – \dfrac{{37}}{9}\\
 \Leftrightarrow x = \left( { – \dfrac{{37}}{9}} \right):\dfrac{1}{2}\\
 \Leftrightarrow x =  – \dfrac{{74}}{9}\\
b,\\
{x^3} = {x^2}\\
 \Leftrightarrow {x^3} – {x^2} = 0\\
 \Leftrightarrow {x^2}\left( {x – 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
x – 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
c,\\
{27^x}:{3^x} = 9\\
 \Leftrightarrow {\left( {27:3} \right)^x} = 9\\
 \Leftrightarrow {9^x} = 9\\
 \Leftrightarrow x = 1\\
d,\\
{64^x}:{4^{x + 2}} = 16\\
 \Leftrightarrow {\left( {{4^3}} \right)^x}:{4^{x + 2}} = {4^2}\\
 \Leftrightarrow {4^{3x}}:{4^{x + 2}} = {4^2}\\
 \Leftrightarrow {4^{3x – \left( {x + 2} \right)}} = {4^2}\\
 \Leftrightarrow {4^{2x – 2}} = {4^2}\\
 \Leftrightarrow 2x – 2 = 2\\
 \Leftrightarrow x = 2\\
e,\\
\dfrac{{81}}{{{3^x}}} = 9\\
 \Leftrightarrow {3^x} = 81:9\\
 \Leftrightarrow {3^x} = 9\\
 \Leftrightarrow {3^x} = {3^2}\\
 \Leftrightarrow x = 2
\end{array}\)