a) Ta có: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{4}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{-1}{4}\\x+\dfrac{3}{4}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{8}\cdot2^3\cdot2^x=2^6\)
\(\Leftrightarrow2^x=2^6\)
hay x=6
c) Ta có: \(\dfrac{1}{5}x+\dfrac{2}{3}=\dfrac{3}{5}\)
\(\Leftrightarrow x\cdot\dfrac{1}{5}=\dfrac{3}{5}-\dfrac{2}{3}=\dfrac{-1}{15}\)
hay \(x=-\dfrac{1}{3}\)
d) Ta có: \(2^x+2^{x+4}=544\)
\(\Leftrightarrow2^x\cdot17=544\)
\(\Leftrightarrow2^x=32\)
hay x=5
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