pt \(\Leftrightarrow\left(2^x\right)^3-\left(\frac{1}{2^{x-1}}\right)^3-6\left(2^x-\frac{1}{2^{x-1}}\right)=1\)
\(\Leftrightarrow\left(2^x-\frac{1}{2^{x-1}}\right)\left(2^{2x}+2^x\cdot\frac{1}{2^{x-1}}+\left(\frac{1}{2^{x-1}}\right)^2\right)-6\left(2^x-\frac{1}{2^{x-1}}\right)=1\)
\(\Leftrightarrow\left(2^x-\frac{1}{2^{x-1}}\right)\left(2^{2x}+\frac{1}{2^{2x-2}}-4\right)=1\)
\(\Leftrightarrow\frac{2^{2x-1}-1}{2^{x-1}}\cdot\frac{2^{4x-2}-4\cdot2^{2x-2}+1}{2^{2x-1}}=1\)
\(\Leftrightarrow\frac{2^{2x-1}-1}{2^{x-1}}\cdot\frac{2^{2\left(2x-1\right)}-2\cdot2^{2x-1}+1}{2^{2x-2}}=1\)
\(\Leftrightarrow\frac{2^{2x-1}-1}{2^{x-1}}\cdot\left(\frac{2^{2x-1}-1}{2^{x-1}}\right)^2=1\)
\(\Leftrightarrow\left(\frac{2^{2x-1}-1}{2^{x-1}}\right)^3=1\)
\(\Leftrightarrow\frac{2^{2x-1}-1}{2^{x-1}}=1\Leftrightarrow2^{2x-1}-1=2^{x-1}\Leftrightarrow\frac{\left(2^x\right)^2}{2}-\frac{2^x}{2}-1=0\)
Giải pt bậc hai được 2x = 2 ↔ x = 1