Gọi a,b (a { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " Gọi a,b (a
Gọi a,b (a
By aikhanh
By aikhanh
By aikhanh
Gọi a,b (a { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " Gọi a,b (a
Áp dụng công thức mũ ta có
$2^{x^2-1} .(2^3)^x = (2.2^{\frac{1}{2}})^{3x}$
$<-> 2^{x^2-1} . 2^{3x} = (2^{\frac{3}{2}})^{3x}$
$<-> 2^{x^2-1+3x} = 2^{\frac{3}{2}.3x}$
$<-> x^2+3x-1 = \dfrac{9}{2}x$
$<-> x^2 -\dfrac{3}{2}x-1 = 0$
Vậy $x = 2$, $x = -\dfrac{1}{2}$.
Do đó $a = 2$, $b = -\dfrac{1}{2}$.