Đáp án:
$A = 1$
Giải thích các bước giải:
Ta có:
$+)\quad b^2 = ac$
$\to \dfrac ab = \dfrac bc$
$+)\quad c^2 = bd$
$\to \dfrac bc = \dfrac cd$
$\to \dfrac ab =\dfrac bc =\dfrac cd$
Đặt $\dfrac ab =\dfrac bc =\dfrac cd= k \qquad (k\ne 0)$
$\to \begin{cases}a = kb\\b = kc\\c = kd\end{cases}$
Khi đó:
$+)\quad \dfrac ab\cdot\dfrac bc\cdot\dfrac cd = k^3$
$\to \dfrac ad = k^3$
$\to \dfrac da =\dfrac{1}{k^3}$
$+)\quad \dfrac{a - 2b + 3c}{b - 2c + 3d}$
$=\dfrac{kb - 2kc + 3kd}{b - 2c + 3d}$
$=\dfrac{k(b - 2c + 3d)}{b - 2c + 3d}$
$= k$
$\to \left(\dfrac{a - 2b + 3c}{b - 2c + 3d}\right)^3 = k^3$
Ta được:
$A =\dfrac da\cdot\left(\dfrac{a - 2b + 3c}{b - 2c + 3d}\right)^3$
$\to A =\dfrac{1}{k^3}\cdot k^3$
$\to A = 1$
