Giải hộ em vs a nhanh hộ em vs em đang cần gấp lắm ạ . nghiệm của pt (2+căn 3)^cosx + (2-căn3)^cosx=4 28/08/2021 Bởi Sadie Giải hộ em vs a nhanh hộ em vs em đang cần gấp lắm ạ . nghiệm của pt (2+căn 3)^cosx + (2-căn3)^cosx=4
$(2 + \sqrt{3})^{cosx} + (2 – \sqrt{3})^{cosx} = 4$ $\Leftrightarrow (2 +\sqrt{3})^{cosx} – (2 + \sqrt{3}) + (2-\sqrt{3})^{cosx} – (2 -\sqrt{3}) = 0$ $\Leftrightarrow (2 + \sqrt{3})\left[(2 + \sqrt{3})^{cosx – 1} – 1\right] = -(2 – \sqrt{3})\left[(2 – \sqrt{3})^{cosx-1} – 1\right]$ $\Leftrightarrow ln\left\{(2 + \sqrt{3})\left[(2 + \sqrt{3})^{cosx – 1} – 1\right]\right\}= ln\left\{-(2 – \sqrt{3})\left[(2 – \sqrt{3})^{cosx-1} – 1\right]\right\}$ $\Leftrightarrow ln(2 + \sqrt{3}) + ln\left[(2 + \sqrt{3})^{cosx – 1} – 1\right] = -ln(2 – \sqrt{3}) + ln\left[(2 – \sqrt{3})^{cosx – 1} – 1\right]$ $\Leftrightarrow \left[ln(2 + \sqrt{3}) + ln(2 – \sqrt{3})\right] + ln\left[(2 + \sqrt{3})^{cosx – 1} – 1\right] = ln\left[(2 – \sqrt{3})^{cosx – 1} – 1\right]$ $\Leftrightarrow ln\left[(2 + \sqrt{3})^{cosx – 1} – 1\right] = ln\left[(2 – \sqrt{3})^{cosx – 1} – 1\right]$ $\Leftrightarrow (2 + \sqrt{3})^{cosx-1} = (2 – \sqrt{3})^{cosx – 1}$ $\Leftrightarrow (cosx-1)ln(2 + \sqrt{3}) = (cosx – 1)ln(2 -\sqrt{3}) = 0$ $\Leftrightarrow (cosx – 1)ln\left(\dfrac{2 + \sqrt{3}}{2 – \sqrt{3}}\right) = 0$ $\Leftrightarrow cosx = 1$ $\Leftrightarrow x = k2\pi \,\, (k \in \Bbb Z)$ Bình luận
$(2 + \sqrt{3})^{cosx} + (2 – \sqrt{3})^{cosx} = 4$
$\Leftrightarrow (2 +\sqrt{3})^{cosx} – (2 + \sqrt{3}) + (2-\sqrt{3})^{cosx} – (2 -\sqrt{3}) = 0$
$\Leftrightarrow (2 + \sqrt{3})\left[(2 + \sqrt{3})^{cosx – 1} – 1\right] = -(2 – \sqrt{3})\left[(2 – \sqrt{3})^{cosx-1} – 1\right]$
$\Leftrightarrow ln\left\{(2 + \sqrt{3})\left[(2 + \sqrt{3})^{cosx – 1} – 1\right]\right\}= ln\left\{-(2 – \sqrt{3})\left[(2 – \sqrt{3})^{cosx-1} – 1\right]\right\}$
$\Leftrightarrow ln(2 + \sqrt{3}) + ln\left[(2 + \sqrt{3})^{cosx – 1} – 1\right] = -ln(2 – \sqrt{3}) + ln\left[(2 – \sqrt{3})^{cosx – 1} – 1\right]$
$\Leftrightarrow \left[ln(2 + \sqrt{3}) + ln(2 – \sqrt{3})\right] + ln\left[(2 + \sqrt{3})^{cosx – 1} – 1\right] = ln\left[(2 – \sqrt{3})^{cosx – 1} – 1\right]$
$\Leftrightarrow ln\left[(2 + \sqrt{3})^{cosx – 1} – 1\right] = ln\left[(2 – \sqrt{3})^{cosx – 1} – 1\right]$
$\Leftrightarrow (2 + \sqrt{3})^{cosx-1} = (2 – \sqrt{3})^{cosx – 1}$
$\Leftrightarrow (cosx-1)ln(2 + \sqrt{3}) = (cosx – 1)ln(2 -\sqrt{3}) = 0$
$\Leftrightarrow (cosx – 1)ln\left(\dfrac{2 + \sqrt{3}}{2 – \sqrt{3}}\right) = 0$
$\Leftrightarrow cosx = 1$
$\Leftrightarrow x = k2\pi \,\, (k \in \Bbb Z)$