Bài 77 trang 127 SGK giải tích 12 nâng cao

LG a

\({2^{{{\sin }^2}x}} + {4.2^{{{\cos }^2}x}} = 6\)

Lời giải chi tiết:

Ta có: \(\,{2^{{{\sin }^2}x}} + {4.2^{{{\cos }^2}x}} = 6\)

\(\Leftrightarrow {2^{1 - {{\cos }^2}x}} + {4.2^{{{\cos }^2}x}} = 6\)

\( \Leftrightarrow \frac{2}{{{2^{{{\cos }^2}x}}}} + {4.2^{{{\cos }^2}x}} = 6\)

Đặt \(t = {2^{{{\cos }^2}x}}\)

Vì \(0 \le {\cos ^2}x \le 1 \) \(\Rightarrow {2^0} \le {2^{{{\cos }^2}x}} \le {2^1}\) \( \Rightarrow 1 \le t \le 2\)

Ta có: 

\({2 \over t} + 4t = 6 \)\(\Leftrightarrow 4{t^2} - 6t + 2 = 0\)

\(\Leftrightarrow \left[ \matrix{
t = 1 \hfill \cr 
t = {1 \over 2}\,\,\left( \text{loại} \right) \hfill \cr} \right.\)

\( \Leftrightarrow {2^{{{\cos }^2}x}} = 1  \Leftrightarrow {\cos ^2}x = 0\)

\(\Leftrightarrow \cos x = 0 \Leftrightarrow x = {\pi  \over 2} + k\pi ,\,k \in \mathbb Z\)

LG b

\({4^{3 + 2\cos 2x}} - {7.4^{1 + \cos 2x}} = {4^{{1 \over 2}}}\)

Lời giải chi tiết:

\({4^{3 + 2\cos 2x}} - {7.4^{1 + \cos 2x}} = {4^{{1 \over 2}}}\)

\( \Leftrightarrow {4}{.4^{2 + 2\cos 2x}} - {7.4^{1 + \cos 2x}} = 2\)

\( \Leftrightarrow {4.4^{2\left( {1 + \cos 2x} \right)}} - {7.4^{1 + \cos 2x}} = 2\)

Đặt \(t = {4^{1 + \cos 2x}}\,\left( {t > 0} \right)\)

Ta có:

\(\eqalign{
& 4{t^2} - 7t - 2 = 0 \Leftrightarrow \left[ \matrix{
t = 2 \hfill \cr 
t = - {1 \over 4}\,\left( \text {loại} \right) \hfill \cr} \right. \cr 
& \Leftrightarrow {2^{2 + 2\cos 2x}} = 2 \Leftrightarrow 2 + 2\cos 2x = 1 \cr 
& \Leftrightarrow \cos 2x = - {1 \over 2} = \cos {{2\pi } \over 3} \cr 
& \Leftrightarrow 2x = \pm {2\pi \over 3} + k2\pi ,\,k \in \mathbb Z \cr} \)

\( \Leftrightarrow x =  \pm \frac{\pi }{3} + k\pi ,k \in Z\)

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