Gtln

L

lan_phuong_000

1.
$y = cos^2 + \dfrac{1}{cos^2x} + cosx + \dfrac{1}{cosx}$ (1)
ĐK: $x \ne \dfrac{\pi}{2} + k\pi$
Đặt: $ t = cosx + \dfrac{1}{cosx}$ \Rightarrow $t^2 = cos^2 + \dfrac{1}{cos^2x} +2$
(1) trở thành : $y = t^2 + t - 2 = (t + \dfrac{1}{2})^2 - \dfrac{9}{4}$
\Rightarrow $Min_y = - \dfrac{9}{4}$
Dấu "=" xảy ra khi và chỉ khi $t + \dfrac{1}{2} = 0$
\Rightarrow $cosx + \dfrac{1}{cosx} = - \dfrac{1}{2}$
\Leftrightarrow $2cos^2x + cosx - 2 =0$
Giải ptlg tìm x. Đơn giản :)

2.hình như đề sai bạn à
 
C

conga222222

$\eqalign{
& y = {\cos ^2}x + {1 \over {{{\cos }^2}x}} + \cos x + {1 \over {\cos x}} = {{{{\left( {\cos x + 1} \right)}^2}} \over 2} + {1 \over 2}{\left( {{1 \over {\cos x}} + 1} \right)^2} + {{{{\cos }^2}x} \over 2} + {1 \over {2{{\cos }^2}x}} - 1 \cr
& \ge {{{{\cos }^2}x} \over 2} + {1 \over {2{{\cos }^2}x}} - 1 \ge 1 - 1 = 0 \cr
& dau = \leftrightarrow \cos x = - 1 \cr} $
 
C

conga222222

$\eqalign{
& {\sin ^2}x + 2\sin x + {3 \over {{{\sin }^2}x}} + 3\sin x + 4\;?\;\left( {de\;nay\;thi\;cung\;tim\;duoc\;min\;thoi\;nhung\;chac\;de\;bai\;la} \right) \cr
& y = {\sin ^2}x + 2\sin x + {3 \over {{{\sin }^2}x}} + {3 \over {\sin x}} + 4\; \cr
& dat\;t = sinx\;\left( { - 1 \le t \le 1} \right) \cr
& y = {t^2} + 2t + {3 \over {{t^2}}} + {3 \over t} + 4 \cr
& {y^/} = 2t + 2 - {6 \over {{t^3}}} - {3 \over {{t^2}}} \cr
& {y^/}\;xac\;dinh \leftrightarrow t \ne 0 \cr
& {y^/} = 0 \leftrightarrow 2{t^4} + 2{t^3} - 3t - 6 = 0 \leftrightarrow \overline \exists \;t \cr
& do\;neu\; - 1 \le t \le 0 \to 2{t^4} + 2{t^3} - 3t - 6 \le 2{t^4} - 3t - 6 \le - 1 < 0 \cr
& neu\;0 \le t \le 1 \to 2{t^4} + 2{t^3} - 3t - 6 \le 2{t^4} + 2{t^3} - 6 \le - 2 < 0 \cr
& y( - 1) = ... \cr
& y(1) = ... \cr
& \mathop {\lim y}\limits_{x \to 0} = + \infty \cr
& \to \min = ... \cr
& \cr} $
 
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