[Toán 11] Lượng giác

T

thuytrong

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T

truongduong9083

Câu 1.
$\bullet$ Do $0 < a, b \leq 1$
Nên $\dfrac{a}{b+1}+\dfrac{b}{a+1} \leq \dfrac{a}{a+b}+\dfrac{b}{a+b} = 1$
$\bullet$ $ \dfrac{a}{b+1}+\dfrac{b}{a+1} \geq \dfrac{(a+b)^2}{a(1+b)+b(1+a)} = \dfrac{1}{1+2ab}$
Mà ta có
$\dfrac{1}{1+2ab} \geq \dfrac{2}{3}$
$\Leftrightarrow 1 \geq 4ab$
$\Leftrightarrow (a+b)^2\geq 4ab$ (Luôn đúng)
 
N

nguyenbahiep1

câu 2

[TEX]cosx \geq 0 \\ sin x = \frac{1}{2} \\ x = \frac{\pi}{6} + k.2.\pi \\ x = \frac{5.\pi}{6} + k.2.\pi ( L) [/TEX]
 
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