Đáp án :

In\(\frac{7}{3}\)

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Cám ơn bạn

Đường thẳng được viết lại \(y=\sqrt{3}.x-1\)

                               \(\Rightarrow\)\(tan\alpha=\sqrt{3}\)

                               \(\Rightarrow\alpha=60^o\)

=> Chọn c

\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)

\(=\frac{b^2c^2}{ab+ca}+\frac{c^2a^2}{bc+ab}+\frac{a^2b^2}{ca+bc}\)

\(\ge\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\frac{1}{2}\left(ab+bc+ca\right)\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel(hoặc áp dụng BĐT quen thuộc: \(\frac{p^2}{m}+\frac{q^2}{n}\ge\frac{\left(p+q\right)^2}{m+n}\) 2 lần),ta có:

\(VT=\frac{\left(\frac{1}{a^2}\right)}{a\left(b+c\right)}+\frac{\left(\frac{1}{b^2}\right)}{b\left(c+a\right)}+\frac{\left(\frac{1}{c^2}\right)}{c\left(a+b\right)}\)

\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}\) (thay abc = 1 vào)

\(=\frac{ab+bc+ca}{2}=\frac{1}{2}\left(ab+bc+ca\right)^{\left(đpcm\right)}\)

\(sin\left(\frac{9\pi}{2}+\alpha\right)=sin\left(4\pi+\frac{\pi}{2}+\alpha\right)=sin\left(\frac{\pi}{2}+\alpha\right)=cos\alpha\)