\(y=\dfrac{2\left(sinx+1\right)-5}{sinx+1}=2-\dfrac{5}{sinx+1}\ge2-\dfrac{5}{1+1}=-\dfrac{1}{2}\)

\(y_{max}=-\dfrac{1}{2}\) khi \(sinx=1\)

\(y_{min}\) không tồn tại

Do M là trung điểm SA, H là trung điểm AB \(\Rightarrow HM\) là đường trung bình tam giác SAB

\(\Rightarrow HM||SB\Rightarrow HM||\left(SBC\right)\)

\(\Rightarrow d\left(M;\left(SBC\right)\right)=d\left(H;\left(SBC\right)\right)\)

Trong mp (ABC) từ H kẻ \(HD\perp BC\), trong mp (SHD) từ H kẻ \(HE\perp SD\)

\(\Rightarrow HE\perp\left(SBC\right)\Rightarrow HE=d\left(H;\left(SBC\right)\right)\)

\(HD=HB.sinB=\dfrac{a}{2}.sin60^0=\dfrac{a\sqrt{3}}{4}\)

Hệ thức lượng trong tam giác vuông SHD:

\(\dfrac{1}{HE^2}=\dfrac{1}{SH^2}+\dfrac{1}{HD^2}\Rightarrow HE=\dfrac{SH.HD}{\sqrt{SH^2+HD^2}}=\dfrac{a\sqrt{51}}{17}\)

\(\Rightarrow d\left(M;\left(SBC\right)\right)=HE=\dfrac{a\sqrt{51}}{17}\)

best hỏi bài ak :)

1.

\(sinx+sin5x+sin3x=0\)

\(\Leftrightarrow2sin3x.cos2x+sin3x=0\)

\(\Leftrightarrow sin3x\left(2cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=k\pi\\2x=\pm\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{3}\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

2.

\(cos7x-cos3x+sin8x+sin2x=0\)

\(\Leftrightarrow-2sin5x.sin2x+2sin5x.cos3x=0\)

\(\Leftrightarrow sin5x\left(sin2x-cos3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin5x=0\\cos3x=sin2x=cos\left(\dfrac{\pi}{2}-2x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=k\pi\\3x=\dfrac{\pi}{2}-2x+k2\pi\\3x=2x-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{5}\\x=\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

1.

Gọi O là giao điểm AC và BD

\(\Rightarrow SO=\left(SAC\right)\cap\left(SBD\right)\)

Trong mp (SBD), nối BI cắt SO tại J

\(J\in SO\Rightarrow J\in\left(SAC\right)\)

\(\Rightarrow J=BI\cap\left(SAC\right)\)

b.

Áp dụng định lý Menelaus cho tam giác BID:

\(\dfrac{OB}{OD}.\dfrac{SD}{SI}.\dfrac{JI}{JB}=1\Leftrightarrow1.\dfrac{3}{2}.\dfrac{JI}{JB}=1\)

\(\Rightarrow IJ=\dfrac{2}{3}JB=\dfrac{2}{3}\left(IB-IJ\right)\Rightarrow IJ=\dfrac{2}{5}IB\)

\(\Rightarrow\dfrac{IJ}{IB}=\dfrac{2}{5}\)

Hình vẽ bài 1:

a.

ĐKXĐ: \(\left\{{}\begin{matrix}\dfrac{2+sinx}{1+cosx}\ge0\left(\text{luôn đúng}\right)\\1+cosx\ne0\end{matrix}\right.\) 

\(\Rightarrow cosx\ne-1\)

\(\Rightarrow x\ne\pi+k2\pi\)

b.

\(\left\{{}\begin{matrix}cos2x\ne0\\sinx+1\ne0\\sin\left(3x+\dfrac{\pi}{6}\right)\ne0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x\ne\dfrac{\pi}{2}+k\pi\\x\ne-\dfrac{\pi}{2}+k2\pi\\3x+\dfrac{\pi}{6}\ne k\pi\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne-\dfrac{\pi}{2}+k2\pi\\x\ne-\dfrac{\pi}{18}+\dfrac{k\pi}{3}\end{matrix}\right.\)

c.

\(\left\{{}\begin{matrix}cos5x\ne0\\sin4x-cos3x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}cos5x\ne0\\sin4x\ne sin\left(\dfrac{\pi}{2}-3x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x\ne\dfrac{\pi}{2}+k\pi\\4x\ne\dfrac{\pi}{2}-3x+k2\pi\\4x\ne3x-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{10}+\dfrac{k\pi}{5}\\x\ne\dfrac{\pi}{14}+\dfrac{k2\pi}{7}\\x\ne-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

a: Hàm số này tuần hoàn theo chu kì T=2pi/5

b: sin2x tuần hoàn theo chu kì T=2pi/2=pi

cosx tuần hoàn theo chu kì T=2pi

=>f(x) tuần hoàn theo chu kì 2pi

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