c.

ĐKXĐ: \(sinx\ne0\Rightarrow x\ne k\pi\)

\(1-\dfrac{\sqrt{3}cosx}{sinx}-4cosx=0\)

\(\Rightarrow sinx-\sqrt{3}cosx-4sinx.cosx=0\)

\(\Leftrightarrow sinx-\sqrt{3}cosx=2sin2x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin2x\)

\(\Leftrightarrow sin2x=sin\left(x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=x-\dfrac{\pi}{3}+k2\pi\\2x=\dfrac{4\pi}{3}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

a.

\(\Leftrightarrow3sin3x-4sin^34x-\sqrt{3}cos9x=2sin2x\)

\(\Leftrightarrow sin9x-\sqrt{3}cos9x=2sin2x\)

\(\Leftrightarrow\dfrac{1}{2}sin9x-\dfrac{\sqrt{3}}{2}cos9x=sin2x\)

\(\Leftrightarrow sin\left(9x-\dfrac{\pi}{3}\right)=sin2x\)

\(\Leftrightarrow\left[{}\begin{matrix}9x-\dfrac{\pi}{3}=2x+k2\pi\\9x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{21}+\dfrac{k2\pi}{7}\\x=\dfrac{4\pi}{33}+\dfrac{k2\pi}{11}\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{x+6}-\sqrt{x+2}}{x^2-4}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{x+6}-2+2-\sqrt{x+2}}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+\dfrac{4-x-2}{2+\sqrt{x+2}}}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}-\dfrac{1}{2+\sqrt{x+2}}\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\dfrac{1}{2+\sqrt{x+2}}}{x+2}\)

\(=\dfrac{\dfrac{1}{\sqrt[3]{\left(2+6\right)^2}+2\cdot\sqrt[3]{2+6}+4}-\dfrac{1}{2+\sqrt{2+2}}}{2+2}\)

\(=\dfrac{\dfrac{1}{\sqrt[3]{64}+2\cdot\sqrt[3]{8}+4}-\dfrac{1}{2+2}}{4}\)

\(=\dfrac{\dfrac{1}{4+2\cdot2+4}-\dfrac{1}{4}}{4}=\left(\dfrac{1}{16}-\dfrac{1}{4}\right):4=\left(\dfrac{1}{16}-\dfrac{4}{16}\right)\cdot\dfrac{1}{4}=\dfrac{-3}{64}\)

1.

\(-1\le sin\left(x-\dfrac{\pi}{2}\right)\le1\Rightarrow1\le y\le5\)

\(y_{min}=1\) khi \(sin\left(x-\dfrac{\pi}{2}\right)=-1\)

\(y_{max}=5\) khi \(sin\left(x-\dfrac{\pi}{2}\right)=1\)

2.

\(-1\le cos2x\le1\Rightarrow\dfrac{5}{2}\le y\le\dfrac{7}{2}\)

\(y_{min}=\dfrac{5}{2}\) khi \(cos2x=1\)

\(y_{max}=\dfrac{7}{2}\) khi \(cos2x=-1\)

3.

\(0\le cos^2\left(2x+\dfrac{\pi}{3}\right)\le1\Rightarrow-2\le y\le-1\)

\(y_{min}=-2\) khi \(cos\left(2x+\dfrac{\pi}{3}\right)=\pm1\)

\(y_{max}=-1\) khi \(cos\left(2x+\dfrac{\pi}{3}\right)=0\)

4.

\(-1\le cos\left(4x^2\right)\le1\Rightarrow-2\le y\le\sqrt{2}-2\)

\(y_{min}=-1\) khi \(cos\left(4x^2\right)=-1\)

\(y_{max}=\sqrt{2}-2\) khi \(cos\left(4x^2\right)=1\)

giải giúp mình bài 2 thôi ạ

dạ thôi mình ko cần nữa ạ. Cảm ơn pạn nào có ý định giúp mik nhe

a.

Kẻ \(AE\perp SD\) 

Do \(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp CD\\CD\perp AD\end{matrix}\right.\) \(\Rightarrow CD\perp\left(SAD\right)\Rightarrow CD\perp AE\)

\(\Rightarrow AE\perp\left(SCD\right)\Rightarrow AE=d\left(A;\left(SCD\right)\right)\)

\(AE=\dfrac{SA.AD}{\sqrt{SA^2+AD^2}}=\dfrac{4a\sqrt[]{5}}{5}\)

\(\left\{{}\begin{matrix}AM\cap\left(SCD\right)=C\\MC=\dfrac{3}{4}AC\end{matrix}\right.\) \(\Rightarrow d\left(M;\left(SCD\right)\right)=\dfrac{3}{4}d\left(A;\left(SCD\right)\right)=\dfrac{3a\sqrt{5}}{5}\)

\(\left\{{}\begin{matrix}MN\cap\left(SCD\right)=S\\NS=\dfrac{1}{2}MS\end{matrix}\right.\) \(\Rightarrow d\left(N;\left(SCD\right)\right)=\dfrac{1}{2}d\left(M;\left(SCD\right)\right)=\dfrac{3a\sqrt{5}}{6}\)

b.

Qua S kẻ tia Sx song song cùng chiều tia DC, trên Sx lấy F sao cho \(SF=DC\)

\(\Rightarrow CDSF\) là hình bình hành \(\Rightarrow CF||SD\Rightarrow\left(SAD\right)||\left(BCF\right)\Rightarrow CD\perp\left(BCF\right)\)

Qua B kẻ \(BG\perp CF\Rightarrow BG\perp\left(SCD\right)\Rightarrow\widehat{BDG}\) là góc giữa BD và (SCD)

SF song song và bằng CD nên SF song song và bằng AB \(\Rightarrow SABF\) là hbh

\(\Rightarrow FB||SA\Rightarrow FB\perp\left(ABCD\right)\) \(\Rightarrow FB\perp BC\)

\(BF=SA=2a\Rightarrow BG=\dfrac{BF.BC}{\sqrt{BF^2+BC^2}}=\dfrac{4a\sqrt{5}}{5}\) 

 \(BD=\sqrt{AB^2+AD^2}=5a\)

\(\Rightarrow sin\widehat{BDG}=\dfrac{BG}{BD}=\dfrac{4\sqrt{5}}{25}\)

c.

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp AD\\AD\perp AB\end{matrix}\right.\) \(\Rightarrow AD\perp\left(SAB\right)\)

\(\Rightarrow\widehat{DBA}\) là góc giữa BD và (SAB)

\(tan\widehat{DBA}=\dfrac{AD}{AB}=\dfrac{4}{3}\Rightarrow\widehat{DBA}\)

d.

Từ B kẻ \(BH\perp AC\) (H thuộc AC)

\(SA\perp\left(ABCD\right)\Rightarrow SA\perp BH\)

\(\Rightarrow BH\perp\left(SAC\right)\Rightarrow\widehat{BSH}\) là góc giữa SB và (SAC)

\(BH=\dfrac{AB.BC}{\sqrt{AB^2+BC^2}}=\dfrac{12a}{5}\)

\(\Rightarrow sin\widehat{BSH}=\dfrac{BH}{SB}=\dfrac{12\sqrt{13}}{65}\Rightarrow\widehat{BSH}\)

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