Xác suất:

\(P=C_9^7.\left(\dfrac{1}{4}\right)^7.\left(\dfrac{3}{4}\right)^2+C_9^8.\left(\dfrac{1}{4}\right)^8.\left(\dfrac{3}{4}\right)^1+C_9^9.\left(\dfrac{1}{4}\right)^9=\dfrac{11}{8192}\)

\(4\sin^22x-4\cos2x-1=0\)

\(\Leftrightarrow4\left(1-\cos^22x\right)-4\cos2x-1=0\)

\(\Leftrightarrow4-4\cos^22x-4\cos2x-1=0\)

\(\Leftrightarrow-4\cos^22x-4\cos2x+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\cos2x=\dfrac{-3}{2}\left(L\right)\end{matrix}\right.\Leftrightarrow\cos2x=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}-k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}-k\pi\end{matrix}\right.\left(k\in Z\right)\) 

Chọn A

4.(1-cos²2x) - 4cos2x -1 =0

-4cos²2x - 4cos2x + 3 = 0

bấm máy tính giải phương trình bậc 2

cos x = \(\dfrac{1}{2}\) => cos x = cos \(\dfrac{\pi}{3}\)  => x = \(\dfrac{\pi}{3}\) + k2\(\pi\)

                                                    x = - \(\dfrac{\pi}{3}\)+ k2\(\pi\)

cos x = \(\dfrac{-3}{2}\) ( vô nghiệm)

a.

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

b.

Ta có \(CD||AB\) (do ABCD là hcn)

Mà \(AB\perp\left(SAD\right)\Rightarrow CD\perp\left(SAD\right)\)

Lại có \(CD\in\left(SCD\right)\)

\(\Rightarrow\left(SCD\right)\perp\left(SAD\right)\)

c.

\(SA\perp\left(ABCD\right)\Rightarrow AD\) là hình chiếu vuông góc của SD lên (ABCD)

\(\Rightarrow\widehat{SDA}\) là góc giữa SD và (ABCD)

\(tan\widehat{SDA}=\dfrac{SA}{AD}=\dfrac{\sqrt{6}}{2}\Rightarrow\widehat{SDA}\approx50^046'\)

d.

Ta có: \(SA\perp\left(ABCD\right)\Rightarrow SA\perp\left(ACD\right)\)

Mà \(SA\in\left(SAD\right)\Rightarrow\left(SAD\right)\perp\left(ACD\right)\) (1)

Theo câu b ta có: \(\left(SAD\right)\perp\left(SCD\right)\) (2)

(1);(2) \(\Rightarrow\widehat{SDA}\) là góc giữa (SCD) và (ACD)

Theo câu c ta có: \(\widehat{SDA}=50^046'\)

10.

\(y=3cos2x-5sinx-1\)

\(=3-6sin^2x-5sinx-1\)

\(=-6sin^2x-5sinx+2\)

\(\Rightarrow y=f\left(t\right)=-6t^2-5t+2\) \(\left(t\in\left[-1;1\right]\right)\)

\(\Rightarrow y\in\left[-9;\dfrac{73}{24}\right]\)

\(\Rightarrow y\in\left\{-9;-8;-7;-6;-5;-4;-3;-2;-1;0;1;2;3\right\}\)

....

A,d của câu nào vậy ah tại mình hỏi đến 4 câu hỏi lận🥲🥲

\(\left\{{}\begin{matrix}S\in\left(SAC\right)\\S\in\left(SBD\right)\end{matrix}\right.\) \(\Rightarrow S=\left(SAC\right)\cap\left(SBD\right)\)

\(\left\{{}\begin{matrix}O\in AC\in\left(SAC\right)\\O\in BD\in\left(SBD\right)\end{matrix}\right.\) \(\Rightarrow O=\left(SAC\right)\cap\left(SBD\right)\)

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

37.

\(y=cos^2x+sinx+1\)

\(=1-sin^2x+sinx+1\)

\(=-sin^2x+sinx+2\in\left[0;\dfrac{9}{4}\right]\)

\(\Rightarrow T=\left\{0;1;2\right\}\)

\(\Rightarrow S_T=0+1+2=3\)

\(y=\dfrac{sinx+2cosx+1}{sinx+cosx+2}\) 

Thấy : \(sinx+cosx+2\ge-1-1+2=0\)  . " = " ko xảy ra nên : \(sinx+cosx+2>0\) 

Suy ra : \(\left(y-1\right)sinx+\left(y-2\right)cosx=1-2y\)  (*)

(*) có no \(\Leftrightarrow\left(y-1\right)^2+\left(y-2\right)^2\ge\left(1-2y\right)^2\Leftrightarrow2y^2-6y+5\ge4y^2-4y+1\Leftrightarrow-2y^2-2y+4\ge0\)

\(\Leftrightarrow-y^2-y+2\ge0\)  \(\Leftrightarrow-2\le y\le1\)

Suy ra : Max y = 1 . Chọn B 

21 : \(cosx-\sqrt{3}sinx=0\) 

cos x = 0 thay vào : sin x = 0 ( L ) 

cos x khác 0 \(\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\); ta có : \(1-\sqrt{3}tanx=0\Leftrightarrow tanx=\dfrac{1}{\sqrt{3}}\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\left(k\in Z\right)\)