ĐK: \(\left\{{}\begin{matrix}x>0\\\log_2x+m\ge0\end{matrix}\right.\)

\(pt\Leftrightarrow\log_2^2x+6\log_2x+9=4\log_2x+4m+4\sqrt{\log_2x+m}+1\)

\(\Leftrightarrow\left(\log_2x+3\right)^2=\left(2\sqrt{\log_2x+m}+1\right)^2\)\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{\log_2x+m}+1=\log_2x+3\\2\sqrt{\log_2x+m}+1=-\log_2x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\sqrt{\log_2x+m}=\log_2x+2\left(1\right)\\2\sqrt{\log_2x+m}=-\log_2x-4\left(2\right)\end{matrix}\right.\)

Xét pt(1) <=> \(4\log_2x+4m=\log_2^2x+4\log_2x+4\Leftrightarrow\log_2^2x+4-4m=0\)

pt có nghiệm thì \(\Delta\ge0\Leftrightarrow-4\left(4-4m\right)\ge0\Leftrightarrow m\ge1\)

Xét pt(2) <=> \(4\log_2x+4m=\log_2^2x+8\log_2x+16\Leftrightarrow\log_2^2x+4\log_2x+16-4m=0\)

pt có nghiệm thì \(\Delta\ge0\Leftrightarrow4^2-4\left(16-4m\right)\ge0\Leftrightarrow m\ge3\)

Để pt đề bài có nghiệm thì ít nhất một trong hai pt (1) và (2) có nghiệm

Từ đề bài suy ra \(1\le m\le2023;m\in Z\) => m thuộc {1;2;....;2023} => Có 2023 giá trị m nguyên thỏa mãn

a: Chọn mp(SAC) có chứa AN

Trong mp(ABCD), gọi O là giao điểm của AC và BD

=>\(O\in\left(SAC\right)\cap\left(SBD\right)\)

mà \(S\in\left(SAC\right)\cap\left(SBD\right)\)

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

Gọi K là giao điểm của AN với SO

=>K là giao điểm của AN với mp(SBD)

b: Chọn mp(SMC) có chứa MN

Trong mp(ABCD), gọi I là giao điểm của MC và BD

=>\(I\in\left(SMC\right)\cap\left(SBD\right)\)

mà \(S\in\left(SMC\right)\cap\left(SBD\right)\)

nên \(\left(SMC\right)\cap\left(SBD\right)=SI\)

Gọi L là giao điểm của SI với MN

=>L là giao điểm của MN với mp(SBD)

\(E=cos\left(x-\dfrac{\Omega}{3}\right)\cdot cos\left(x+\dfrac{\Omega}{4}\right)+cos\left(x+\dfrac{\Omega}{6}\right)\cdot cos\left(x+\dfrac{3}{4}\Omega\right)\)

\(=\dfrac{1}{2}\left[cos\left(x-\dfrac{\Omega}{3}+x+\dfrac{\Omega}{4}\right)+cos\left(x-\dfrac{\Omega}{3}-x-\dfrac{\Omega}{4}\right)\right]+\dfrac{1}{2}\cdot\left[cos\left(x+\dfrac{\Omega}{6}+x+\dfrac{3}{4}\Omega\right)+cos\left(x+\dfrac{\Omega}{6}-x-\dfrac{3}{4}\Omega\right)\right]\)

\(=\dfrac{1}{2}\cdot\left[cos\left(2x-\dfrac{\Omega}{12}\right)+cos\left(-\dfrac{7}{12}\Omega\right)\right]+\dfrac{1}{2}\left[cos\left(2x+\dfrac{11}{12}\Omega\right)+cos\left(-\dfrac{7}{12}\Omega\right)\right]\)

\(=\dfrac{1}{2}\left[cos\left(2x-\dfrac{\Omega}{12}\right)+cos\left(2x+\dfrac{11}{12}\Omega\right)+cos\left(-\dfrac{7}{12}\Omega\right)+cos\left(-\dfrac{7}{12}\Omega\right)\right]\)

\(=\dfrac{1}{2}\left[cos\left(2x+\dfrac{11}{12}\Omega-\Omega\right)+cos\left(2x+\dfrac{11}{12}\Omega\right)+\dfrac{-\sqrt{6}+\sqrt{2}}{2}\right]\)

\(=\dfrac{1}{2}\left[cos\left[\Omega-\left(2x+\dfrac{11}{12}\Omega\right)\right]+cos\left(2x+\dfrac{11}{12}\Omega\right)+\dfrac{-\sqrt{6}+\sqrt{2}}{2}\right]\)

\(=\dfrac{1}{2}\left[-cos\left(2x+\dfrac{11}{12}\Omega\right)+cos\left(2x+\dfrac{11}{12}\Omega\right)\right]+\dfrac{-\sqrt{6}+\sqrt{2}}{4}=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)

a: \(\dfrac{sina+sin2a}{1+cosa+cos2a}=\dfrac{sina+2\cdot sina\cdot cosa}{1+cosa+2\cdot cos^2a-1}\)

\(=\dfrac{sina\left(2\cdot cosa+1\right)}{2\cdot cos^2a+cosa}=\dfrac{sina\left(2\cdot cosa+1\right)}{cosa\left(2\cdot cosa+1\right)}=\dfrac{sina}{cosa}=tana\)

b: \(\dfrac{cos\left(\dfrac{\Omega}{4}+a\right)-cos\left(\dfrac{\Omega}{4}-a\right)}{sin\left(\dfrac{\Omega}{4}+a\right)-sin\left(\dfrac{\Omega}{4}-a\right)}\)

\(=\dfrac{cos\left(\dfrac{\Omega}{4}\right)\cdot cosa-sin\left(\dfrac{\Omega}{4}\right)\cdot sina-cos\left(\dfrac{\Omega}{4}\right)\cdot cosa-sin\left(\dfrac{\Pi}{4}\right)\cdot sina}{sin\left(\dfrac{\Pi}{4}\right)\cdot cosa+sina\cdot cos\left(\dfrac{\Pi}{4}\right)-sin\left(\dfrac{\Pi}{4}\right)\cdot cosa+sina\cdot cos\left(\dfrac{\Pi}{4}\right)}\)

\(=\dfrac{-2\cdot sina\cdot\dfrac{\sqrt{2}}{2}}{2\cdot sina\cdot\dfrac{\sqrt{2}}{2}}=-1\)

c: \(\dfrac{sin^2a}{4-4\cdot sin^2\left(\dfrac{a}{2}\right)}=\dfrac{sin^2a}{4\left(1-sin^2\left(\dfrac{a}{2}\right)\right)}=\dfrac{sin^2a}{4\cdot cos^2\left(\dfrac{a}{2}\right)}\)

\(=\dfrac{\left(sin\left(2\cdot\dfrac{a}{2}\right)\right)^2}{4\cdot cos^2\left(\dfrac{a}{2}\right)}=\dfrac{4\cdot sin^2\left(\dfrac{a}{2}\right)\cdot cos^2\left(\dfrac{a}{2}\right)}{4\cdot cos^2\left(\dfrac{a}{2}\right)}=sin^2\left(\dfrac{a}{2}\right)\)

\(\dfrac{sin4a}{1+cos4a}\cdot\dfrac{cos2a}{1+cos2a}\)

\(=\dfrac{2\cdot sin2a\cdot cos2a}{1+cos\left(2\cdot2a\right)}\cdot\dfrac{cos2a}{1+2\cdot cos^2a-1}\)

\(=\dfrac{2\cdot sin2a\cdot cos2a}{2\cdot cos^2a}\cdot\dfrac{cos2a}{1+2\cdot cos^22a-1}\)

\(=\dfrac{2\cdot sin2a\cdot cos2a}{2\cdot cos^2a}\cdot\dfrac{cos2a}{2\cdot cos^22a}\)

\(=\dfrac{sin2a}{cos2a}\cdot\dfrac{cos2a}{2\cdot cos^2a}=\dfrac{sin2a}{2\cdot cos^2a}=\dfrac{2\cdot sina\cdot cosa}{2\cdot cos^2a}=\dfrac{sina}{cosa}=tana\)

\(6\cdot sin2x-3=-m\)

=>\(6\cdot sin2x=3-m\)

=>\(sin2x=\dfrac{3-m}{6}\)

Để phương trình có nghiệm thì \(-1< =\dfrac{3-m}{6}< =1\)

=>-6<=3-m<=6

=>-6-3<=-m<=6-3

=>-9<=-m<=3

=>9>=m>=-3

Vậy: -3<=m<=9

\(sin^4a+cos^4a=\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a\)

\(=1-2\cdot\left(sina\cdot cosa\right)^2\)

\(=1-2\cdot\left(\dfrac{1}{2}\cdot sin2a\right)^2\)

\(=1-\dfrac{sin^22a}{2}\)

\(=\dfrac{3}{4}+\dfrac{1}{4}-\dfrac{1}{2}\cdot sin^22a=\dfrac{3}{4}+\dfrac{1}{4}\left(1-2\cdot sin^22a\right)\)

\(=\dfrac{3}{4}+\dfrac{1}{4}\cdot cos\left(2\cdot2a\right)=\dfrac{3}{4}+\dfrac{1}{4}\cdot cos4a\)

Ta có: 

\(sin^4a+cos^4a=\left(sin^2a+cos^2a\right)^2-2sin^2a\cdot cos^2a\\ =1^2-2\left(\dfrac{2sina\cdot cosa}{2}\right)^2\\ =1-2\cdot\dfrac{sin^22a}{4}=1-\dfrac{sin^22a}{2}\\ =\dfrac{3}{4}+\left(\dfrac{1}{4}-\dfrac{sin^22a}{2}\right)\\ =\dfrac{3}{4}+\dfrac{1-2sin^22a}{4}\\ =\dfrac{3}{4}+\dfrac{cos4a}{4}=\dfrac{3}{4}+\dfrac{1}{4}cos4a\)