Câu 1: \(u_4=u_1+3k\)

=>\(3k=\dfrac{3}{8}-3=\dfrac{3}{8}-\dfrac{24}{8}=-\dfrac{21}{8}\)

=>\(k=-\dfrac{7}{8}\)

\(u_7=u_1+6k=3+6\cdot\dfrac{-7}{8}=3-\dfrac{42}{8}=\dfrac{24-42}{8}=-\dfrac{18}{8}=-\dfrac{9}{4}\)

Câu 2: 

\(\dfrac{u_5}{u_8}=8\)

=>\(\dfrac{u_1\cdot q^4}{u_1\cdot q^7}=8\)

=>\(\dfrac{1}{q^3}=8\)

=>\(q=\dfrac{1}{2}\)

\(u_{12}=u_1\cdot q^{11}=12\cdot\left(\dfrac{1}{2}\right)^{11}=\dfrac{12}{2^{11}}=\dfrac{3}{2^9}\)

Câu 3:

Tổng của 5 số hạng đầu là:

\(S_5=\dfrac{u_1\cdot\left(1-q^5\right)}{1-q}=\dfrac{2\cdot\left(1-4^5\right)}{1-4}=682\)

=>Chọn D

Câu 25:

\(0< \alpha< \dfrac{\Omega}{2}\)

=>\(0< sin\alpha< 1;0< cos\alpha< 1\)

 \(\sqrt{\dfrac{1+sin\alpha}{1-sin\alpha}}+\sqrt{\dfrac{1-sin\alpha}{1+sin\alpha}}\)

\(=\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}\)

\(=\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{cos^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{cos^2\alpha}}\)

\(=\dfrac{1+sin\alpha+1-sin\alpha}{cos\alpha}=\dfrac{2}{cos\alpha}\)

Câu 28:

a, \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-\dfrac{9}{25}=\dfrac{16}{25}\Leftrightarrow cosx=\dfrac{4}{5}\)

\(tanx=\dfrac{sinx}{cosx}=-\dfrac{3}{5}:\left(\dfrac{4}{5}\right)=-\dfrac{3}{4}\)

\(cotx=-\dfrac{4}{3}\)

c, \(sin^2x+cos^2x=1\Leftrightarrow sin^2x=1-\dfrac{9}{25}=\dfrac{16}{25}\Leftrightarrow sinx=\dfrac{4}{5}\)

\(tanx=\dfrac{sinx}{cosx}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)

\(cotx=\dfrac{3}{4}\)

b, \(cos^2x+sin^2x=1\Leftrightarrow sin^2x=1-\dfrac{1}{16}=\dfrac{15}{16}\Leftrightarrow sinx=\dfrac{\sqrt{15}}{4}\)

\(tanx=\dfrac{\sqrt{15}}{4}:\dfrac{1}{4}=\sqrt{15}\)

\(cotx=\dfrac{1}{\sqrt{15}}\)

d, \(sin^2x+cos^2x=1\Leftrightarrow sin^2x=1-\dfrac{25}{169}=\dfrac{144}{169}\Leftrightarrow sinx=\dfrac{12}{13}\)

\(tanx=\dfrac{12}{13}:\left(-\dfrac{5}{13}\right)=-\dfrac{12}{5}\)

\(cotx=-\dfrac{5}{12}\)

a: \(\Omega< x< \dfrac{3}{2}\Omega\)

=>cosx<0

Ta có: \(sin^2x+cos^2x=1\)

=>\(cos^2x=1-sin^2x=1-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\)

mà cosx<0

nên \(cosx=-\dfrac{4}{5}\)

\(tanx=\dfrac{sinx}{cosx}=\dfrac{-3}{5}:\dfrac{-4}{5}=\dfrac{3}{4}\)

\(cotx=\dfrac{1}{tanx}=\dfrac{4}{3}\)

b: \(0< x< \dfrac{\Omega}{2}\)

=>sin x>0

\(sin^2x+cos^2x=1\)

=>\(sin^2x=1-\left(\dfrac{1}{4}\right)^2=\dfrac{15}{16}\)

mà sin x>0

nên \(sinx=\dfrac{\sqrt{15}}{4}\)

\(tanx=\dfrac{sinx}{cosx}=\dfrac{\sqrt{15}}{4}:\dfrac{1}{4}=\sqrt{15}\)

\(cotx=\dfrac{1}{tanx}=\dfrac{1}{\sqrt{15}}=\dfrac{\sqrt{15}}{15}\)

c: 0<x<90 độ

=>sin x>0

\(sin^2x+cos^2x=1\)

=>\(sin^2x=1-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}=\left(\dfrac{4}{5}\right)^2\)

mà sin x>0

nên \(sinx=\dfrac{4}{5}\)

\(tanx=\dfrac{sinx}{cosx}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)

\(cotx=1:\dfrac{4}{3}=\dfrac{3}{4}\)

d: \(180^0< x< 270^0\)

=>sin x<0

\(sin^2x+cos^2x=1\)

=>\(sin^2x=1-\left(-\dfrac{5}{13}\right)^2=1-\dfrac{25}{169}=\dfrac{144}{169}\)

mà sin x<0

nên \(sinx=-\dfrac{12}{13}\)

\(tanx=\dfrac{sinx}{cosx}=\dfrac{-12}{13}:\dfrac{-5}{13}=\dfrac{12}{5}\)

\(cotx=\dfrac{1}{tanx}=\dfrac{5}{12}\)

\(A=2\cdot cos\left(\dfrac{\Omega}{2}+x\right)+sin\left(5\Omega-x\right)+sin\left(\dfrac{3\Omega}{2}+x\right)+cos\left(\dfrac{\Omega}{2}+x\right)\)

\(=3\cdot cos\left(\dfrac{\Omega}{2}+x\right)+sin\left(\Omega-x\right)+sin\left(\dfrac{\Omega}{2}+\Omega+x\right)\)

\(=-3\cdot sinx+sinx+cos\left(\Omega+x\right)\)

\(=-2\cdot sinx-cosx\)

\(B=sin\left(\Omega+x\right)-cos\left(\dfrac{\Omega}{2}+x\right)+cot\left(2\Omega-x\right)+tan\left(\dfrac{2\Omega}{2}+x\right)\)

\(=-sinx+sinx+cot\left(-x\right)+tan\left(x\right)\)

\(=tanx-cotx=tanx-\dfrac{1}{tanx}=\dfrac{tan^2x-1}{tanx}\)

Giúp mình với ạ mình cảm ơn 

Câu 1:

\(2\sin x-\sqrt{3}=0\\ \Leftrightarrow\sin x=\dfrac{\sqrt{3}}{2}=\sin\dfrac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{\pi}{3}+k_12\pi\\x_2=\pi-\dfrac{\pi}{3}+k_22\pi=\dfrac{2\pi}{3}+k_22\pi\end{matrix}\right.\left(k_1,k_2\inℤ\right)\)

Mà: \(x\in\left[0;2\pi\right]\) do đó nên: \(k_1=0,k_2=0\)

Vậy tập nghiệm pt là: \(S=\left\{\dfrac{\pi}{3};\dfrac{2\pi}{3}\right\}\) (2 nghiệm => D)

Câu 2:

Vì: \(-1\le\cos x\le1\forall x\)

\(\Rightarrow-1\le m+1\le1\\ \Leftrightarrow-2\le m\le0\)

Mà: \(m\inℤ\Rightarrow m\in\left\{-2;-1;0\right\}\) (C)

Câu 1: \(2\cdot sinx-\sqrt{3}=0\)

=>\(sinx=\dfrac{\sqrt{3}}{2}\)

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

Để \(x\in\left[0;2\Omega\right]\) thì \(\left[{}\begin{matrix}\dfrac{\Omega}{3}+k2\Omega\in\left[0;2\Omega\right]\\\dfrac{2}{3}\Omega+k2\Omega\in\left[0;2\Omega\right]\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2k+\dfrac{1}{3}\in\left[0;2\right]\\2k+\dfrac{2}{3}\in\left[0;2\right]\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2k\in\left[-\dfrac{1}{3};\dfrac{5}{3}\right]\\2k\in\left[-\dfrac{2}{3};\dfrac{4}{3}\right]\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k\in\left[-\dfrac{1}{6};\dfrac{5}{6}\right]\\k\in\left[-\dfrac{1}{3};\dfrac{2}{3}\right]\end{matrix}\right.\Leftrightarrow k=0\)

=>Chọn B

Câu 2:

Để phương trình cosx =m+1 có nghiệm thì -1<=m+1<=1

=>-2<=m<=0

mà m nguyên

nên \(m\in\left\{-2;-1;0\right\}\)

=>Chọn C

a: \(cos\left(x-15^0\right)=\dfrac{\sqrt{2}}{2}\)

=>\(\left[{}\begin{matrix}x-15^0=45^0+k\cdot360^0\\x-15^0=-45^0+k\cdot360^0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=60^0+k\cdot360^0\\x=-30^0+k\cdot360^0\end{matrix}\right.\)

b: \(cos\left(2x+\dfrac{\Omega}{3}\right)+cos\left(x-\dfrac{\Omega}{3}\right)=0\)

=>\(cos\left(2x+\dfrac{\Omega}{3}\right)=-cos\left(x-\dfrac{\Omega}{3}\right)\)

=>\(cos\left(2x+\dfrac{\Omega}{3}\right)=cos\left(\Omega-x+\dfrac{\Omega}{3}\right)\)

=>\(cos\left(2x+\dfrac{\Omega}{3}\right)=cos\left(-x+\dfrac{4\Omega}{3}\right)\)

=>\(\left[{}\begin{matrix}2x+\dfrac{\Omega}{3}=-x+\dfrac{4\Omega}{3}+k2\Omega\\2x+\dfrac{\Omega}{3}=x-\dfrac{4}{3}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\Omega+k2\Omega\\x=-\dfrac{5}{3}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{3}+\dfrac{k2\Omega}{3}\\x=-\dfrac{5}{3}\Omega+k2\Omega\end{matrix}\right.\)

c: \(sin\left(3x+1\right)=sin\left(x-2\right)\)

=>\(\left[{}\begin{matrix}3x+1=x-2+k2\Omega\\3x+1=\Omega-x+2+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-3+k2\Omega\\4x=1+\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{3}{2}+k\Omega\\x=\dfrac{1}{4}+\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\end{matrix}\right.\)

a, d(B;SC) = d(B;(SAC)) 

Kẻ BH vuông AC 

Ta có d(B;(SAC)) = BH 

ADHT : \(\dfrac{1}{BH^2}=\dfrac{1}{AB^2}+\dfrac{1}{BC^2}=\dfrac{1}{a^2}+\dfrac{1}{a^2}=\dfrac{2a^2}{a^4}=\dfrac{2}{a^2}\Rightarrow BH=\dfrac{a}{\sqrt{2}}\)

b, 

Ta có AB vuông BC

SA vuông BC; AB; SA chứa (SAB)

=> BC vuông (SAB) 

Kẻ AK vuông SB => AK là kc giứa (A;(SBC)) 

=> AK = a/ căn 2

c, Kẻ CD // AB 

=> d(AB;SC) = d(AB;(SCD)) = d(A;(SCD)) 

Kẻ AM vuông CD; SA vuông CD 

=> CD vuông (SAM) 

Kẻ AG vuông SM => AG là khoảng cách 

Xét tứ giác ABCM có AM// BC; AB//MC 

=> tg ABCM là hbh => AM = BC = a

Xét tam giác SAM vuông tại A 

ADHT \(\dfrac{1}{AG^2}=\dfrac{1}{SA^2}+\dfrac{1}{AK^2}\Rightarrow AG=\dfrac{a}{\sqrt{2}}\)