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

Hh

-_-

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

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

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

Em tham khảo nha:

The illustrations give the number of full-time male and female health-centre members in Miami and Florida throughout the year 2013 and outline the ratio of these members based on their gender and age. Overall, more male were active health centre members than that of females and just over half of them were between 21 to 45 years old.

As the bar chart implies, 500 females were active members of different fitness clubs in Florida and Miami in January 2012 while their male counterparts were slightly higher in number. The membership count to these gyms and health centres remained almost the same till April but afterwards increased rapidly between June and September. During this period, fitness centres in these two areas had more than 800 active male participants and slightly over 700 female members. The number of male health club members had always been higher than that of females. From October until the end of the year, these clubs retained a fewer number of members and this number varied from 500 to 700. 

According to the pie charts, more than half of the male members of these fitness clubs were between 21 to 45 years old while this was exactly 50% for women. Over 65 years old men accounted for one-quarter participants while it was 6% higher in the case of females. Remaining one-fifth fitness club members were from different other age groups.

Theo nguyên tắc bổ sung, trên một gen: %A = %T và %G =%X; %A + %G = 50% ⁽¹⁾

Mà theo bài ra, loại nucleotide không bổ sung với nucleotide loại ademin có thể là G hoặc X ➜ %A - %G = 20% ⁽²⁾

Từ (1) và (2) tính ra được %A = %T = (50 + 20) : 2 = 35%; %G = %X = 50 - 35 = 15%.

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

Cứu 

Cái này mà là lớp 11?????

1 did your father leave

2 had - was sleeping

3 was walking - lost

4 were having - started

5 was Jack doing - broke

6 Were you doing - was talking

#\(yGLinh\)

1. D

2. C

3. B

4. B