41 last

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44 was

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41. last
42. on
43. a
44. is
45. Have
46. am
47. you
48. to
49. I'll
50. opens

41. last

42. on

43. a

44. was

45. have

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47. you

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50. opens

a: Xét ΔAHB vuông tại H và ΔCHA vuông tại H có

góc HAB=góc HCA

=>ΔAHB đồng dạng với ΔCHA

b,c: góc FAE+góc FHE=180 độ

=>FAEH nội tiếp

=>góc HFE=góc HAE=góc C

Xét ΔHFE vuông tại H và ΔHCA vuông tại H có

góc HFE=góc HCA

=>ΔHFE đồng dạng với ΔHCA

=>HF/HC=HE/HA

=>HF*HA=HC*HE

1.1.

\(sinx=\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\dfrac{1}{4}+k2\pi\\x=\pi-arcsin\dfrac{1}{4}+k2\pi\end{matrix}\right.\)

1.2.

\(sinx=-\dfrac{\sqrt{3}}{2}\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

1.3.

\(sinx=-1\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\)

1 B

2 A

3 C

4 B

5 C

6 B

7 C

8 A

9 B

10 D

11 B

12 D

13C

14 D

15 B

Từ pt (E) ta xác định được: \(a=5;b=3;c=4\)

\(F_1F_2=2c=8\Rightarrow\) chu vi tam giác \(MF_1F_2=MF_1+MF_2+F_1F_2=2a+2c=18\)

\(\Rightarrow\) nửa chu vi \(p=9\)

Tam giác \(MF_1F_2\) vuông tại M \(\Rightarrow OM=\dfrac{1}{2}F_1F_2=4\)

Gọi \(M\left(x;y\right)\Rightarrow\overrightarrow{OM}=\left(x;y\right)\Rightarrow OM^2=x^2+y^2=16\)

\(\Rightarrow x^2=16-y^2\)

Thay vào pt (E):

\(\dfrac{16-y^2}{25}+\dfrac{y^2}{9}=1\Rightarrow y^2=\dfrac{81}{16}\Rightarrow\left|y\right|=\dfrac{9}{4}\)

\(S_{MF_1F_2}=\dfrac{1}{2}F_1F_2.d\left(M;F_1F_2\right)=\dfrac{1}{2}.2c.\left|y\right|=9\)

\(\Rightarrow r=\dfrac{S_{MF_1F_2}}{p}=1\)

13.

\(\dfrac{sinx-sin3x+sin5x}{cosx-cos3x+cos5x}\)

\(=\dfrac{2sin3x.cos2x-sin3x}{2cos3x.cos2x-cos3x}\)

\(=\dfrac{\left(2cos2x-1\right)sin3x}{\left(2cos2x-1\right)cos3x}\)

\(=\dfrac{sin3x}{cos3x}=tan3x\)

14)\(\dfrac{1-sin2x}{1+sin2x}=\dfrac{sin\dfrac{\pi}{4}-sin2x}{sin\dfrac{\pi}{4}+sin2x}\)\(=\dfrac{2.sin\left(\dfrac{\pi}{4}-x\right)cos\left(\dfrac{\pi}{4}+x\right)}{2sin\left(\dfrac{\pi}{4}+x\right)cos\left(\dfrac{\pi}{4}-x\right)}\)\(=tan\left(\dfrac{\pi}{4}-x\right)cot\left(\dfrac{\pi}{4}+x\right)\)\(=tan\left[\dfrac{\pi}{2}-\left(\dfrac{\pi}{4}+x\right)\right]cot\left(\dfrac{\pi}{4}+x\right)\)

\(=cot\left(\dfrac{\pi}{4}+x\right).cot\left(\dfrac{\pi}{4}+x\right)\)

\(=cot^2\left(\dfrac{\pi}{4}+x\right)\)

15)\(\dfrac{sinx+sin3x+sin5x}{cosx+cos3x+cos5x}\)\(=\dfrac{\left(sinx+sin5x\right)+sin3x}{\left(cosx+cos5x\right)+cos3x}\)\(=\dfrac{2sin3x.cos\left(-2x\right)+sin3x}{2cos3x.cos\left(-2x\right)+cos3x}=\dfrac{sin3x\left[2cos\left(-2x\right)+1\right]}{cos3x\left[2cos\left(-2x\right)+1\right]}\)\(=tan3x\)

16)\(\dfrac{cos5x-cosx}{sin4x+sin2x}=\dfrac{-2.sin3x.sin2x}{2.sin3x.cosx}\)\(=\dfrac{-2.sinx.cosx}{cosx}\)\(=-2.sinx\)

17)\(\dfrac{sin^4x-cos^4x+cos^2x}{2\left(1-cosx\right)}\)\(=\dfrac{cos^2x-\left(sin^2x+cos^2x\right)\left(cos^2x-sin^2x\right)}{2\left(1-cosx\right)}\)\(=\dfrac{cos^2x-\left(cos^2x-sin^2x\right)}{2\left(1-cosx\right)}\)\(=\dfrac{sin^2x}{2\left(1-cosx\right)}\)

\(=\dfrac{1-cos^2x}{2\left(1-cosx\right)}=\dfrac{1+cosx}{2}\)\(=cos^2\dfrac{x}{2}\)

18) (Xem lại đề)

19)\(\dfrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)\(=\dfrac{\left(1+cos2x\right)+\left(cosx+cos3x\right)}{2cos^2x+cosx-1}\)\(=\dfrac{2cos^2x+2.cos2x.cosx}{2cos^2x+cosx-1}\)

\(=\dfrac{2cosx\left(cosx+cos2x\right)}{2cos^2x+cosx-1}\)\(=\dfrac{2cosx\left(cosx+2cos^2-1\right)}{2cos^2x+cosx-1}\)\(=2cosx\)