Bài 1:

a: Đặt 14,4-3,6t^2=0

=>3,6t^2=14,4

=>t^2=4

=>t=2

b: TXĐ: [0;2]

TGT: [0;14,4]

9D

8B

7A

6A

5A

4D

3B

2B

1C

21D

20A
18D

19A

17A

16A

14C

a.

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

\(=1-\dfrac{1}{2}\left(2sina.cosa\right)^2=1-\dfrac{1}{2}sin^22a\)

\(=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4a\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4a\)

b.

\(sin^6a+cos^6a=\left(sin^2a+cos^2a\right)^2-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)

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

\(=1-\dfrac{3}{4}sin^22a=1-\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4a\right)\)

\(=\dfrac{5}{8}+\dfrac{3}{8}cos4a\)

e.

\(\dfrac{cos2a}{1+sin2a}=\dfrac{cos^2a-sin^2a}{sin^2a+cos^2a+2sina.cosa}=\dfrac{\left(cosa-sina\right)\left(cosa+sina\right)}{\left(sina+cosa\right)^2}=\dfrac{cosa-sina}{cosa+sina}\)

f.

\(cotx+tanx=\dfrac{cosx}{sinx}+\dfrac{sinx}{cosx}\)

\(=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{sinx.cosx}\)

\(=\dfrac{2}{2sinx.cosx}=\dfrac{2}{sin2x}\)

\(\dfrac{\pi}{2}< a< \pi\Rightarrow\left\{{}\begin{matrix}sina>0\\cosa< 0\end{matrix}\right.\)

\(sin2a=-\dfrac{5}{9}\Leftrightarrow sina.cosa=-\dfrac{5}{18}\Rightarrow cosa=-\dfrac{5}{18sina}\)

Thế vào \(sin^2a+cos^2a=1\)

\(sin^2a+\dfrac{25}{324sin^2a}=1\Leftrightarrow sin^4a-sin^2a+\dfrac{25}{324}=0\)

\(\Rightarrow\left[{}\begin{matrix}sin^2a=\dfrac{9-2\sqrt{14}}{8}\\sin^2a=\dfrac{9+2\sqrt{14}}{8}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}sina=\sqrt{\dfrac{9-2\sqrt{14}}{8}};cosa=-\sqrt{\dfrac{9+2\sqrt{14}}{8}}\\sina=\sqrt{\dfrac{9+2\sqrt{14}}{8}};cosa=-\sqrt{\dfrac{9-2\sqrt{14}}{8}}\end{matrix}\right.\)

\(y=\dfrac{2x^2-6x+10}{2\left(x^2+1\right)}=\dfrac{11\left(x^2+1\right)-9x^2-6x-1}{2\left(x^2+1\right)}=\dfrac{11}{2}-\dfrac{\left(3x+1\right)^2}{2\left(x^2+1\right)}\le\dfrac{11}{2}\)

\(y_{max}=\dfrac{11}{2}\) khi \(x=-\dfrac{1}{3}\)

\(y=\dfrac{2x^2-6x+10}{2\left(x^2+1\right)}=\dfrac{x^2+1+x^2-6x+9}{2\left(x^2+1\right)}=\dfrac{1}{2}+\dfrac{\left(x-3\right)^2}{2\left(x^2+1\right)}\ge\dfrac{1}{2}\)

\(y_{min}=\dfrac{1}{2}\) khi \(x=3\)

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.

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

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

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

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