D C D B C C C D A 

Sam puts up the decorations.

Five guests came to the party.

Molly and Sam play video games with their cousins

The family was celebrating too early. Dad's birthday was two months away.

relationship

celebration

married

golden

quietly

celebratory

refer

1. he dislike being called " the liar " => He dislike people..CALLING HIM THE LIAR.

2. The police are following the suspects => The suspects ..ARE BEING FOLLOWED BY THE POLICE.

3. She always expects to be admired by everybody => She always expects everybody..TO ADMIRE HER...

4. Someone stole his car two days ago => He had ..HIS CAR STOLEN BY SOMEONE TWO DAYS AGO..

d d d a d a a d a b c a d c b a

Gọi \(M\left(x;y\right)\) là 1 điểm bất kì trên (E) \(\Rightarrow\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\) (1)

Gọi \(M'\left(x';y'\right)\) là ảnh của M qua phép tịnh tiến \(\overrightarrow{v}\Rightarrow M'\in\left(E'\right)\) với (E') là ảnh của (E) qua phép tịnh tiến nói trên

\(\left\{{}\begin{matrix}x'=x+3\\y'=y-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=x'-3\\y=y'+2\end{matrix}\right.\)

Thế vào (1):

\(\dfrac{\left(x'-3\right)^2}{16}+\dfrac{\left(y'+2\right)^2}{9}=1\)

Hay pt (E') có dạng: \(\dfrac{\left(x-3\right)^2}{16}+\dfrac{\left(y+2\right)^2}{9}=1\)

a.

Đặt \(sinx+cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow1+2sinx.cosx=t^2\Rightarrow2sinx.cosx=t^2-1\)

Phương trình trở thành:

\(3t=2\left(t^2-1\right)\)

\(\Leftrightarrow2t^2-3t-2=0\)

\(\Rightarrow\left[{}\begin{matrix}t=2>\sqrt{2}\left(loại\right)\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow sinx+cosx=-\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{8}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\\x+\dfrac{\pi}{4}=\pi-arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\\x=\dfrac{3\pi}{4}-arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\end{matrix}\right.\)

b.

ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(1+\dfrac{sinx}{cosx}=2\sqrt{2}sinx\)

\(\Rightarrow sinx+cosx=2\sqrt{2}sinx.cosx\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}sin2x\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=sin2x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\)

c.

\(\Leftrightarrow1+sinx+cosx+sinx.cosx=2\)

\(\Leftrightarrow sinx+cosx+sinx.cosx=1\)

Đặt \(sinx+cosx=t\in\left[-\sqrt[]{2};\sqrt{2}\right]\)

\(\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)

Phương trình trở thành:

\(t+\dfrac{t^2-1}{2}=1\)

\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow sinx+cosx=1\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow...\)

a.

\(90^0< a< 180^0\Rightarrow cosa< 0\)

\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\dfrac{2\sqrt{2}}{3}\)

\(tana=\dfrac{sina}{cosa}=-\dfrac{\sqrt{2}}{4}\)

b.

\(0< a< 90^0\Rightarrow cosa>0\)

\(\Rightarrow cosa=\sqrt{1-sin^2a}=\dfrac{4}{5}\)

\(tana=\dfrac{sina}{cosa}=\dfrac{3}{4}\)

\(cota=\dfrac{1}{tana}=\dfrac{4}{3}\)

c.

\(A=\dfrac{\dfrac{sina}{cosa}+\dfrac{3cosa}{sina}}{\dfrac{sina}{cosa}+\dfrac{cosa}{sina}}=\dfrac{sin^2a+3cos^2a}{sin^2a+cos^2a}=1+2cos^2a=\dfrac{17}{8}\)

d.

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

\(=\dfrac{3-2cos^2a}{1+cos^2a}=\dfrac{19}{13}\)

e.

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

\(=3-sin^2a=\dfrac{26}{9}\)

f.

\(C=\dfrac{\dfrac{2sina}{cosa}+\dfrac{3cosa}{cosa}}{\dfrac{sina}{cosa}+\dfrac{cosa}{cosa}}=\dfrac{2tana+3}{tana+1}=\dfrac{7}{3}\)

g.

\(C=\dfrac{\dfrac{3sina}{sina}-\dfrac{4cosa}{sina}}{\dfrac{cosa}{sina}-\dfrac{2sina}{sina}}=\dfrac{3-4cota}{cota-2}=1+\sqrt{5}\)

Anh nghĩ với bài kiểm tra em nên tự làm nhé. 

a.

\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{6}}{3}\)

\(cos\left(x+\dfrac{\pi}{3}\right)=cosx.cos\left(\dfrac{\pi}{3}\right)-sinx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{6}-3}{6}\)

b.

\(\pi< x< \dfrac{3\pi}{2}\Rightarrow sinx< 0\)

\(\Rightarrow sinx=-\sqrt{1-cos^2x}=-\dfrac{5}{13}\)

\(B=sin\left(\dfrac{\pi}{3}-x\right)=sin\left(\dfrac{\pi}{3}\right).cosx-cos\left(\dfrac{\pi}{3}\right).sinx=...\) (bạn tự thay số bấm máy)

c.

\(A=cos^2x+cos^2y+2cosx.cosy+sin^2x+sin^2y+2sinx.siny\)

\(=\left(cos^2x+sin^2x\right)+\left(cos^2y+sin^2y\right)+2\left(cosx.cosy+sinx.siny\right)\)

\(=1+1+2cos\left(x-y\right)\)

\(=2+2cos\left(\dfrac{\pi}{3}\right)=...\)

d.

\(B=cos^2x+sin^2y+2cosx.siny+cos^2y+sin^2x-2sinx.cosy\)

\(=\left(cos^2x+sin^2x\right)+\left(cos^2y+sin^2y\right)-2\left(sinx.cosy-cosx.siny\right)\)

\(=2-2sin\left(x-y\right)=2-2sin\left(\dfrac{\pi}{3}\right)=...\)

1D 2D 3D 4A 5A 6D 7A 8B 9D 10A 11B 12C 13A 14D 15C 16B 17C 18B