Giúp mình bài này đc không ạ, mình cần gấp lắm😥
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KY
15 tháng 7 2021
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..
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
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: \(\dfrac{\left(x-3\right)^2}{16}+\dfrac{\left(y+2\right)^2}{9}=1\)
TQ
0
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2021
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:
\(3t=2\left(t^2-1\right)\)
\(\Leftrightarrow2t^2-3t-2=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2>\sqrt{2}\left(loại\right)\\t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow sinx+cosx=-\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{8}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\\x+\dfrac{\pi}{4}=\pi-arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\\x=\dfrac{3\pi}{4}-arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2021
b.
ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(1+\dfrac{sinx}{cosx}=2\sqrt{2}sinx\)
\(\Rightarrow sinx+cosx=2\sqrt{2}sinx.cosx\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}sin2x\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x+\dfrac{\pi}{4}+k2\pi\\2x=\dfrac{3\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2021
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.
\(0< a< 90^0\Rightarrow cosa>0\)
\(\Rightarrow cosa=\sqrt{1-sin^2a}=\dfrac{4}{5}\)
\(tana=\dfrac{sina}{cosa}=\dfrac{3}{4}\)
\(cota=\dfrac{1}{tana}=\dfrac{4}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2021
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.
\(A=\dfrac{\dfrac{cosa}{sina}+\dfrac{3sina}{cosa}}{\dfrac{2cosa}{sina}+\dfrac{sina}{cosa}}=\dfrac{cos^2a+3sin^2a}{2cos^2a+sin^2a}=\dfrac{cos^2a+3\left(1-cos^2a\right)}{2cos^2a+\left(1-cos^2a\right)}\)
\(=\dfrac{3-2cos^2a}{1+cos^2a}=\dfrac{19}{13}\)
TL
1
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2021
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)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2021
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.
\(B=cos^2x+sin^2y+2cosx.siny+cos^2y+sin^2x-2sinx.cosy\)
\(=\left(cos^2x+sin^2x\right)+\left(cos^2y+sin^2y\right)-2\left(sinx.cosy-cosx.siny\right)\)
\(=2-2sin\left(x-y\right)=2-2sin\left(\dfrac{\pi}{3}\right)=...\)
1 tháng 8 2021
1D 2D 3D 4A 5A 6D 7A 8B 9D 10A 11B 12C 13A 14D 15C 16B 17C 18B
1 Jill reminded John to do the washing up
2 The police ordered his men to search all the shops on that street
3 She blamed me for ignoring the notice about life-saving equipment
4 My aunt advised me not to argut with my father
5 Stella congratulated Jeff on having got an promotion at last
6 Kevin apoligized to Sarah for making her angry
7 The man warn his son to put down the gun
8 Ron denied being in the town on the night of the robbery
9 Ted promise to pay back the money at the end of that month
10 George encouraged Susan to send her story to the magazine
11 Natalie accused Tom of lying to her