help với ạ
help với ạ
I'm going to visit Mui Ne this summer
help với ạ
use
is jane watching
doesn't go
is ating
do you study
am not listening- am playing
is raining- rains
are reading
am often playing=>often plays
is wanting=>wants
doesn't =>don't
have=>are having
skip=>skipping
is having=>has
cook =>am cooking
IV,
1. use
2. Is Jane watching...?
3. doesn't go
4. is eating
5. Do you sutdy...?
6. amn't listening... am playing
7. is going to rain... rains
8. are reading
V,
1. playing=> play
2. is wanting=> want
3. doesn't=> don't
4. have=> are having
5. skip=> skipping
6. is having=> has
7. cook=> am cooking
help với ạ :(((
Help với ạ!!!!!
c.
ĐKXĐ: \(sinx\ne0\Rightarrow x\ne k\pi\)
\(1-\dfrac{\sqrt{3}cosx}{sinx}-4cosx=0\)
\(\Rightarrow sinx-\sqrt{3}cosx-4sinx.cosx=0\)
\(\Leftrightarrow sinx-\sqrt{3}cosx=2sin2x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin2x\)
\(\Leftrightarrow sin2x=sin\left(x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x-\dfrac{\pi}{3}+k2\pi\\2x=\dfrac{4\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
a.
\(\Leftrightarrow3sin3x-4sin^34x-\sqrt{3}cos9x=2sin2x\)
\(\Leftrightarrow sin9x-\sqrt{3}cos9x=2sin2x\)
\(\Leftrightarrow\dfrac{1}{2}sin9x-\dfrac{\sqrt{3}}{2}cos9x=sin2x\)
\(\Leftrightarrow sin\left(9x-\dfrac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}9x-\dfrac{\pi}{3}=2x+k2\pi\\9x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{21}+\dfrac{k2\pi}{7}\\x=\dfrac{4\pi}{33}+\dfrac{k2\pi}{11}\end{matrix}\right.\)
b.
\(\Leftrightarrow\sqrt{3}cosx+sin3x-sinx-sin3x=\sqrt{2}\)
\(\Leftrightarrow\sqrt{3}cosx-sinx=\sqrt{2}\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{6}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{6}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{5\pi}{12}+k2\pi\end{matrix}\right.\)
Help với ạ!!!!!
Help với ạ
\(\dfrac{2}{7}=\dfrac{10}{35}=\dfrac{12}{42}=\dfrac{16}{56}\)
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