1.spent

 2.has asked

 4.wanted

 6.taking

 7.have thought

 8.has worked

A(x_A,y_A)

\(Q_{\left(O,\dfrac{\pi}{2}\right)}\left(A\right)=C\)       \(\Rightarrow C\left(-y_A,x_A\right)\)

\(FeCl_3+3Ca\left(OH\right)_2\rightarrow2Fe\left(OH\right)_3\downarrow+3CaCl\)

\(Fe^{3+}+3OH^-\rightarrow Fe\left(OH\right)_3\downarrow\)

4000-5=3995-11=3984

a) \(sinx=\dfrac{1}{2}\Rightarrow sinx=sin\dfrac{\pi}{6}\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

b) Pt: \(\Rightarrow cos\left(x+\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{3}}{2}=cos\dfrac{5\pi}{6}\)

....bạn tự tìm x nhé!

c)Pt: \(\Rightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-4\end{matrix}\right.\)  (loại sinx=-4 vì sinx\(\in[-1,1]\)

Để lực tổng hợp tác dụng lên điện tích bằng q₀=0 thì \(\overrightarrow{F_{10}}+\overrightarrow{F_{20}}=\overrightarrow{0}\)

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{F_{10}}\uparrow\downarrow\overrightarrow{F_{20}}\\F_{10}=F_{20}\end{matrix}\right.\)

Ta có \(MB-MA=r_1-r_2=10\)                              (1)

Mà \(F_{10}=F_{20}\Rightarrow k\cdot\dfrac{\left|q_1q_0\right|}{r^2_1}=k\cdot\dfrac{\left|q_2q_0\right|}{r^2_2}\)\(\Rightarrow\dfrac{r_1}{r_2}=2\)    (2)

Từ 1 và 2 \(\Rightarrow\left\{{}\begin{matrix}r_1=20cm\\r_2=10cm\end{matrix}\right.\)

\(\Rightarrow\) M cách A 10cm và cách B 20cm

Pt: \(\Rightarrow-3\left(cos^2x-sin^2x\right)-\sqrt{3}sin2x=0\)

      \(\Rightarrow-3cos2x-\sqrt{3}sin2x=0\)

      \(\Rightarrow sin2x+\sqrt{3}cos2x=0\)

      \(\Rightarrow2sin\left(2x+\dfrac{\pi}{3}\right)=0\)  \(\Rightarrow sin\left(2x+\dfrac{\pi}{3}\right)=0\)

                                            \(\Rightarrow2x+\dfrac{\pi}{3}=k\pi\left(k\in Z\right)\)

                                            \(\Rightarrow x=-\dfrac{\pi}{6}+k\dfrac{\pi}{2}\)

\(2sin^2x+3cosx-3=0\Rightarrow2\left(1-cos^2x\right)+3cosx-3=0\)

                                       \(\Rightarrow-2cos^2x+3cosx-1=0\)

                                       \(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

....Bạn tự tìm x nhé!

\(\sqrt{3}cot^2x+\left(1-\sqrt{3}\right)cotx-1=0\)

Đk: \(sinx\ne0\Rightarrow x\ne m\pi\)

Pt: \(\Rightarrow\left[{}\begin{matrix}cotx=1\\cotx=-\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)(tmđk \(x\ne m\pi\))

a) \(cos\left(4x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\Rightarrow cos\left(4x+\dfrac{\pi}{3}\right)=cos\dfrac{\pi}{6}\)

                                      \(\Rightarrow\left[{}\begin{matrix}4x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\4x+\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

                                            ..... bạn tự tìm x nhé!

b)\(sin^2x-3sin3x+2=0\)\(\Rightarrow sin^2x-3\left(3sinx-4sin^3x\right)+2=0\)

\(\Rightarrow12sin^3x+sin^2x-9sinx+2=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\dfrac{2}{3}\\sinx=\dfrac{1}{4}\end{matrix}\right.\)    \(\Rightarrow\).... bạn tự tìm x nhé!

c)\(tan\left(2x+10^o\right)=\sqrt{3}\Rightarrow tan\left(2x+10^o\right)=tan60^o\)

                                     \(\Rightarrow2x+10^o=60^o+k180^o\)

                                     \(\Rightarrow x=25^o+k90^o\left(k\in Z\right)\)

d) \(tanx\cdot cot2x=1\)

Đk: \(\left\{{}\begin{matrix}cosx\ne0\\sin2x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+m\pi\\x\ne m\dfrac{\pi}{2}\end{matrix}\right.\)

Pt: \(\Rightarrow tanx=tan2x\Rightarrow x=2x+k\pi\)

                                 \(\Rightarrow x=k\pi\)

  Đối chiếu với đk trên thỏa mãn đk\(\Rightarrow x=k\pi\)