t cũng ra y như c :D

chả ra đáp án A còn gì nữa :3 :3 

d/

\(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}cos\left(\frac{x}{5}-\frac{\pi}{12}\right)-\frac{\sqrt{3}}{2}sin\left(\frac{x}{5}-\frac{\pi}{12}\right)\right)=sin\left(\frac{x}{5}+\frac{2\pi}{3}\right)-sin\left(\frac{3x}{5}+\frac{\pi}{6}\right)\)

\(\Leftrightarrow\sqrt{2}cos\left(\frac{x}{5}-\frac{\pi}{12}+\frac{\pi}{3}\right)=2cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)sin\left(\frac{\pi}{4}-\frac{x}{5}\right)\)

\(\Leftrightarrow cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=\sqrt{2}cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)cos\left(\frac{x}{5}-\frac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=0\\cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{5}-\frac{\pi}{4}=\frac{\pi}{2}+k\pi\\\frac{2x}{5}+\frac{5\pi}{12}=\frac{\pi}{4}+k2\pi\\\frac{2x}{5}+\frac{5\pi}{12}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{15\pi}{4}+k5\pi\\x=-\frac{5\pi}{12}+k5\pi\\x=-\frac{5\pi}{3}+k5\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow\sqrt{3}sin\left(x-\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}-x\right)=2sin1972x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin\left(x-\frac{\pi}{3}\right)+\frac{1}{2}cos\left(x-\frac{\pi}{3}\right)=sin1972x\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}+\frac{\pi}{6}\right)=sin1972x\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin1972x\)

\(\Leftrightarrow\left[{}\begin{matrix}1972x=x-\frac{\pi}{6}+k2\pi\\1972x=\frac{7\pi}{6}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{11826}+\frac{k2\pi}{1971}\\x=\frac{7\pi}{11838}+\frac{k2\pi}{1973}\end{matrix}\right.\)

a, Ta có:

Hai hàm sóng trực giao nhau khi  \(I=\int\psi_{1s}.\psi_{2s}d\psi=0\) \(\Leftrightarrow I=\iiint\psi_{1s}.\psi_{2s}dxdydz=0\)

Chuyển sang tọa độ cầu ta có:  \(\begin{cases}x=r.\cos\varphi.sin\theta\\y=r.\sin\varphi.sin\theta\\z=r.\cos\theta\end{cases}\)

\(\Rightarrow\)\(I=\frac{a^3_o}{4.\sqrt{2.\pi}}\int\limits^{\infty}_0\left(2-\frac{r}{a_o}\right).e^{-\frac{3.r}{2.a_o}}.r^2.\sin\theta dr\int\limits^{2\pi}_0d\varphi\int\limits^{\pi}_0d\theta\)

       \(=a^3_o.\sqrt{\frac{\pi}{2}}\)(.\(2.\int\limits^{\infty}_0r^2.e^{-\frac{3.r}{2.a_o}}dr-\frac{1}{a_o}.\int\limits^{\infty}_0r^3.e^{-\frac{3.r}{2.a_o}}dr\))

         \(=a_o.\sqrt{\frac{\pi}{2}}.\left(2.I_1-\frac{1}{a_o}.I_2\right)\)  

Tính \(I_1\):

Đặt \(r^2=u\); \(e^{-\frac{3r}{2a_o}}dr=dV\)

\(\Rightarrow\begin{cases}2.r.dr=du\\-\frac{2a_o}{3}.e^{-\frac{3r}{2a_o}}=V\end{cases}\)    \(\Rightarrow I_1=-r^2.\frac{2a_o}{3}.e^{-\frac{3r}{2a_o}}+\frac{4.a_o}{3}.\int\limits^{\infty}_0r.e^{-\frac{3r}{2a_o}}dr\)\(=0+\frac{4a_o}{3}.I_{11}\)

Tính \(I_{11}\):

Đặt r=u; \(e^{-\frac{3r}{2a_o}}dr=dV\)\(\Rightarrow\begin{cases}dr=du\\-\frac{2a_o}{3}.e^{-\frac{3r}{2a_o}}=V\end{cases}\)\(\Rightarrow I_{11}=0+\frac{2a_0}{3}.\int\limits^{\infty}_0e^{-\frac{3r}{2a_o}}dr=\frac{4a^2_o}{9}\)

\(\Rightarrow2.I_1=2.\frac{4a_o}{3}.\frac{4a_o^2}{9}=\frac{32a^3_o}{27}\)

Tính \(I_2\):

Đặt \(r^2=u;e^{-\frac{3r}{2a_o}}dr=dV\) \(\Rightarrow\)\(3r^2dr=du;-\frac{2a_o}{3}.e^{-\frac{3r}{2a_o}}=V\)

\(\Rightarrow I_2=0+2.a_o.\int\limits^{\infty}_0r^2.e^{-\frac{3r}{2a_o}}dr\)\(\Rightarrow\frac{1}{a_o}.I_2=2a_o.\frac{16a^3_o}{27}.\frac{1}{a_o}=\frac{32a^3_o}{27}\)

\(\Rightarrow I=a^3_o.\sqrt{\frac{\pi}{2}}.\left(\frac{32a^3_o}{27}-\frac{32a^3_o}{27}\right)=0\)

Vậy hai hàm sóng này trực giao với nhau.

b,

Xét hàm \(\Psi_{1s}\):

Hàm mật độ sác xuất là: \(D\left(r\right)=\Psi^2_{1s}=\frac{1}{\pi}.a^3_o.e^{-\frac{2r}{a_o}}\)

\(\Rightarrow D'\left(r\right)=-\frac{2.a_o^2}{\pi}.e^{-\frac{2r}{a_o}}=0\)

\(\Rightarrow\)Hàm đạt cực đại khi \(r\rightarrow o\) nên hàm sóng có dạng hình cầu.

Xét hàm \(\Psi_{2s}\):

Hàm mật độ sác xuất: \(D\left(r\right)=\Psi_{2s}^2=\frac{a^3_o}{32}.\left(2-\frac{r}{a_o}\right)^2.e^{-\frac{r}{a_0}}\)\(\Rightarrow D'\left(r\right)=\left(2-\frac{r}{a_o}\right).e^{-\frac{r}{a_o}}.\left(-4+\frac{r}{a_o}\right)=0\)

\(\Rightarrow r=2a_o\Rightarrow D\left(r\right)=0\); \(r=4a_o\Rightarrow D\left(r\right)=\frac{a^3_o}{8}.e^{-4}\)

Vậy hàm đạt cực đại khi \(r=4a_o\), tại \(D\left(r\right)=\frac{a^3_o}{8}.e^{-4}\)

hai hàm trực giao: I=\(\int\)\(\Psi\)*\(\Psi\)d\(\tau\)=0

Ta có: I=\(\int\limits^{ }_x\)\(\int\limits^{ }_y\)\(\int\limits^{ }_z\)\(\Psi\)*\(\Psi\)dxdydz=0

           =\(\int\limits^{ }_r\)\(\int\limits^{ }_{\theta}\)\(\int\limits^{ }_{\varphi}\)\(\Psi\)_1s\(\Psi\)_2sr²sin\(\theta\)drd\(\theta\)d\(\varphi\)

          =\(\int\limits^{\infty}_0\)\(\int\limits^{\pi}_0\)\(\int\limits^{2\pi}_0\)(2-\(\frac{r}{a_0}\)).e^-3r/a₀r²sin\(\theta\)drd\(\theta\)d\(\varphi\)

          =C.\(\int\limits^{\infty}_0\)(2-\(\frac{r}{a_0}\)).e^-3r/a₀r2dr.\(\int\limits^{\pi}_0\)sin\(\theta\)\(\int\limits^{2\pi}_0\)d\(\varphi\)

 với C=\(\frac{1}{4\sqrt{2\pi}}\)a₀^-3

 Xét tích phân: J=\(\int\limits^{\infty}_0\)(2-\(\frac{r}{a_0}\)).e^-3r/a₀r²dr

 =\(\int\limits^{\infty}_0\)(2r²- \(\frac{r^3}{a_0}\)).e^-3r/a₀dr

 =\(\int\limits^{\infty}_0\)(2r²- \(\frac{r^3}{a_0}\)).\(\frac{-2a_0}{3}\)de^-3r/a₀

  =\(\frac{-2a_0}{3}\).((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀\(-\)\(\int\)(4r-\(\frac{3r^2}{a_0}\))e^-3r/adr)

 =\(\frac{-2a_0}{3}\)((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ - \(\int\)(4r-\(\frac{3r^2}{a_0}\)).\(\frac{-2a_0}{3}\)de^-3r/a)

 =\(\frac{-2a_0}{3}\)((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ +\(\frac{2a_0}{3}\).((4r-\(\frac{3r^2}{a_0}\))e^-3r/a₀ - \(\int\)(4 - \(\frac{6r}{a_0}\))e^-3r/a₀dr))

 =\(\frac{-2a_0}{3}\)((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ +\(\frac{2a_0}{3}\).((4r-\(\frac{3r^2}{a_0}\))e^-3r/a₀- \(\int\)(4 - \(\frac{6r}{a_0}\))\(\frac{-2a_0}{3}\).de^-3r/a₀))

 =\(\frac{-2a_0}{3}\)(((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ +\(\frac{2a_0}{3}\).((4r-\(\frac{3r^2}{a_0}\))e^-3r/a₀+\(\frac{2a_0}{3}\)((4-\(\frac{6r}{a_0}\)).e^-3r/a₀ + \(\int\)(\(\frac{6}{a_0}\)e^-3r/a₀dr)))

=\(\frac{-2a_0}{3}\)(((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ +\(\frac{2a_0}{3}\).((4r-\(\frac{3r^2}{a_0}\))e^-3r/a₀+\(\frac{2a_0}{3}\)((4-\(\frac{6r}{a_0}\)).e^-3r/a₀ + \(\int\)(\(\frac{6}{a_0}\).\(\frac{-2a_0}{3}\)de^-3r/a₀)))

=\(\frac{-2a_0}{3}\)((((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ +\(\frac{2a_0}{3}\).((4r-\(\frac{3r^2}{a_0}\))e^-3r/a₀+\(\frac{2a_0}{3}\)((4-\(\frac{6r}{a_0}\)).e^-3r/a₀ - 4.e^-3r/a₀))))

chịu thua

giải ko ra hay sao ạ

a)      

\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) =  - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} =  - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi  + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x =  - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

b)     \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

c)       

\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x =  - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x =  - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x =  - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)

d)      

\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x =  - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x =  - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x =  - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)

e)      

\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)

f)       

\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x =  - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)

Thể tích khối cầu là: \(\frac{4}{3}\pi R^3\)

Độ dài cạnh hình vuông là: \(R\sqrt{2}\).

Thể tích của khối trụ là: \(\left(\frac{R\sqrt{2}}{2}\right)^2\pi\left(R\sqrt{2}\right)=\frac{\pi R^3\sqrt{2}}{2}\)

Phần thể tích khối cầu nằm ngoài khối trụ là: \(\frac{\pi R^3}{6}\left(8-3\sqrt{2}\right)\).

V(tru) =π.R^2/4.R√3
V(cau)=4/3.π.R^3
Vnktru=πR^3(4/3-√3/4)
=πR^3/12.(16-3√3)
Chọn (B).

Vtru =π.R^2/4.R√3
 

Vcau=4/3.π.R^3
 

Vnktru=πR^3(4/3-√3/4)
 

=πR^3/12.(16-3√3)
 

Chọn (B).

Mà \(\Delta\)p_x.\(\Delta\)x=m.\(\Delta\)V_x.\(\Delta\)_x =\(\frac{h}{2\pi}\)

=> \(\Delta\)x = \(\frac{6.625.10^{-34}}{2\pi.10^6.9,1.10^{-31}}\)=1,16.10^-10

TXĐ: `D=RR\\{π/2+kπ ; -π/4 +kπ}`

Mà `-π/2+k2π` và `π/2+k2π \in π/2 +kπ`

`=>` Không nằm trong TXĐ.