\(AA=0,25+0,4.\dfrac{1-\left(\dfrac{1}{2}\right)^3}{2}=0,425\)

\(Aa=0,4.\dfrac{1}{2^3}=0,05\)

\(aa=0,35+0,4.\dfrac{1-\left(\dfrac{1}{2}\right)^3}{2}=0,525\)

Vậy tỉ lệ kiểu gen của quần thể sau 3 thế hệ tự thụ phấn là: 0,425AA:0,05Aa:0,525aa

c. \(5cos2x-12sinx=13\)

\(\Leftrightarrow5-10sin^2x-12sinx-13=0\)

\(\Leftrightarrow-10sin^2x-12sinx-8=0\)

Phương trình vô nghiệm

Vậy.....

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

\(\Leftrightarrow\frac{1}{\sqrt{2}}.sinx+\frac{1}{\sqrt{2}}cosx=1\)

\(\Leftrightarrow cos\frac{\pi}{4}.sinx+sin\frac{\pi}{4}.cosx=1\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi,k\in Z\)

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

Vậy....

Câu 1:

b. \(sin7x+\sqrt{3}cos7x=\sqrt{2}\)

\(\Leftrightarrow\frac{1}{2}sin7x+\frac{\sqrt{3}}{2}cos7x=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow cos\frac{\pi}{3}.sin7x+sin\frac{\pi}{3}.cos7x=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(7x+\frac{\pi}{3}\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{3}=\frac{\pi}{4}+k2\pi,k\in Z\\7x+\frac{\pi}{3}=\frac{3}{4}\pi+k2\pi,k\in Z\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{84}+k\frac{2}{7}\pi\\x=\frac{5\pi}{84}+k\frac{2}{7}\pi\end{matrix}\right.k\in Z}\)

Vậy....

N=\(\frac{3910.2}{3,4}=2300\)

A=\(2300.23\%=529\)

a. \(l=408nm\) \(\Rightarrow N=\frac{l.2}{0,34}=\frac{408.2}{0,34}=2400\) (nu)

b. \(N_1=N_2=\frac{N}{2}=\frac{2400}{2}=1200\left(nu\right)\)

\(A=T=22\%N=22\%.2400=528\left(nu\right)\)

\(G=X=\frac{N-\left(A+T\right)}{2}=\frac{2400-\left(528+528\right)}{2}=672\left(nu\right)\)

\(A_1=T_2=20\%.N_1=20\%.1200=240\left(nu\right)\)

\(T_1=A_2=A-A_1=528-240=288\left(nu\right)\)

\(\Rightarrow\%T_1=\frac{288}{1200}.100=24\%\)

\(G_1=X_2=35\%.N_2=35\%.1200=420\left(nu\right)\)

\(\Rightarrow\%G_1=\frac{420}{1200}.100=35\%\)

\(X_1=G_2=G-G_1=672-420=252\left(nu\right)\)

\(\Rightarrow\%X_1=\frac{252}{1200}.100=21\%\)