Cho P: A a B b D d M E m e × a a B B D d M e m e  hoán vị gen diễn ra với tần số f = 0.2 Theo lí thuyết, tỉ lệ F có 6 gen lặn là

A.  1/4.                           

B. 15/256.                      

C. 1/36.                          

D. 3/64.

\(a.b=4500\left\{\begin{matrix}BCNN\left(a;b\right)=300\\UCLN\left(a;b\right)=15\end{matrix}\right.\)

\(UCLN\left(a;b\right)=15\Rightarrow\left\{\begin{matrix}a=15.m\\b=15.n\end{matrix}\right.\)

\(\)( \(m;n\in\) N*, \(UCLN\left(m;n\right)=1,\left(m< n\right)\))

\(a.b=4500\)

\(\Leftrightarrow15.m.15.n=4500\)

\(\Leftrightarrow m.n=4500:\left(15.15\right)\)

\(\Leftrightarrow m.n=20\)

Vì \(a+15=b\) nên \(a=60\) và \(b=75\)

\(\left(2n-5\right)⋮n\)

\(\Rightarrow n+n-5⋮n\)

Vì \(n+n⋮n\) nên \(-5⋮n\)

\(\Leftrightarrow2n-5\inƯ\left(-5\right)=\left\{-5;-1;1;5\right\}\)

Vậy \(2n-5\in\left\{0;2;3;5\right\}\)

\(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)

\(\Rightarrow\frac{B}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(\Rightarrow\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)

\(\Rightarrow\frac{B}{7}=\frac{1}{2}-\frac{1}{28}\)

\(\Rightarrow\frac{B}{7}=\frac{14}{28}-\frac{1}{28}\)

\(\Rightarrow\frac{B}{7}=\frac{13}{28}\)

\(\Rightarrow B=\frac{13}{28}.7\)

\(\Rightarrow B=\frac{13}{4}\)

\(S=1.2+2.3+3.4+4.5+5.6+...+99.100\)

\(3S=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3\)

\(3S=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)\(1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101+98.99.100\)

\(3S=\left(1.2.3-1.2.3\right)+\left(2.3.4-2.3.4\right)+...+\left(98.99.100-98.99.100\right)+99.100.101\)

\(3S=99.100.101=9999000\)

\(S=9999000:3=3333000\)

\(\Rightarrow S=3333000\)

Ta có:

\(Q=\frac{2010+2011+2012}{2011+2012+2013}=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+1013}+\frac{2012}{2011+2012+2013}\)

Vì \(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)

\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)

\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)

\(\Rightarrow P>Q\)

CONAN và KUDO SHINICHI tớ thấy cậu ấy làm đúng rồi đó

Bài 13:

a) \(n-1\inƯ\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

Vậy \(n-1\in\left\{-7;-3;-1;0;2;3;5;9\right\}\)

b) \(n+3\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)

Vậy \(n+3\in\left\{-15;-9;-7;-6;-5;-4;-2;-1;0;1;3;9\right\}\)