Chọn đáp án D

Ko hiểu đề bài

Đề yêu cầu gì nhỉ?

a: ƯCLN(a,b)=45

=>\(\left\{{}\begin{matrix}a=45k\\b=45c\end{matrix}\right.\)

Ta có: a+b=270

=>45k+45c=270

=>45(k+c)=270

=>k+c=6

=>\(\left(k;c\right)\in\left\{\left(1;5\right);\left(2;4\right);\left(3;3\right);\left(4;2\right);\left(5;1\right)\right\}\)

=>\(\left(a,b\right)\in\left\{\left(45;225\right);\left(90;180\right);\left(135;135\right);\left(180;90\right);\left(225;45\right)\right\}\)

mà ƯCLN(a,b)=45

nên \(\left(a,b\right)\in\left\{\left(45;225\right);\left(225;45\right)\right\}\)

b: \(ƯCLN\left(a,b\right)=5\)

=>\(\left\{{}\begin{matrix}a=5k\\b=5c\end{matrix}\right.\)

\(5k\cdot5c=300\)

=>\(25\cdot k\cdot c=300\)

=>\(k\cdot c=\dfrac{300}{25}=12\)

=>\(\left(k;c\right)\in\left\{\left(1;12\right);\left(12;1\right);\left(2;6\right);\left(6;2\right);\left(3;4\right);\left(4;3\right)\right\}\)

=>\(\left(a,b\right)\in\left\{\left(5;60\right);\left(60;5\right);\left(10;30\right);\left(30;10\right);\left(15;20\right);\left(20;15\right)\right\}\)

mà ƯCLN(a,b)=5

nên \(\left(a,b\right)\in\left\{\left(5;60\right);\left(60;5\right);\left(15;20\right);\left(20;15\right)\right\}\)

Đáp án : ........................................................................................................................................................................................................................................................................................................................................................................................................................................................

:)))

Tích 2 số: a.b = BCNN(a,b).ƯCLN(a,b) = 40500
Vì ƯCLN(a,b) = 15 => a = 15k; b = 15q (với (k,q) = 1)
=> 15k.15q = 40500 => k.q = 180
Vì (k, q) = 1 => (k, q) ∈ {(4,45); (5,36), (9,20); (20;9); (36;5);(45;4)}
Vậy (a, b) ∈ {(60;675);(75;540);(135;300);(300;135);(540;75);(675;60)}

a+b=1200

\(A+G=50\%N\left(1\right)\\ M\text{à}:\dfrac{A+T}{G+X}=\dfrac{3}{2}\left(2\right)\\ \left(1\right),\left(2\right)\Rightarrow A=T=30\%N;G=X=20\%N\\ H=2A+3G\\ \Leftrightarrow2700=120\%N\\\Leftrightarrow N=2250\left(Nu\right)\\ a,L_{genB}=\dfrac{N}{2}.3,4=\dfrac{2250}{2}.3,4=3825\left(A^o\right)\\ A=T=30\%N=30\%.2250=675\left(Nu\right)\\ G=X=20\%N=20\%.2250=450\left(Nu\right)\\ b,A_{con}=T_{con}=2^3.A=8.675=5400\left(Nu\right)\\ G_{con}=X_{con}=G.2^3=450.8=3600\left(Nu\right)\)

\(c,N_b=\dfrac{6744.10^2}{300}=2248\left(Nu\right)\)

=> Dạng ĐB gen: Mất 1 cặp Nu

a) \(\left\{{}\begin{matrix}2A+3G=2700\\\dfrac{A+T}{G+X}=\dfrac{3}{2}\Leftrightarrow\dfrac{A}{G}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=T=675\left(nu\right)\\G=X=450\left(nu\right)\end{matrix}\right.\)

Tổng số nu : \(N=2A+2G=2250\left(nu\right)\)

Chiều dài : \(L=\dfrac{3,4N}{2}=3825\left(A^o\right)\)

b) \(\left\{{}\begin{matrix}A=T=675\cdot2^3=5400\left(nu\right)\\G=X=450\cdot2^3=3600\left(nu\right)\end{matrix}\right.\)

c) Tổng số nu của gen b

\(N=\dfrac{M}{300}=2248\left(nu\right)\)

=> Dạng đột biến mất một cặp nucleotit