Vì \(BCNN\left(a,b\right)=300\) và \(ƯCLN\left(a,b\right)=15\)

\(\Leftrightarrow a.b=300.15=4500\)

Vì \(ƯCLN\left(a,b\right)=15\) nên \(a=15m\) và \(b=15n\) với \(ƯCLN=\left(m,n\right)=1\)

Vì \(a+15=b\Rightarrow15m+15=15n\Rightarrow15\left(m+1\right)=15n\)

\(\Leftrightarrow m+1=n\)

Mà \(a.b=4500\Rightarrow15m.15n=4500\Rightarrow15.15.m.n=4500\)

\(\Leftrightarrow m.n=20\)

\(\Leftrightarrow m=1\) và \(n=20\) hoặc \(m=4\) và \(n=5\)

Vì BCNH A ; B) = 300 ; ƯCLN ( A ; B ) = 15 \(\Rightarrow\) AB = 4500

Ta có : ƯCLN ( A , B ) = 15 \(\Rightarrow\) A = 15k ; B = 15q Với ƯCLN ( k ; q ) = 1

\(\Rightarrow\) 15k x 15q = 4500

\(\Rightarrow\) 225kq = 4500

\(\Rightarrow\) kq = 20

Mà ƯCLN ( k ; q ) = 1 

Vậy từ đó ta có bảng sau :

Mà theo đề bài A + 15 = B \(\Rightarrow\)  A = 60 b = 75 

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

M.n giúp mk với, mk sẽ tick hết. Mk cần gấp lắm luôn á

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