Cho (a,b)=1.Tìm:
a. (a+b,a-b)
b.(7a+9b,3a+8b)
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Cho (a,b)=1. Tìm (7a+9b,3a+8b)
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cho(a,b)=1.tìm
a,(a+b,a-b)
b,(7a+9b,3a+8b)
a) Đặt \(\left(a+b,a-b\right)=d\)
\(\Rightarrow\left\{{}\begin{matrix}a+b⋮d\\a-b⋮d\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a⋮d\\2b⋮d\end{matrix}\right.\). Do \(\left(a,b\right)=1\) nên từ đây suy ra \(d\in\left\{1,2\right\}\)
b) Đặt \(\left(7a+9b,3a+8b\right)=d\)
\(\Rightarrow\left\{{}\begin{matrix}7a+9b⋮d\\3a+8b⋮d\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}21a+27b⋮d\\21a+56b⋮d\end{matrix}\right.\) \(\Rightarrow29b⋮d\)
Lại có \(\left\{{}\begin{matrix}56a+72b⋮d\\27a+72b⋮d\end{matrix}\right.\Rightarrow29a⋮d\)
Mà \(\left(a,b\right)=1\) \(\Rightarrow d\in\left\{1,29\right\}\)
cho (a,b)=1 tìm (a+b,a-b)
tìm (7a+9b,3a+8b)
cho(a,b)=1
a)(a+b,a-b)
b)(7a+9b,3a+8b)
cho (a;b) =1 tim :
a)(a=b:a-b)
b)(7a+9b;3a+8b)
Cho (a,b)=1.Tìm:
a, (a+b,a-b).
b, (7a+9b,3a+8b).
cho (a,b)=1 tìm
a) (a+b,a-b)
b) (7a+9b,3a+8b)
Cho (a,b)=1.Tìm:
a, (a+b,a-b).
b, (7a+9b,3a+8b).
Cho (a,b)=1 . Tìm (7a+9b,3a+8b)