Cho a,b là các số nguyên tố , CMR 5a + 2b : 17 khi và chỉ khi 9a + 7b :17
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5a+2b⋮17\)
\(\Rightarrow60a+24b⋮17\)
\(\Rightarrow\left(51a+17b\right)+\left(9a+7b\right)⋮17\)
Do \(51a+17b⋮17\Rightarrow9a+7b⋮17\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(5\cdot\left(5a+2b\right)+\left(9a+7b\right)=25a+10b+9a+7b=34a+17b\)
\(\Rightarrow34a+17b=17\left(2a+b\right)⋮17\)
Do đó: \(\left(5a+2b\right)⋮17\Rightarrow\left(9a+7b\right)⋮17\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9a+7b⋮17\Rightarrow3\left(9a+7b\right)=27a+21b⋮17\)
\(17a+17b⋮17\)
\(\Rightarrow27a+21b-17a-17b=10a+4b=2\left(5a+2b\right)⋮17\)
\(\Rightarrow5a+2b⋮17\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Ta có:
\(b^2+c^2-a^2+2bc=(b^2+2bc+c^2)-a^2\)
\(=(b+c)^2-a^2=(2p-a)^2-a^2\) (do \(a+b+c=2p\) )
\(=4p^2-4pa+a^2-a^2=4p^2-4pa=4p(p-a)\)
Do đó ta có đpcm.
Bài 2:
Dấu \(\Leftrightarrow \) thể hiện bài toán đúng trong cả 2 chiều.
Ta có: \(5a+2b\vdots 17\)
\(\Leftrightarrow 2(5a+2b)\vdots 17\)
\(\Leftrightarrow 10a+4b\vdots 17\)
\(\Leftrightarrow 10a+4b+17a+17b\vdots 17\)
\(\Leftrightarrow 27a+21b\vdots 17\)
\(\Leftrightarrow 3(9a+7b)\vdots 17\)
\(\Leftrightarrow 9a+7b\vdots 17\) (do 3 và 17 nguyên tố cùng nhau)
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
chịuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu