\(5a+2b⋮17\)

\(\Rightarrow60a+24b⋮17\)

\(\Rightarrow\left(51a+17b\right)+\left(9a+7b\right)⋮17\)

Do \(51a+17b⋮17\Rightarrow9a+7b⋮17\Rightarrowđpcm\)

Hay quá coolki

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: \(\left(5a+2b\right)⋮17\Rightarrow\left(9a+7b\right)⋮17\)

\(9a+7b⋮17\Rightarrow3\left(9a+7b\right)=27a+21b⋮17\)

\(17a+17b⋮17\)

\(\Rightarrow27a+21b-17a-17b=10a+4b=2\left(5a+2b\right)⋮17\)

\(\Rightarrow5a+2b⋮17\)

$5a+2b⋮17$

$\Rightarrow 60a+24b⋮17$

$\Rightarrow \left(51a+17b\right)+\left(9a+7b\right)⋮17$

Do $51a+17b⋮17\Rightarrow 9a+7b⋮17\Rightarrow đpcm$

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.

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