[Toán 9] Chia hết

V

vngocvien97

Ta có:
$7(a+b)\vdots \ 7$
hay$(3a+b)+(4a+6b)\vdots \ 7$
Mà $(3a+b)\vdots \ 7$
\Rightarrow$(4a+6b)\vdots \ 7$
hay$(2a+3b)\vdots \ 7$
\Rightarrow$7a(2a+3b)\vdots \ 49$
\Leftrightarrow$(14a^2+21ab)\vdots \ 49$
Lại có:$(3a+b)^2=(9a^2+6ab+b^2)\vdots \ 49$$(1)$
Mà$14a^2+21ab=(9a^2+6ab+b^2)+(5a^2+15ab-b^2)\vdots \ 49$$(2)$
Từ (1) và (2) \Rightarrow$(5a^2+15ab-b^2)\vdots \ 49$$(dpcm)$
 
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