(Toán 8) Chứng minh

M

manh550

1.Ta có: $(a+b+c+d) (a-b-c+d) = (a-b+c-d) (a+b-c-d)$
<=> $[(a+d)+(b+c)] [(a+d)-(b+c)] = [(a-d)-(b-c)] [(a-d)+(b-c)]$
<=> $(a+d)^2 - (b+c)^2 = (a-d)^2 - (b-c)^2 (hđt (a+b)(a-b)=a^2-b^2)$
<=> $(a+d)^2 - (a-d)^2 = (b+c)^2 - (b-c)^2$
<=> $(a+d+a-d)(a+d-a+d) = (b+c+b-c)(b+c-b+c)$
<=> $2a . 2d = 2b . 2c$
<=> $a . d = b . c$
 
You must log in or register to reply here.
Top Bottom