Toán 8 [Đại số]

N

nguyennguyen1108

1
Ta có: x+y=3
$=> (x+y)^2=9$
$=> x^2+2xy +y^2=9 mà x^2+y^2=5$ nên
$2xy=4$
$xy=2=> -xy=-2$
Ta lại có: $x^3+y^3 =(x+y)(x^2-xy+y^2)$
$= 3 (5-2)$
$=3.3=9$

2
$(5a-3b+8c)(5a-3b-8c)= (3a-5b)^2$
$=(5a-3b)^2 –(8c)^2$
$=25a^2 -30ab +9b^2 -64c^2$
$=25a^2-30ab +9b^2 -16 (a^2 –b^2)$
$=9a^2 -30ab +25b^2$
$=(3a-5b)^2$
 
Top Bottom