a) $\frac{(a+b)2 – (a-b)2}{4}$ = ab b) 2($x^{2}$+$y^{2}$ )=(x+y)$^{2}$ +(x-y)$^{2}$ c) $x^{2}$ +y$^{2}$ =(x+y)$^{2}$ -2xy d) (x+y) $^{2}$ – (x-y)

0 bình luận về “a) $\frac{(a+b)2 – (a-b)2}{4}$ = ab b) 2($x^{2}$+$y^{2}$ )=(x+y)$^{2}$ +(x-y)$^{2}$ c) $x^{2}$ +y$^{2}$ =(x+y)$^{2}$ -2xy d) (x+y) $^{2}$ – (x-y)”

1. Đáp án:

   a) ` ((a+b)^2 – (a-b)^2)/4 = (a^2+ 2ab +b^2 – a^2 + 2ab – b^2)/4`

   ` = ((a^2 – a^2) + (b^2-b^2) + 4ab)/4`

   ` = 4(ab)/4`

   ` = ab`

   b) ` (x+y)^2 + (x-y)^2 = x^2 + 2xy + y^2 + x^2 -2xy + y^2`

   ` = (x^2 +x^2) + (y^2 +y^2) + (2xy -2xy)`

   ` = 2x^2 + 2y^2 + 0`

   ` = 2(x^2 +y^2)`

   `c) `

   ` (x+y)^2 – 2xy = x^2  +2xy + y^2 – 2xy`

   ` = x^2 + y^2 + (2xy – 2xy)`

   ` = x^2 + y^2`

   `d)`

   ` ( x+y)^2  -(x-y)(x+y) = x^2 + 2xy +y^2 – (x^2  -y^2)`

   ` = (x^2 -x^2) + 2xy  + (y^2 + y^2)`

   ` = 0 + 2xy + 2y^2`

   ` = 2y(x+y)`

   Bình luận

2. a)2(a^2+b^2)=0 => a^2+b^2=0 => a=0,b=0

   b)2(x^2+y^2)=2(x^2+y^2)

   c)x^2+y^2=x^2+y^2

   d)x^2+y^2+2xy-x^2+y^2=2xy+2y^2

   => 0=0

   Bình luận