Bài 2 : Hằng đẳng thức dạng ( A – B)^2 = A^2 – 2AB + B^2 a) -( 2y – 5 )^2 b) ( 1 – 5a )^2 c) ( 1/3x – 3y )^2 25/07/2021 Bởi Delilah Bài 2 : Hằng đẳng thức dạng ( A – B)^2 = A^2 – 2AB + B^2 a) -( 2y – 5 )^2 b) ( 1 – 5a )^2 c) ( 1/3x – 3y )^2
a) $-(2y-5)^2=-(4y^2-20y+25)=-4y^2+20y-25$ b) $(1-5a)^2=1-10a+25a^2$ c) `(1/3 x-3y)^2=1/9 x^2- 2xy+9y^2` Bình luận
`a,-(2y-5)^2=-(4y^2-20y+25)=-4y^2+20y-25` `b,(1-5a)^2=1-10a+25a^2` `c,(1/3x-3y)^2=1/9x^2-2xy+9y^2` Bình luận
a) $-(2y-5)^2=-(4y^2-20y+25)=-4y^2+20y-25$
b) $(1-5a)^2=1-10a+25a^2$
c) `(1/3 x-3y)^2=1/9 x^2- 2xy+9y^2`
`a,-(2y-5)^2=-(4y^2-20y+25)=-4y^2+20y-25`
`b,(1-5a)^2=1-10a+25a^2`
`c,(1/3x-3y)^2=1/9x^2-2xy+9y^2`