a>(x+y+4).(x+y-4) b>(x-y+6).(x+y-6) c>(y+2z-3).(y-2z-3) d>(x+2y+3z).(2y+3z-x) 11/08/2021 Bởi Sarah a>(x+y+4).(x+y-4) b>(x-y+6).(x+y-6) c>(y+2z-3).(y-2z-3) d>(x+2y+3z).(2y+3z-x)
a) (x+y+4)(x+y-4) =[(x+y)+4][(x+y)-4]= (x+y)^2-4^2= x^2+2xy+y^2-16 b)(x-y+6)(x+y-6) = [x-(y-6)][x+(y-6)]= x^2-(y-6)^2 = x^2-y^2+12y-36 c)(y+2z-3)(y-2z-3) = [(y-3)+2z][(y-3)-2z]= (y-3)^2-4z^2 = y^2-6y+9-4z^2 d>(x+2y+3z).(2y+3z-x) = [(2y+3z)+x][(2y+3z)-x]= (2y+3z)^2-x^2 = 4y^2+12yz+9z^2-x^2 Bình luận
Đáp án: Ta có : a,$(x+y+4).(x+y-4)$ $ = ( x + y)^2 – 4^2$ $ =( x + y)^2 – 16$ b, $(x-y+6).(x+y-6)$ $ = [ x – ( y + 6)][ x + (y – 6)]$ $ = x^2 – (y + 6)^2$ c, $(y+2z+ 3).(y-2z-3)$ $ = [y + (2z + 3)].[y – (2z + 3)]$ $ = y^2 – (2z + 3)^2$ d, $(x+2y+3z).(2y+3z-x)$ $ = [(2y + 3z) + x][(2y + 3z) – x]$ $ = (2y + 3z)^2 – x^2$ Giải thích các bước giải: Bình luận
a) (x+y+4)(x+y-4)
=[(x+y)+4][(x+y)-4]
= (x+y)^2-4^2
= x^2+2xy+y^2-16
b)(x-y+6)(x+y-6)
= [x-(y-6)][x+(y-6)]
= x^2-(y-6)^2
= x^2-y^2+12y-36
c)(y+2z-3)(y-2z-3)
= [(y-3)+2z][(y-3)-2z]
= (y-3)^2-4z^2
= y^2-6y+9-4z^2
d>(x+2y+3z).(2y+3z-x)
= [(2y+3z)+x][(2y+3z)-x]
= (2y+3z)^2-x^2
= 4y^2+12yz+9z^2-x^2
Đáp án:
Ta có :
a,$(x+y+4).(x+y-4)$
$ = ( x + y)^2 – 4^2$
$ =( x + y)^2 – 16$
b, $(x-y+6).(x+y-6)$
$ = [ x – ( y + 6)][ x + (y – 6)]$
$ = x^2 – (y + 6)^2$
c, $(y+2z+ 3).(y-2z-3)$
$ = [y + (2z + 3)].[y – (2z + 3)]$
$ = y^2 – (2z + 3)^2$
d, $(x+2y+3z).(2y+3z-x)$
$ = [(2y + 3z) + x][(2y + 3z) – x]$
$ = (2y + 3z)^2 – x^2$
Giải thích các bước giải: