Dùng hàng đẳng thức a) xᒾ/4 + 2xy + 4yᒾ b) 9.(2x+3)ᒾ – 4.(x+1)ᒾ 02/07/2021 Bởi Clara Dùng hàng đẳng thức a) xᒾ/4 + 2xy + 4yᒾ b) 9.(2x+3)ᒾ – 4.(x+1)ᒾ
`a)` `x^2/4+2xy+4y^2` `=(x/2)^2+2.x/2. 2y+(2y)^2` `=(x/2+2y)^2` `b)` `9(2x+3)^2-4(x+1)^2` `=[3(2x+3)]^2-[2(x+1)]^2` `=[3(2x+3)+2(x+1)][3(2x+3)-2(x+1)]` `=(6x+9+2x+2)(6x+9-2x-2)` `=(8x+11)(4x+7)` Bình luận
`a,` `x^2/4+2xy+4y^2` `=(x/2)^2+2.x/2.2y+(2y)^2` `=(x/2+2y)^2` `⇒Hđt` thứ nhất `b,` `9(2x+3)^2-4(x+1)^2` `=[3(2x+3)]^2-[2(x+1)]^2` `=[3(2x+3)+2(x+1)][3(2x+3)-2(x+1)]` `=(6x+9+2x+2)(6x+9-2x-2)` `=(8x+11)(4x+7)` `⇒Hđt` thứ ba Bình luận
`a)` `x^2/4+2xy+4y^2`
`=(x/2)^2+2.x/2. 2y+(2y)^2`
`=(x/2+2y)^2`
`b)` `9(2x+3)^2-4(x+1)^2`
`=[3(2x+3)]^2-[2(x+1)]^2`
`=[3(2x+3)+2(x+1)][3(2x+3)-2(x+1)]`
`=(6x+9+2x+2)(6x+9-2x-2)`
`=(8x+11)(4x+7)`
`a,` `x^2/4+2xy+4y^2`
`=(x/2)^2+2.x/2.2y+(2y)^2`
`=(x/2+2y)^2`
`⇒Hđt` thứ nhất
`b,` `9(2x+3)^2-4(x+1)^2`
`=[3(2x+3)]^2-[2(x+1)]^2`
`=[3(2x+3)+2(x+1)][3(2x+3)-2(x+1)]`
`=(6x+9+2x+2)(6x+9-2x-2)`
`=(8x+11)(4x+7)`
`⇒Hđt` thứ ba