4(2x+y)^2+4(2x+y)+1
要步驟+答案
F.2因式分解
2007-11-01 8:11 am
回答 (5)
2007-11-01 8:15 am
✔ 最佳答案
4(2x+y)^2+4(2x+y)+1= [2(2x + y)]^2 + 2(2)(2x + y) + 1^2
= [2(2x + y) + 1]^2
= (4x + 4y + 1)^2
2007-11-04 12:03 am
4(2x+y)^2+4(2x+y)+1
=4(4x^2+y^2)+4(2x+y)+1
=16x^2+4y^2+8x+4y+1
=8x(2x+1)+4y(y+1)+1
=4(4x^2+y^2)+4(2x+y)+1
=16x^2+4y^2+8x+4y+1
=8x(2x+1)+4y(y+1)+1
2007-11-03 7:42 am
4(2x+y)^2+4(2x+y)+1
= [2(2x + y)]^2 + 2(2)(2x + y) + 1^2----------- using the identies
= [2(2x + y) + 1]^2-----------------------------------simpify
= (4x + 4y + 1)^2 //
= [2(2x + y)]^2 + 2(2)(2x + y) + 1^2----------- using the identies
= [2(2x + y) + 1]^2-----------------------------------simpify
= (4x + 4y + 1)^2 //
2007-11-01 8:59 pm
4(2x+y)^2+4(2x+y)+1
= [2(2x + y)]^2 + 2(2)(2x + y) + 1^2-----------------應用恆等式:a^2+2ab+b^2
= [2(2x + y) + 1]^2----------------------------------------十字相乘
= (4x + 4y + 1)^2 -----------------------------------------答案
十字相乘的過程:
2(2x+y)↘1
2(2x+y)↗1
－－－－－－－－
2(2x+y)+2(2x+y)=4(2x+y)=2(2)(2x+y)
= [2(2x + y)]^2 + 2(2)(2x + y) + 1^2-----------------應用恆等式:a^2+2ab+b^2
= [2(2x + y) + 1]^2----------------------------------------十字相乘
= (4x + 4y + 1)^2 -----------------------------------------答案
十字相乘的過程:
2(2x+y)↘1
2(2x+y)↗1
－－－－－－－－
2(2x+y)+2(2x+y)=4(2x+y)=2(2)(2x+y)
2007-11-01 8:16 am
4(2x+y)^2+4(2x+y)+1
= [2(2x+y)+1]^2
因為 a^2+2ab+b^2 = (a+b)^2
= [2(2x+y)+1]^2
因為 a^2+2ab+b^2 = (a+b)^2
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