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Question

phân tích thành nhân tử

$a. 4x^4 + 4x^3-x^2-x$

$b. x^6 - x^4 - 9x^3 + 9x^2$

$c. x^4 - 4x^3 + 8x^2 - 16x + 16$

$d.(xy +4)^2 - 4(x+y)^2$

hoaphonglan101 · Answer 1 · 8/27/2014 1:10:35 PM

a)

4x3(x+1)−x(x+1)=(x+1)[4x3−x]=x(x+1)(4x2−1)=x(x+1)(2x+1)(2x−1)

d)

(xy+4)2−(2x+2y)2=(xy+4+2x+2y)(xy+4−2x−2y)=(x+2)(y+2)(x−2)(y−2)

b)

x4(x2−1)−9x2(x−1)=(x−1)[x4(x+1)−9x2]=x2(x−1)[x2(x+1)−9]

c)

x4−4x3+8x2−16x+16=(x−2)2(x2+4)

Trả lời 27-08-14 01:10 PM

hoaphonglan101
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score 0 · Answer 2

a) $4x^3 (x+1) -x(x+1) = (x+1) [4x^3 -x]=x(x+1)(4x^2 -1) = x(x+1)(2x+1)(2x-1)$

d) $(xy+4)^2 -(2x+2y)^2 = (xy+4 +2x+2y)(xy+4-2x-2y)=(x+2)(y+2)(x-2)(y-2)$

b) $x^4 (x^2 -1) -9x^2 (x-1)=(x-1)[ x^4(x+1) -9x^2]=x^2(x-1)[x^2 (x+1)-9]$

c) $x^4−4x^3+8x^2−16x+16=(x-2)^2 (x^2+4)$

2 Đáp án