[Toán 8] Phân tích đa thức thành nhân tử

T

thaokute9x

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N

nguyenbahiep1

Bài 1.......................................................

a)

[TEX](3-x)^2[/TEX]

b)

[TEX](5+x)^2[/TEX]

c)

[TEX](\frac{1}{2}- 2x)( \frac{1}{4}+ 4x^2 + x)[/TEX]

d)

[TEX](\frac{x}{8}-5y)(\frac{x}{8}+5y)[/TEX]

e) sai đề

f)

[TEX](x-\frac{1}{2})^2[/TEX]


câu 2

[TEX]4x(4y^2-4y +1) = 4x(2y-1)^2[/TEX]
 
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N

nvk1997bn

a)(X-3)^2
b)(X+5)^2
c)(1/2-2X)(1/4+X+4X^2)
d)(1/8X-5y)*(1/8X+5Y)
e)(1/3+2X)*(1/9-2/3X+4X^2)
f)(X-1/2)^2
BAI 2 xem lại đề hộ mình
 
H

harrypham

a, [TEX]9-6x+x^2=x^2- 2 \cdot 3x+3^2=(x-3)^2[/TEX].

b, [TEX]10x+25+x^2=(x+5)^2[/TEX]

c, [TEX]\frac{1}{8}-8x^3= \left( \frac{1}{2} \right)^3- (2x)^3= \left( \frac{1}{2}-2x \right) \left( \frac{1}{4}+x+4x^2 \right)[/TEX]

d, [TEX]\frac{1}{64}x^2-25y^2= \left( \frac{x}{8} \right)^2-(5y)^2= \left( \frac{x}{8}-5y \right) \left( \frac{x}{8}+5y \right)[/TEX]

e, [TEX]\frac{1}{27}+8x^3= \left( \frac{1}{3} \right)^3+(2x)^3 = \left( \frac{1}{3}+2x \right) \left( \frac{1}{9}- \frac{2x}{3}+ 4x^2 \right)[/TEX]

f, [TEX]\frac{1}{4}-x+x^2=x^2- 2 \cdot \frac{1}{2}x+ \frac{1}{4}= \left(x- \frac{1}{2} \right)^2[/TEX]
 
M

mituotroile

phân tích đa thức sau thành nhân tử:

bài 1:
a,9-6x+x^2=3^2 - 2.3x +x^2
=(3 - x)^2

b, 10x +25+x^2 = x^2 +10x +25
= x^2 +2.5x+5^2
= (x + 5)^2

d, 1/64.x^2 - 25y^2 = (1/8.x)^2 -(5y)^2
=(1/8.x + 5y)(1/8x-5y)

c, 1/8-8x^3=(1/2)^3-(2x)^3
= (1/2-2x)[(1/2)^2 +1/2.2x +(2x)^2]
= (1/2-2x)(1/4 +x+4x^2)

e, chep sai đề bài: 1/27+8x^3=(1/3)^3+(2x)^3
=(1/3+2x)[(1/3)^2 -1/3.2x+(2x)^2]
=(1/3 +2x)(1/9 - 2/3.x+4x^2)


f, 1/4 -x +x^2= (1/2)^2-x+x^2
=(1/2-x)^2




Bài 2:12xy^2-12xy +3x=3x(4y^2-4y+1)
=3x[(2y)^2 -2.2y+1^2]
=3x(2y-1)^2
 
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