[Đại 8]Phân tích đa thức thành nhân tử

T

transformers123

b/ $x^3+8+x(x+2)$
$=(x+2)(x^2-2x+4)+x(x+2)$
$=(x+2)(x^2-x+4)$
 
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T

transformers123

d)$\frac{1}{4}(a+1)^2-\frac{4}{9}(a-2)^2$
$\dfrac{1}{4}(a+1)^2-\dfrac{4}{9}(a-2)^2$
$=(\dfrac{a}{2}+\dfrac{1}{2})^2-(\dfrac{2a}{3}-\dfrac{4}{3})^2$
$=(\dfrac{a}{2}+\dfrac{1}{2}-\dfrac{2a}{3}+\dfrac{4}{3})(\dfrac{a}{2}+\dfrac{1}{2}+\dfrac{2a}{3}-\dfrac{4}{3})$
$=(\dfrac{-a}{6}+\dfrac{11}{6})(\dfrac{7a}{6}-\dfrac{5}{6})$
 
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D

deadguy

1.
a)
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b)
gif.latex
 
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