Question

BT: Phân tích thành nhân tử

a, \(3-\sqrt{3}+\sqrt{15}-3\sqrt{15}\)

b, \(\sqrt{1-a}+\sqrt{1-a^2}\) ( với 1 > a > -1 )

c, \(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}\) ( với a,b > 0 )

d, \(x-y+\sqrt{xy^2}-\sqrt{y^3}\) ( với x,y > 0 )

Answer 1

a)

$3-\sqrt{3}+\sqrt{15}-3\sqrt{15}$

$=\sqrt{3}(\sqrt{3}-1)-\sqrt{15}(3-1)$

$=(\sqrt{3}(\sqrt{3}-1)-\sqrt{15}(\sqrt{3}+1)(\sqrt{3}-1)$

$=(\sqrt{3}-1)[\sqrt{3}-\sqrt{15}(\sqrt{3}+1)]$

$=(\sqrt{3}-1)(\sqrt{3}-\sqrt{45}-\sqrt{15})$

b)

$\sqrt{1-a}+\sqrt{1-a^2}=\sqrt{1-a}+\sqrt{(1-a)(1+a)}$

$=\sqrt{1-a}+\sqrt{1-a}.\sqrt{1+a}=\sqrt{1-a}(1+\sqrt{1+a})$

Answer 2

c)

$\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}$

$=(\sqrt{a^3}+\sqrt{a^2b})-(\sqrt{b^3}+\sqrt{ab^2})$

$=\sqrt{a^2}(\sqrt{a}+\sqrt{b})-\sqrt{b^2}(\sqrt{b}+\sqrt{a})$

$=a(\sqrt{a}+\sqrt{b})-b(\sqrt{b}+\sqrt{a})$

$=(\sqrt{a}+\sqrt{b})(a-b)=(\sqrt{a}+\sqrt{b})^2(\sqrt{a}-\sqrt{b})$

d)

$x-y+\sqrt{xy^2}-\sqrt{y^3}$

$=(x-y)+(\sqrt{xy^2}-\sqrt{y^3})$

$=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})+y(\sqrt{x}-\sqrt{y})$

$=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}+y)$