Question

phân tích thành nhân tử

\(a-6\sqrt{a}+9-b^2\)

\(x-9\)

\(x-7\sqrt{x}+12\)

\(x\sqrt{x}-64\)

làm chi tiết xíu giúp em ạ.

Answer 1

a: =(căn a-3)^2-b^2

=(căn a-3-b)(căn a-3+b)

b: \(x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)

c: \(x-7\sqrt{x}+12=x-3\sqrt{x}-4\sqrt{x}+12=\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)\)

d: x*căn x-64

=(căn x)^3-4^3

=(căn x-4)(x+4căn x+16)

Answer 2

\(a-6\sqrt{a}+9-b^2\\ =\left(\sqrt{a}+3\right)^2-b^2\\ =\left(\sqrt{a}+3-b\right)\left(\sqrt{a}+3+b\right)\)

 

\(x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)

 

\(x-7\sqrt{x}+12\\ =x-4\sqrt{x}-3\sqrt{x}+12\\ =\sqrt{x}\left(\sqrt{x}-4\right)-3\left(\sqrt{x}-4\right)\\ =\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)\)

 

\(x\sqrt{x}+64\\ =\sqrt{x^3}+4^3\\ =\left(\sqrt{x}\right)^3+4^3\\ =\left(\sqrt{x}+4\right)\left(x-4\sqrt{x}+16\right)\)