a/ \(x^2+5\sqrt{x}+6=x^2+2\sqrt{x}+3\sqrt{x}+6\)

   \(=\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)=\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\)

b/ \(x^2+4\sqrt{x}+3=x^2+\sqrt{x}+3\sqrt{x}+3\)

     \(=\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\)

c/ k bik làm

Bạn trả lời sai rồi. \(\sqrt{x}\cdot\sqrt{x}\)chỉ được \(x\)mà thôi, ko được \(x^2\)đâu!

a, \(x-\sqrt{x}\)= \(\sqrt{x}.\left(\sqrt{x}-1\right)\)

b, 3x+6\(\sqrt{x}\)= \(\sqrt{x}.\left(3\sqrt{x}+6\right)\)

c, x+2\(\sqrt{x}+1\)= \(\left(\sqrt{x}\right)^2+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)

d, \(3x-5\sqrt{x}+2=3x-3\sqrt{x}-2\sqrt{x}+2\)

=\(3\sqrt{x}.\left(\sqrt{x}-1\right)-2.\left(\sqrt{x}-1\right)\)

=\(\left(3\sqrt{x}-2\right).\left(\sqrt{x}-1\right)\)

\(=x+2\sqrt{xy}+y-9\)

\(=\left(\sqrt{x}+\sqrt{y}\right)^2-3^2\)

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

Cảm ơn bạn nha

Mình làm được rồi

a.\(\sqrt{x}\left(\sqrt{x}-2\right)\)

b.\(\left(3-\sqrt{x}\right)\left(2+\sqrt{x}\right)\)

Đề sai vc

 uk t cx thấy sai

1/ \(x-6\sqrt{x}-8=\left(\sqrt{x}-3+\sqrt{17}\right)\left(\sqrt{x}-3-\sqrt{17}\right)\)

2/ Bài này làm gì còn phân tích được nữa.

a,\(x\sqrt{y}+y\sqrt{x}=\sqrt{x}\sqrt{y}.\left(\sqrt{x}+\sqrt{y}\right).\)

c,\(\sqrt{a}-a^2=\sqrt{a}.\left(1-a\sqrt{a}\right)\)

d,\(x-5\sqrt{x}+6=x-3\sqrt{x}-2\sqrt{x}+6\)

\(=\sqrt{x}.\left(\sqrt{x}-3\right)-2.\left(\sqrt{x}-3\right)\)\(=\left(\sqrt{x}-3\right).\left(\sqrt{x}-2\right)\)