\(a\sqrt{a}+2a+\sqrt{a}+2=\left(a\sqrt{a}+2a\right)+\left(\sqrt{a}+2\right)\)

\(=a\left(\sqrt{a}+2\right)+\left(\sqrt{a}+2\right)=\left(\sqrt{a}+2\right)\left(a+1\right)\)

\(a\sqrt{a}+2a+\sqrt{a}+2=a\left(\sqrt{a}+2\right)+\left(\sqrt{a}+2\right)=\left(a+1\right)\left(\sqrt{a}+2\right)\)

\(a\sqrt{a}+2a+\sqrt{a}+2=\left(\sqrt{a}+2\right)\left(a+1\right)\)

\(=\sqrt{a-b}\left(\sqrt{a+b}+1\right)\)

\(=\sqrt{\left(a+b\right)\left(a-b\right)}+\sqrt{a-b }\)
\(=\sqrt{a-b}\cdot\sqrt{a+b}+\sqrt{a-b}\)
\(=\sqrt{a-b}\cdot\left(\sqrt{a+b}+1\right)\)

\(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

\(a\sqrt{b}+\sqrt{ab}+\sqrt{a}+1\)

\(=\sqrt{ab}\cdot\sqrt{a}+\sqrt{ab}+\sqrt{a}+1\)

\(=\left(\sqrt{ab}\cdot\sqrt{a}+\sqrt{ab}\right)+\left(\sqrt{a}+1\right)\)

\(=\sqrt{ab}\cdot\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{ab}+1\right)\left(\sqrt{a}+1\right)\)

a√b + √(ab) + √a + 1

= [a√b + √(ab)] + (√a + 1)

= √(ab)(√a + 1) + (√a + 1)

= (√a + 1)[√(ab) + 1]

a) 

Đa thức bậc nhất không phân tích được nhân tử :v

b)

Đặt \(\sqrt{x}=a;\sqrt{y}=b\) Khi đó:

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

c)

Tương tự câu b) thì ta sẽ có:

\(x\sqrt{x}-27=a^3-27=\left(a-3\right)\left(a^2+3a+9\right)\)

Bài làm:

a) \(9x-5=\left(3\sqrt{x}\right)^2-\sqrt{5}==\left(3\sqrt{x}-\sqrt{5}\right)\left(3\sqrt{x}+\sqrt{5}\right)\)

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

c) \(x\sqrt{x}-27=\left(\sqrt{x}\right)^3-3^3=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)\)

dễ mà tự làm đi

giúp mình với 

\(=\sqrt{a-b}-\sqrt{\left(a-b\right)\left(a+b\right)}=\sqrt{a-b}\left(1-\sqrt{a+b}\right)\)