\(a,=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)=\left(b\sqrt{a}+1\right)\left(\sqrt{a}+1\right)\\ b,a< 0\text{ nên biểu thức không phân tích đc}\)

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

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

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

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

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

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

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

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

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

a√b + √(ab) + √a + 1

= [a√b + √(ab)] + (√a + 1)

= √(ab)(√a + 1) + (√a + 1)

= (√a + 1)[√(ab) + 1]

\(\left(ab-1\right)^2+\left(a+b\right)^2\)

\(=a^2b^2-2ab+1+a^2+2ab+b^2\)

\(=a^2+a^2b^2+b^2+1\)\(>0\)

biểu thức không phân tích thành nhân tử được nhé

Ta có :

\(\left(ab-1\right)^2+\left(a+b\right)^2=a^2b^2-2ab+1+a^2+2ab\) \(+b^2\)

\(=a^2b^2+a^2+b^2+1\)\(=a^2\left(b^2+1\right)+b^2+1=\left(b^2+1\right)\left(a^2+1\right)\)

bc(b+c)+ca(c-a)-ab(a+b)

=bc(b+c)+ca(b+c)-ca(a+b)-ab(a+b)

=c(a+b)(b+c)-a(a+b)(b+c)

=(c-a)(a+b)(b+c)

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

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

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

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

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

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

ab + b√a + √a + 1 = [(√a)2b + b√a] + (√a + 1)

= b√a(√a + 1) + (√a + 1) = (√a + 1)(b√a + 1)