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

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

b, \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)

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

a) $(\sqrt{a}+1)(b\sqrt{a}+1)$.
b) $(x-y)(\sqrt{x}+\sqrt{y})$

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

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

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

với a,b,x,y không âm ta có

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

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

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

b, \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)

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

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

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

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

\(a,B=4\sqrt{x=1}-3\sqrt{x+1}+2\)\(\sqrt{x+1}+\sqrt{x+1}\)

\(=4\sqrt{x+1}\)

\(b,\)đưa về \(\sqrt{x+1}=4\Rightarrow x=15\)

a, Với \(x\ge-1\)

\(\Rightarrow B=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)

\(=4\sqrt{x+1}\)

b, Ta có B = 16 hay 

\(4\sqrt{x+1}=16\Leftrightarrow\sqrt{x+1}=4\)bình phương 2 vế ta được 

\(\Leftrightarrow x+1=16\Leftrightarrow x=15\)

a) B = 4√x+1                                   b) x = 15

xy - y√x + √x - 1

= (√x)2.y - y√x + √x - 1

= y√x(√x - 1) + √x - 1

= (√x - 1)(y√x + 1) với x ≥ 1

\(a)\) \(xy-y\sqrt{x}+\sqrt{x}-1\)

= \(y\sqrt{x}.(\sqrt{x}-1)+\sqrt{x}-1\)

=\((\sqrt{x}-1).(y\sqrt{x}+1)\).

\(b)\)\(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)

=\(\sqrt{a}.\sqrt{x}-\sqrt{b}.\sqrt{y}+\sqrt{b}.\sqrt{x}-\sqrt{a}.\sqrt{y}\)

=\(\sqrt{a}.\sqrt{x}+\sqrt{b}.\sqrt{x}-\sqrt{a}.\sqrt{y}-\sqrt{b}.\sqrt{y}\)

=\(\sqrt{x}.(\sqrt{a}+\sqrt{b})-\sqrt{y}.(\sqrt{a}+\sqrt{b})\)

=\((\sqrt{x}-\sqrt{y}).(\sqrt{a}+\sqrt{b})\).

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

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

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

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

\(d)\) \(12-\sqrt{x}-x\)

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

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

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

a) \(xy-y\sqrt{x}+\sqrt{x}-1=\left(\sqrt{x}\right)^2.y-y\sqrt{x}+\sqrt{x}-1\)

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

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

b) \(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}=\left(\sqrt{ax}+\sqrt{bx}\right)-\left(\sqrt{ay}+\sqrt{by}\right)\)

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

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

c) \(\sqrt{a+b}+\sqrt{a^2-b^2}=\sqrt{a+b}+\sqrt{\left(a-b\right)\left(a+b\right)}\)

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

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

d) \(12-\sqrt{x}-x=12-\sqrt{4x}+\sqrt{3x}-x\)

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

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

a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)

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

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

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

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

\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)

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

do \(x\ge0;y\ge0\)

\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)

b) Ta có :

\(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2.2a+2^2.a^2\right)}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left[1^2-2.1.2a+\left(2a\right)^2\right]}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2a\right)^2}\)

\(=\frac{2}{2a-1}.\sqrt{5}.\sqrt{a^2}.\sqrt{\left(1-2a\right)^2}\)

\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|\)

Vì a > 0,5 <=> a > 0 <=> | a | = a

Vì a > 0,5 <=> 2a > 2 . 0,5 <=> 2a > 1 hay 1 < 2a

<=> 1 - 2a < 0 <=> | 1 - 2a | = - ( 1 - 2a )

= -1 + 2a = 2a - 1

Thay vào trên ta được : 

\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|=\frac{2}{2a-1}\sqrt{5}.a.\left(2a-1\right)=2a\sqrt{5}\)

Vậy : \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=2a\sqrt{5}\)