19.
$\sqrt[]{3+x}$ + $\sqrt[]{6-x}$ -$\sqrt[]{(3+x)(6-x)}$ =3 (-3$\leq$x $\leq$ 6)
đặt t=$\sqrt[]{3+x}$ + $\sqrt[]{6-x}$ (t$\geq$ 0)
-> t$^{2}$= 9+ 2$\sqrt[]{(3+x)(6-x)}$
<-> $\sqrt[]{(3+x)(6-x)}$= $\frac{t^2-9}{2}$
=> t-$\frac{t^2-9}{2}$ =3
<-> 2t- t$^{2}$ +9=6
<-> t$^{2}$-2t-3=0
<-> t=-1 (loại) hoặc t=3 (tm)
<-> $\sqrt[]{(3+x)(6-x)}$= $\frac{3^2-9}{2}$ =0
<-> x=-3 hoặc x=6
20.
$\sqrt[]{4x+1}$ - $\sqrt[]{3x-2}$ = $\frac{x+3}{2}$ (x$\geq$ $\frac{2}{3}$)
$\frac{(\sqrt[]{4x+1} - \sqrt[]{3x-2})(\sqrt[]{4x+1} +\sqrt[]{3x-2})}{\sqrt[]{4x+1} +\sqrt[]{3x-2}}$ = $\frac{x+3}{2}$
$\frac{4x+1-3x+2}{\sqrt[]{4x+1} +\sqrt[]{3x-2}}$ = $\frac{x+3}{2}$
$\frac{x+3}{\sqrt[]{4x+1} +\sqrt[]{3x-2}}$ = $\frac{x+3}{2}$
x+3=0 <-> x=-3(loại)
hoặc
$\sqrt[]{4x+1} +\sqrt[]{3x-2}$=2
đặt $\sqrt[]{4x+1}$=a $\sqrt[]{3x-2}$=b
-> 3$a^{2}$ -4$b^{2}$ =11
mà a+b=2 <-> a=2-b
-> 3$(2-b)^{2}$ -4$b^{2}$ =11
<-> b=-6-$\sqrt[]{37}$ (loại) hoặc b=-6+$\sqrt[]{37}$ -> a=8-$\sqrt[]{37}$
-> 4x+1=(8-$\sqrt[]{37}$)$^{2}$
<-> x=$\frac{(8-\sqrt[]{37})^{2}-1}{4}$ (tm)
