a.
$\sqrt[]{7-x}$+ $\sqrt[]{2+x}$ - $\sqrt[]{(7-x)(2+x)}$=3 (-2$\leq$ x$\leq$ 7)
đặt $\sqrt[]{7-x}$+ $\sqrt[]{2+x}$=t (t$\geq$ 0)
$t^{2}$=9+2 $\sqrt[]{(7-x)(2+x)}$ <-> $\sqrt[]{(7-x)(2+x)}$=$\frac{t^2-9}{2}$
-> t-$\frac{t^2-9}{2}$ = 3
2t-$t^{2}$+9=6
$t^{2}$-2t-3=0
t=-1(loại) t=3
-> $\sqrt[]{(7-x)(2+x)}$=$\frac{3^2-9}{2}$ =0
<-> x=7 hoặc x=-2
b.
$\sqrt[]{x+1}$+ $\sqrt[]{4-x}$ + $\sqrt[]{(x+1)(4-x)}$=5 (-1$\leq$ x$\leq$ 4)
đặt $\sqrt[]{x+1}$+ $\sqrt[]{4-x}$=t (t$\geq$ 0)
$t^{2}$=5+2 $\sqrt[]{(x+1)(4-x)}$ <-> $\sqrt[]{(x+1)(4-x)}$=$\frac{t^2-5}{2}$
t+$\frac{t^2-5}{2}$ =5
2t+$t^{2}$ -5=10
$t^{2}$+2t-15=0
t=-5 (loại) hoặc t=3
-> $\sqrt[]{(x+1)(4-x)}$=$\frac{3^2-5}{2}$ =2
4x-$x^{2}$+4-x=4
- $x^{2}$ +3x=0
x=0 hoặc x=3
c.
$\sqrt[3]{5x+7}$ - $\sqrt[3]{5x+3}$=1
đặt $\sqrt[3]{5x+7}$=a $\sqrt[3]{5x+3}$=b
-> $a^{3}$ -$b^{3}$=4
mà a-b=1 <-> a=b+1
-> $(b+1)^{3}$ -$b^{3}$=4
3$b^{2}$ +3b-3=0
b=$\frac{-1+\sqrt[]{5}}{2}$ <-> x=$\frac{-5+\sqrt[]{5}}{5}$
hoặc b=$\frac{-1-\sqrt[]{5}}{2}$ <-> x=$\frac{-5-\sqrt[]{5}}{5}$
