a, $\sqrt[]{15-x}$= $\sqrt[]{3-x}$+$2$
⇔ $\sqrt[]{15-x}$ $-$ $\sqrt[]{3-x}$=$2$
⇔ $15-x+3-x$ + 2.$\sqrt[]{(15-x).(3-x)}$= $2$
⇔ $\sqrt[]{(15-x).(3-x)}$= $7-x$ ( $x≤7$)
⇔ $(15-x).(3-x)$=$(7-x)^{2}$
⇔ $45-15x-3x+x^{2}$=$49-14x+x^{2}$
⇔ $4x+4=0$
⇔ $x=-1 (tm)$
b, $(x+3)$.$\sqrt[]{10-x²}$ = $x²-x-17$ $(x≤ \sqrt[]{10}, x≥-\sqrt[]{10}$)
⇔ $(x+3)$.$\sqrt[]{10-x²}$= $(x-4).(x+3)$
⇔ $(x+3)$.($\sqrt[]{10-x²}$+$4-x$)= 0
⇔ \(\left[ \begin{array}{l}x+3=0\\\sqrt[]{10-x²}+4-x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-3(tm)\\\sqrt[]{10-x²}+4-x=0\end{array} \right.\)
Ta có: $\sqrt[]{10-x²}$+$4-x$= 0
⇔ $\sqrt[]{10-x²}$=$x-4$ ( $x≥4$)
⇔ $10-x²=x²-8x+16$
⇔ $2x²-8x+6=0$
⇔ \(\left[ \begin{array}{l}x=3(ktm)\\x=1(ktm)\end{array} \right.\)
