a)√(15-x)=√(3-x)+2 b)(x+3).√(10-x ²)=x ²-x-17

a)√(15-x)=√(3-x)+2
b)(x+3).√(10-x ²)=x ²-x-17

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  1. a, $\sqrt[]{15-x}$= $\sqrt[]{3-x}$+$2$ 
    ⇔ $\sqrt[]{15-x}$ $-$ $\sqrt[]{3-x}$=$2$ 
    ⇔ $15-x+3-x$ + 2.$\sqrt[]{(15-x).(3-x)}$= $2$
    ⇔ $\sqrt[]{(15-x).(3-x)}$= $7-x$ ( $x≤7$)
    ⇔ $(15-x).(3-x)$=$(7-x)^{2}$
    ⇔ $45-15x-3x+x^{2}$=$49-14x+x^{2}$
    ⇔ $4x+4=0$
    ⇔ $x=-1 ™$

    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
    ⇔ $\sqrt[]{10-x²}$=$x-4$ ( $x≥4$)
    ⇔ $10-x²=x²-8x+16$
    ⇔ $2x²-8x+6=0$
    ⇔ \(\left[ \begin{array}{l}x=3(ktm)\\x=1(ktm)\end{array} \right.\) 

     

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