2x-x²+4 √(3-x)(x+1)=9
⇔ 2x-x²-9+$4\sqrt[]{2x}$ -x²+3=0
⇔ 2x-x²+3-12+$4\sqrt[]{2x}$ -x²+3=0
Đặt t=$\sqrt[]{2x}$ -x²+3 (t≥0) pt trở thành:
⇒ t²-12+4t=0
Δ'=$(-6)^{2}$ -1.4=32
⇒$\sqrt[]{Δ'}$=$4\sqrt[]{2}$
Δ'>0⇒ pt có 2 no pb
t1=$\frac{6+4\sqrt[]{2}}{1}$ =$6+4\sqrt[]{2}$ (tmđk)
t2=$\frac{6-4\sqrt[]{2}}{1}$=$6-4\sqrt[]{2}$ (tmđk)
+nếu t=$6+4\sqrt[]{2}$
⇔$x^{2}$=$6+4\sqrt[]{2}$
⇔x= $\sqrt[]{6+4\sqrt[]{2}}$
⇔x=$\sqrt[]{(\sqrt[]{2}^{})^{2}+2.2.\sqrt[]{2}+2^{2}}$
⇒x=±$\sqrt[]{2}$+2
+nếu t=$6-4\sqrt[]{2}$
⇔$x^{2}$=$6-4\sqrt[]{2}$
⇔x= $\sqrt[]{6-4\sqrt[]{2}}$
⇒x=±$\sqrt[]{2}$-2
Vây pt có 4 no....
