Toán Lớp 9: $\sqrt{3x + 1}$ – $\sqrt{6 – x}$ + 3$x^{2}$ – 14 x – 8 = 0
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Toán Lớp 9: $\sqrt{3x + 1}$ – $\sqrt{6 – x}$ + 3$x^{2}$ – 14 x – 8 = 0
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$\textit{Giải đáp + Lời giải và giải thích chi tiết:}$
\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0 (-1/3 \le x \le 6)
<=>(\sqrt{3x+1}-4)+(1-\sqrt{6-x})+3x^2-14x-5=0
<=>((\sqrt{3x+1})^2-4^2)/(\sqrt{3x+1}+4)+(1^2-(\sqrt{6-x})^2)/(1+\sqrt{6-x})+(3x+1)(x-5)=0
<=>(3x-15)/(\sqrt{3x+1}+4)+(x-5)/(1+\sqrt{6-x})+(3x+1)(x-5)=0
<=>(x-5)(3/(\sqrt{3x+1}+4)+1/(1+\sqrt{6-x})+3x+1)=0
Dễ thấy 3/(\sqrt{3x+1}+4)+1/(1+\sqrt{6-x})+3x+1>0 \forall x tmđk nên x-5=0<=>x=5 $(tm)$
Vậy S={5}