a.
[tex]DK:x<1\\2\sqrt{1-x}=\frac{x^2-3x+1}{x-1}\\\Leftrightarrow 2\sqrt{1-x}(x-1)=(x^2-2x+1)-x+1-1\\\Leftrightarrow 2\sqrt{1-x}(x-1)=(x-1)^2-(x-1)-1\\\sqrt{1-x}=t>0\\\Leftrightarrow 2t.(-t^2)=t^4-(-t^2)-1\\\Leftrightarrow t^4+2t^3+t^2-1=0\\\Leftrightarrow (t^2+t-1)(t^2+t+1)=0\\\Leftrightarrow t^2+t-1=0(t^2+t+1>0)\\\Leftrightarrow t=\frac{\sqrt{5}-1}{2}(Do:t>0)\\\Leftrightarrow \sqrt{1-x}=\frac{\sqrt{5}-1}{2}\\\Leftrightarrow x=\frac{\sqrt{5}-1}{2}(TM)[/tex]
[tex]2cos^2\frac{x}{2}+ 4cos^2\frac{x}{3}=1\\\Leftrightarrow 2cos^23.\frac{x}{6}+ 4cos^22.\frac{x}{6}=1\\\Leftrightarrow 2(4cos^3\frac{x}{6}-3cos\frac{x}{6})^2+4(2cos^2\frac{x}{6}-1)^2-1=0\\\Leftrightarrow 2(16cos^6\frac{x}{6}-24cos^4\frac{x}{6}+9cos^2\frac{x}{6})+4(4cos^4\frac{x}{6}-4cos^2\frac{x}{6}+1)-1=0\\cos\frac{x}{6}=t\\\Leftrightarrow 32t^6-48t^4+18t^2+16t^4-16t^2+4-1=0\\\Leftrightarrow 32t^6-32t^4+2t^2+3=0\\\Leftrightarrow (4t^2-3)(4t^2+1)(2t^2-1)=0\\\Leftrightarrow cos\frac{x}{6}=+-\frac{\sqrt{3}}{2}\\or\\cos\frac{x}{6}=+-\frac{\sqrt{2}}{2}[/tex]
Còn lại đơn giản rồi, anh xử nốt nhé
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