Cách 3 đi ==":Từ giả thiết suy ra: $b=\frac{a+c}{1-ac}.$Đặt $a=tanA;b=....c=....$Khi đó: $tanB=\frac{tanA+tanC}{1-tanA.tanC}=tan(A+C)$ $\Rightarrow B=A+C$Do đó: $P=2cos^2A-2cos^2B+3cos^2C=cos2A-cos2C+3cos^2C$ $P=2sinC.sin(A+B)-3sin^2C+3$ $P=-[\sqrt{3}sinC-\frac{1}{\sqrt{3}}sin(A+B)]^2+\frac{1}{3}sin^2(A+B)+3$$\rightarrow P\leq 3+\frac{1}{3}=\frac{10}{3}$Vậy $maxP=\frac{10}{3}$Đẳng thức khi: $sin(A-B)=1$ $sinC=\frac{1}{3}.sin(A-B)$ $B=A+C$ $\Leftrightarrow a=\frac{\sqrt{2}}{2};b=\sqrt{2};c=\frac{\sqrt{2}}{4}./$ Note:

Cách 3 đi ==":Từ giả thiết suy ra: $b=\frac{a+c}{1-ac}.$Đặt $a=tanA;b=....c=....$Khi đó: $tanB=\frac{tanA+tanC}{1-tanA.tanC}=tan(A+C)$ $\Rightarrow B=A+C$Do đó: $P=2cos^2A-2cos^2B+3cos^2C=cos2A-cos2C+3cos^2C$ $P=2sinC.sin(A+B)-3sin^2C+3$ $P=-[\sqrt{3}sinC-\frac{1}{\sqrt{3}}sin(A+B)]^2+\frac{1}{3}sin^2(A+B)+3$$\rightarrow P\leq 3+\frac{1}{3}=\frac{10}{3}$Vậy $maxP=\frac{10}{3}$Đẳng thức khi: $sin(A+B)=1$ $sinC=\frac{1}{3}.sin(A-B)$ $B=A+C$ $\Leftrightarrow a=\frac{\sqrt{2}}{2};b=\sqrt{2};c=\frac{\sqrt{2}}{4}./$ Note: