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Từ gthiêt suy ra: $ac+a.\frac{1}{b}+c.\frac{1}{b}=1$suy ra tồn tại $\Delta ABC$ sao cho $a=tan\frac{A}{2};\frac{1}{b}=tan\frac{B}{2};c=tan\frac{C}{2}$Khi đó:$P=2.cos^2\frac{A}{2}-2sin^2$$\frac{B}{2}+3cos^2\frac{C}{2}$$P=cosA+cosB+3-3sin^2\frac{C}{2}$ $=2sin\frac{C}{2}.cos\frac{A-B}{2}+3-sin^2\frac{C}{2}$ $=\frac{10}{3}-3(sin\frac{C}{2}-\frac{1}{3}.cos\frac{A-B}{2})^2-\frac{1}{3}.cos^2\frac{A-B}{2}\leq \frac{10}{3}$Vậy max$P=\frac{10}{3}$khi$a= \frac{1}{\sqrt{2}};b=\sqrt{2};c=\frac{1}{2\sqrt{2}}$

Từ gthiêt suy ra: $ac+a.\frac{1}{b}+c.\frac{1}{b}=1$suy ra tồn tại $\Delta ABC$ sao cho $a=tan\frac{A}{2};\frac{1}{b}=tan\frac{B}{2};c=tan\frac{C}{2}$Khi đó:$P=2.cos^2\frac{A}{2}-2sin^2$$\frac{B}{2}+3cos^2\frac{C}{2}$$P=cosA+cosB+3-3sin^2\frac{C}{2}$ $=2sin\frac{C}{2}.cos\frac{A-B}{2}+3-3sin^2\frac{C}{2}=3-3(sin\frac{C}{2}-\frac{1}{3}.cos\frac{A-B}{2})^2+\frac{1}{3}.cos^2\frac{A-B}{2}\leq \frac{10}{3}$Vậy max$P=\frac{10}{3}$khi$a= \frac{1}{\sqrt{2}};b=\sqrt{2};c=\frac{1}{2\sqrt{2}}$P/s : Tui chỉnh nv k chắc :D nếu như của bà thì nó ra $\leq 3$

Từ gthiêt suy ra: $ac+a.\frac{1}{b}+c.\frac{1}{b}=1$suy ra tồn tại $\Delta ABC$ sao cho $a=tan\frac{A}{2};\frac{1}{b}=tan\frac{B}{2};c=tan\frac{C}{2}$Khi đó:$P=2.cos^2\frac{A}{2}-2sin^2$$\frac{B}{2}+3cos^2\frac{C}{2}$$P=cosA+cosB+3-3sin^2\frac{C}{2}$ $=2sin\frac{C}{2}.cos\frac{A-B}{2}+3-sin^2\frac{C}{2}$ $=\frac{10}{3}-3(sin\frac{C}{2}-\frac{1}{3}.cos\frac{A-B}{2})^2-\frac{1}{3}.cos^2\frac{A-B}{2}\leq \frac{10}{3}$Vậy max$P=\frac{10}{3}$ khi.....

Từ gthiêt suy ra: $ac+a.\frac{1}{b}+c.\frac{1}{b}=1$suy ra tồn tại $\Delta ABC$ sao cho $a=tan\frac{A}{2};\frac{1}{b}=tan\frac{B}{2};c=tan\frac{C}{2}$Khi đó:$P=2.cos^2\frac{A}{2}-2sin^2$$\frac{B}{2}+3cos^2\frac{C}{2}$$P=cosA+cosB+3-3sin^2\frac{C}{2}$ $=2sin\frac{C}{2}.cos\frac{A-B}{2}+3-sin^2\frac{C}{2}$ $=\frac{10}{3}-3(sin\frac{C}{2}-\frac{1}{3}.cos\frac{A-B}{2})^2-\frac{1}{3}.cos^2\frac{A-B}{2}\leq \frac{10}{3}$Vậy max$P=\frac{10}{3}$ khi $a= \frac{1}{\sqrt{2}};b=\sqrt{2};c=\frac{1}{2\sqrt{2}}$

Từ gthiêt suy ra: $ac+a.\frac{1}{b}+c.\frac{1}{b}=1$suy ra tồn tại $\Delta ABC$ sao cho $a=tan\frac{A}{2};\frac{1}{b}=tan\frac{B}{2};c=tan\frac{C}{2}$Khi đó:$P=2.cos^2\frac{2}{2}-2sin^2 $$\frac{B}{2}+3cos^2\frac{C}{2}$$ P=cosA+cosB+3-3sin^2\frac{C}{2}=2sin\frac{C}{2}.cos\frac{A-B}{2}+3-sin^2\frac{C}{2}=\frac{10}{3}-3(sin\frac{C}{2}-\frac{1}{3}.cos\frac{A-B}{2})^2-\frac{1}{3}.cos^2\frac{A-B}{2}\leq \frac{10}{3}$Vậy max$P=\frac{10}{3}$ khi.....

Từ gthiêt suy ra: $ac+a.\frac{1}{b}+c.\frac{1}{b}=1$suy ra tồn tại $\Delta ABC$ sao cho $a=tan\frac{A}{2};\frac{1}{b}=tan\frac{B}{2};c=tan\frac{C}{2}$Khi đó:$P=2.cos^2\frac{A}{2}-2sin^2 $$\frac{B}{2}+3cos^2\frac{C}{2}$$ P=cosA+cosB+3-3sin^2\frac{C}{2}=2sin\frac{C}{2}.cos\frac{A-B}{2}+3-sin^2\frac{C}{2}=\frac{10}{3}-3(sin\frac{C}{2}-\frac{1}{3}.cos\frac{A-B}{2})^2-\frac{1}{3}.cos^2\frac{A-B}{2}\leq \frac{10}{3}$Vậy max$P=\frac{10}{3}$ khi.....