Lời giải bài 6 chuyên đề Ứng dụng nghiệm phương trình bậc hai.
a. $\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\leq \frac{1}{8}$ .
<=> $8\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\leq 1$ .
Xét $T=8\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}$ .
<=> $T=4\sin \frac{C}{2}(\cos \frac{A-B}{2}-\cos \frac{A+B}{2})$ .
<=> $T=4\sin \frac{C}{2}(\cos \frac{A-B}{2}-\sin \frac{C}{2})$ .
=> $4\sin ^{2}\frac{C}{2}-4\cos \frac{A-B}{2}.\sin \frac{C}{2}+T=0$ . (1)
Phương trình (1) có nghiệm <=> $\Delta {}' \geq 0$
<=> $4\cos ^{2}\frac{A-B}{2}-4T\geq 0$
=> $T\leq \cos ^{2}\frac{A-B}{2}\leq 1$
=> $\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\leq \frac{1}{8}$ . (đpcm)
b. Đặt X = $\cos A+\cos B+\cos C$
<=> $X=2\cos \frac{A+B}{2}.\cos \frac{A-B}{2}+1-2\sin ^{2}\frac{C}{2}$
=> $2\sin ^{2}\frac{C}{2}-2\cos \frac{A-B}{2}.\sin \frac{C}{2}+X-1=0$ (2)
Phương trình (2) có nghiệm <=> $\Delta {}' \geq 0$
<=> $\cos ^{2}\frac{A-B}{2}-2(X-1)\geq 0$
<=> $X\leq 1+\frac{1}{2}\cos ^{2}\frac{A-B}{2}\leq 1+\frac{1}{2}=\frac{3}{2}$
=> $\cos A+\cos B+\cos C\leq \frac{3}{2}$ . (đpcm)
c. Đặt V = $\sin ^{2}A+\sin ^{2}B+\sin ^{2}C$
<=> $V=\frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}+1-\cos ^{2}C$
<=> $V=2-\frac{1}{2}(\cos 2A+\cos 2B)-\cos ^{2}C$
<=> $V=2-\cos (A+B)\cos (A-B)-\cos ^{2}C$
<=> $V=2+\cos C.\cos (A+B)-\cos ^{2}C$
=> $\cos ^{2}C-\cos (A-B)\cos C+V-2=0$ (3)
Phương trình (3) có nghiệm <=> $\Delta \geq 0$
<=> $\cos ^{2}(A-B)-4(V-2)\geq 0$
<=> $V\leq 2+\frac{1}{4}\cos ^{2}(A-B)$
<=> $V\leq 2+\frac{1}{4}=\frac{9}{4}$ .
=> $\sin ^{2}A+\sin ^{2}B+\sin ^{2}C\leq \frac{9}{4}$ . (đpcm)
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