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)

 - - - - - Hết ! - - - -