Question

[Umi]

Answer 1

[Nguyễn Ngọc Anh Minh]

\(PT\Leftrightarrow1-2\sin^2\dfrac{A}{2}+2\sqrt{2}\cos\dfrac{B+C}{2}\cos\dfrac{B-C}{2}=2\)

Ta có

\(\cos\dfrac{B+C}{2}=\cos\dfrac{180^o-A}{2}=\cos\left(90^o-\dfrac{A}{2}\right)=\sin\dfrac{A}{2}\)

\(\Rightarrow PT\Leftrightarrow2\sin^2\dfrac{A}{2}-2\sqrt{2}\cos\dfrac{B-C}{2}\sin\dfrac{A}{2}+1=0\) (1)

(PT bậc 2 ẩn là \(\sin\dfrac{A}{2}\) )

\(\Delta=8\cos^2\dfrac{B-C}{2}-8=8\left(\cos^2\dfrac{B-C}{2}-1\right)=\)

\(=8\left(\cos^2\dfrac{B-C}{2}-\cos^2\dfrac{B-C}{2}-\sin^2\dfrac{B-C}{2}\right)=-8\sin^2\dfrac{B-C}{2}\)

Phương trình có nghiệm khi

\(\Delta=-8\sin^2\dfrac{B-C}{2}\ge0\Leftrightarrow\sin^2\dfrac{B-C}{2}\le0\)

\(\Rightarrow\sin^2\dfrac{B-C}{2}=0\Rightarrow\sin\dfrac{B-C}{2}=0\Rightarrow\dfrac{B-C}{2}=0\Rightarrow B=C\)

=> tg ABC cân tại A

Thay \(\dfrac{B-C}{2}=0\) vào (1)

\(\left(1\right)\Leftrightarrow2\sin^2\dfrac{A}{2}-2\sqrt{2}\sin\dfrac{A}{2}+1=0\)

Giải PT được \(\sin\dfrac{A}{2}=\dfrac{\sqrt{2}}{2}\Rightarrow\dfrac{A}{2}=45^o\Rightarrow A=90^o\)

=> tg ABC vuông cân tại A (đpcm)