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Question

$1)$ Cho tam giác

$ABC$ thỏa mãn điều kiện

$\tan\frac{B}{2}\tan\frac{C}{2}=\frac{1}{3}$ . Giải phương trình

$x^{2}+x-\cos A-\frac{1}{4}cos(B-C)=0$

NguyễnTốngKhánhLinh · Answer 1 · 8/22/2013 7:06:57 PM

$tan\frac{B}2.tan\frac{C}2 = \frac{1}3 \Leftrightarrow 3sin\frac{B}2.sin\frac{C}2 = cos\frac{B}2.cos\frac{C}2$

$\Leftrightarrow sin\frac{B}2.sin\frac{C}2 = cos\frac{B}2.cos\frac{C}2 - 2sin\frac{B}2.sin\frac{C}2$

$\Leftrightarrow cos\frac{B}2.cos\frac{C}2 + sin\frac{B}2.sin\frac{C}2 = 2cos\frac{B}2.cos\frac{C}2 - 2sin\frac{B}2.sin\frac{C}2$

$\Leftrightarrow cos(\frac{B}2 - \frac{C}2) = 2cos(\frac{B}2+\frac{C}2)$

Thế vào PT có

$x^2+x+cos(B+C)-\frac1{4}cos(B-C)=0$

$\Leftrightarrow x^2+x+2cos^2(\frac{B+C}2)-1-\frac{1}4[2cos^2(\frac{B-C}2)-1]=0$

$\Leftrightarrow x^2+x+2cos^2(\frac{B+C}2)-1-\frac{1}2.[4cos^2(\frac{B+C}2)]+\frac{1}4=0$

$\Leftrightarrow x^2+x-1+\frac{1}4=0$

$\Leftrightarrow x^2+x-\frac{3}4=0$

$\Leftrightarrow x=\frac{1}2;-\frac{3}2$

1 Đáp án