Quay lại đáp án

	2	bớt 420 ký tự trong đáp án		Sửa 04-02-16 08:57 AM ๖ۣۜTQT☾♋☽ 133 4
ta có: 2abc(cosA+cosB) = $2abc(\frac{b^2+c^2-a^2}{a^2+c^2-b^2})$ = ab2 + ba2 + ac2 + bc2 - a3 - b3 = (a + b) (c2 – (a + b)2 = (a + b)(c + b – a)(c + a – b) (đpcm). ta có: 2abc(cosA+cosB) = $2abc(\frac{b^2+c^2-a^2}{a^2+c^2-b^2})$ = ab2 + ba2 + ac2 + bc2 - a3 - b3 = (a + b) (c2 – (a + b)2 = (a + b)(c + b – a)(c + a – b) (đpcm). ta có: 2abc(cosA+cosB) = $2abc(\frac{b^2+c^2-a^2}{a^2+c^2-b^2})$ = ab2 + ba2 + ac2 + bc2 - a3 - b3 = (a + b) (c2 – (a + b)2 = (a + b)(c + b – a)(c + a – b) (đpcm).

	1	tạo mới		Trả lời 22-01-16 08:16 PM ๖ۣۜTQT☾♋☽ 133 4
ta có: 2abc(cosA+cosB) = $2abc(\frac{b^2+c^2-a^2}{a^2+c^2-b^2})$ = ab2 + ba2 + ac2 + bc2 - a3 - b3 = (a + b) (c2 – (a + b)2 = (a + b)(c + b – a)(c + a – b) (đpcm).

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