❄Jewish Hải❄
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camcon
Nguyễn Việt Lâm
13 giờ trước (22:04)

\(\Leftrightarrow m\left(2^{x^2-5x+6}-1\right)+2^{1-x^2}\left(1-2^{x^2-5x+6}\right)=0\)

\(\Leftrightarrow\left(2^{x^2-5x+6}-1\right)\left(m-2^{1-x^2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+6=0\\2^{1-x^2}=m\end{matrix}\right.\)

\(\Rightarrow2^{1-x^2}=m\) có 2 nghiệm pb khác 2 và 3

\(1-x^2\le1;\forall x\Rightarrow0< 2^{1-x^2}\le2\)

\(\Rightarrow0< m< 2\)

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camcon
gurp ble
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Nguyên Nguyễn Hồ Xuân
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Ẩn danh
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Big City Boy
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Nguyễn Đức Trí
20 giờ trước (14:16)

Ta có :

\(S=cos2A+\sqrt[]{3}.\left(cos2B+cos2C\right)\)

\(\Leftrightarrow S=2cos^2A-1+2\sqrt[]{3}.\left[cos\left(B+C\right).cos\left(B-C\right)\right]\)

\(\Leftrightarrow2S=4cos^2A-4\sqrt[]{3}.cosA.cos\left(B-C\right)-2\)

\(\Leftrightarrow2S=4cos^2A-4\sqrt[]{3}.cosA.cos\left(B-C\right)+3cos^2\left(B-C\right)+3-3cos^2\left(B-C\right)-5\)

\(\Leftrightarrow2S=\left[2cosA-\sqrt[]{3}.cos\left(B-C\right)\right]^2+3\left[1-cos^2\left(B-C\right)\right]-5\)

\(\Leftrightarrow2S=\left[2cosA-\sqrt[]{3}.cos\left(B-C\right)\right]^2+3sin^2\left(B-C\right)-5\)

\(\Leftrightarrow S=\dfrac{1}{2}\left[2cosA-\sqrt[]{3}.cos\left(B-C\right)\right]^2+\dfrac{3}{2}sin^2\left(B-C\right)-\dfrac{5}{2}\)

mà \(\left\{{}\begin{matrix}\dfrac{1}{2}\left[2cosA-\sqrt[]{3}.cos\left(B-C\right)\right]^2\ge0\\\dfrac{3}{2}sin^2\left(B-C\right)\ge0\end{matrix}\right.\)

\(\Rightarrow S\ge-\dfrac{5}{2}\)

\(\Rightarrow GTNN\left(S\right)=-\dfrac{5}{2}\)

Dấu "=" xảy ra khi và chỉ khi

\(\left\{{}\begin{matrix}sin\left(B-C\right)=0\\cosA=\dfrac{\sqrt[]{3}}{2}\left(B-C\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}B-C=0\\cosA=\dfrac{\sqrt[]{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=30^o\\B=C=75^o\end{matrix}\right.\)

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Hoàng Văn
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Hồ Quỳnh Anh
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