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Lời giải:
Vì $\tan a=\frac{\sin a}{\cos a}$ xác định nên $\cos a\neq 0$. Do đó:
\(A=\frac{\sin a+\cos a}{\cos a-\sin a}=\frac{\frac{\sin a+\cos a}{\cos a}}{\frac{\cos a-\sin a}{\cos a}}=\frac{\frac{\sin a}{\cos a}+1}{1-\frac{\sin a}{\cos a}}=\frac{\tan a+1}{1-\tan a}=\frac{\frac{1}{2}+1}{1-\frac{1}{2}}=3\)
\(A=\left(\sin\alpha+\cos\alpha+\sin\alpha-\cos\alpha\right)^2-2\left(\sin\alpha+\cos\alpha\right)\left(\sin\alpha-\cos\alpha\right)\)
\(=4\sin^2\alpha-2\sin^2\alpha+2\cos^2\alpha=2\left(\sin^2\alpha+\cos^2\alpha\right)=2\)
\(B=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2-1=0\)
\(C=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\)
\(=3\left(\sin^2\alpha+\cos^2\alpha-\frac{1}{9}\right)^2-\frac{1}{9}=\frac{61}{27}\)
\(tana=\frac{1}{2}\)
\(\Rightarrow\frac{sina}{cosa}=\frac{1}{2}\)
\(2sina=cosa\)
\(A=\frac{sina+cosa}{cosa-sina}\)
\(=\frac{sina+2sina}{2sina-sina}\)
\(=\frac{3sina}{sina}=3\)
Sửa lại đề: CMR $P=\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\leq 1$
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Lời giải:
Do $abc=1$ nên tồn tại $x,y,z>0$ sao cho $(a,b,c)=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})$
Bài toán đã cho trở thành:
Cho $x,y,z>0$. CMR $P=\frac{y}{x+2y}+\frac{z}{y+2z}+\frac{x}{z+2x}\leq 1$
Thật vậy:
$P=\frac{1}{2}(\frac{1-\frac{x}{x+2y})+\frac{1}{2}(1-\frac{y}{y+2z})+\frac{1}{2}(1-\frac{z}{z+2x})$
$=\frac{3}{2}-\frac{1}{2}\left(\frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x}\right)(*)$
Áp dụng BĐT Cauchy-Schwarz:
$\frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x}=\frac{x^2}{x^2+2xy}+\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}\geq \frac{(x+y+z)^2}{x^2+2xy+y^2+2yz+z^2+2zx}=\frac{(x+y+z)^2}{(x+y+z)^2}=1(**)$
Từ $(*); (**)\Rightarrow P\leq \frac{3}{2}-\frac{1}{2}.1=1$
Ta có đpcm.
Dấu "=" xảy ra khi $x=y=z$ hay $a=b=c=1$
Đặt \(\left(a;b;c\right)\rightarrow\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)\Rightarrow abc=1\left(TMGT\right)\)
Ta có:
\(\frac{1}{a+2}=\frac{1}{\frac{x}{y}+2}=\frac{1}{\frac{x+2y}{y}}=\frac{y}{x+2y}=\frac{y^2}{xy+2y^2}\)
Tương tự:
\(\frac{1}{b+2}=\frac{z^2}{yz+z^2};\frac{1}{c+2}=\frac{x^2}{zx+x^2}\)
Ta có:
\(\frac{x^2}{xz+2x^2}+\frac{y^2}{xy+2y^2}+\frac{z^2}{yz+2z^2}\ge\frac{\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+xy+yz+zx}\)
Mặt khác \(xy+yz+zx\le x^2+y^2+z^2\)
\(\Rightarrow2\left(x^2+y^2+z^2\right)+xy+yz+zx\le3\left(x^2+y^2+z^2\right)\)
Rồi OK.Đến đây tịt r:( GOD nào vào thông não hộ ạ:(
\(=\frac{\sqrt{2}}{2}-\frac{3}{2}.\frac{3\sqrt{2}}{2}+\frac{2}{5}.5\sqrt{2}=\frac{\sqrt{2}}{2}-\frac{9}{2\sqrt{2}}+2\sqrt{2}\)
\(=\frac{2-9+8}{2\sqrt{2}}=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}\)
\(A=\frac{3x}{4}+\frac{x}{4}+\frac{1}{x}\ge\frac{3x}{4}+2\sqrt{\frac{x}{4x}}\ge\frac{3.2}{4}+1=\frac{5}{2}\)
\(A_{min}=\frac{5}{2}\) khi \(x=2\)
\(B=\frac{24x}{25}+\frac{x}{25}+\frac{1}{x}\ge\frac{24x}{25}+2\sqrt{\frac{x}{25x}}\ge\frac{24.5}{25}+\frac{2}{5}=\frac{26}{5}\)
\(B_{min}=\frac{26}{5}\) khi \(x=5\)
Câu C bạn coi lại đề, nếu đúng thế này thì ko tồn tại min
\(a,VT=cot\alpha+\dfrac{sin\alpha}{1+cos\alpha}\\ =\dfrac{cos\alpha}{sin\alpha}+\dfrac{sin\alpha}{1+cos\alpha}\\ =\dfrac{cos\alpha\left(1+cos\alpha\right)+sin^2\alpha}{sin\alpha\left(1+cos\alpha\right)}\\ =\dfrac{cos\alpha+cos^2\alpha+sin^2\alpha}{sin\alpha\left(1+cos\alpha\right)}\\ =\dfrac{cos\alpha+1}{sin\alpha\left(1+cos\alpha\right)}\\ =\dfrac{1}{sin\alpha}=VP\left(dpcm\right)\)
\(b,VT=\dfrac{1}{1-sin\alpha}+\dfrac{1}{1+sin\alpha}\\ =\dfrac{1+sin\alpha+1-sin\alpha}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}\\ =\dfrac{2}{1-sin^2\alpha}\\ =\dfrac{2}{cos^2\alpha}=VP\left(dpcm\right)\)