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Lời giải:
\(\frac{1+\cos B}{\sin B}=\frac{2a+c}{\sqrt{(2a-c)(2a+c)}}\)
\(\Rightarrow \frac{(1+\cos B)^2}{\sin ^2B}=\frac{2a+c}{2a-c}\) (bình phương 2 vế)
\(\Leftrightarrow \frac{1+\cos ^2B+2\cos B}{\sin ^2B}=\frac{2a-c+2c}{2a-c}\)
\(\Leftrightarrow \frac{\sin ^2B+2\cos ^2B+2\cos B}{\sin ^2B}=1+\frac{2c}{2a-c}\)
\(\Leftrightarrow \frac{\cos ^2B+\cos B}{\sin ^2B}=\frac{c}{2a-c}\)
\(\Rightarrow (2a-c)(\cos ^2B+\cos B)=c\sin ^2B\)
\(\Leftrightarrow 2a\cos ^2B+(2a-c)\cos B=c\sin ^2B+c\cos ^2B=c(\sin ^2B+\cos ^2B)=c\)
\(\Leftrightarrow 2a(\cos ^2B+\cos B)=c(\cos B+1)\)
\(\Leftrightarrow (\cos B+1)(2a\cos B-c)=0\)
Với mọi \(\widehat{B}< 180^0\Rightarrow \cos B+1\neq 0\). Suy ra \(2a\cos B-c=0\Rightarrow \cos B=\frac{c}{2a}\)
Kẻ đường cao $CH$ xuống $AB$
\(\cos B=\frac{BH}{BC}=\frac{BH}{a}=\frac{c}{2a}\)
\(\Rightarrow BH=\frac{c}{2}\) hay $H$ là trung điểm của $AB$. Vậy $CH$ đồng thời là đường cao và đường trung tuyến, suy ra tam giác $ABC$ cân tại $C$
Bài 1:
a)
\(\sin ^2x+\sin ^2x\cot^2x=\sin ^2x(1+\cot^2x)=\sin ^2x(1+\frac{\cos ^2x}{\sin ^2x})\)
\(=\sin ^2x.\frac{\sin ^2x+\cos^2x}{\sin ^2x}=\sin ^2x+\cos^2x=1\)
b)
\((1-\tan ^2x)\cot^2x+1-\cot^2x\)
\(=\cot^2x(1-\tan^2x-1)+1=\cot^2x(-\tan ^2x)+1=-(\tan x\cot x)^2+1\)
\(=-1^2+1=0\)
c)
\(\sin ^2x\tan x+\cos^2x\cot x+2\sin x\cos x=\sin ^2x.\frac{\sin x}{\cos x}+\cos ^2x.\frac{\cos x}{\sin x}+2\sin x\cos x\)
\(=\frac{\sin ^3x}{\cos x}+\frac{\cos ^3x}{\sin x}+2\sin x\cos x=\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin x\cos x}=\frac{(\sin ^2x+\cos ^2x)^2}{\sin x\cos x}=\frac{1}{\sin x\cos x}\)
\(=\frac{1}{\frac{\sin 2x}{2}}=\frac{2}{\sin 2x}\)
Bài 2:
Áp dụng BĐT Cauchy Schwarz ta có:
\(P=\frac{a^2}{\sqrt{a(2c+a+b)}}+\frac{b^2}{\sqrt{b(2a+b+c)}}+\frac{c^2}{\sqrt{c(2b+c+a)}}\)
\(\geq \frac{(a+b+c)^2}{\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}}(*)\)
Tiếp tục áp dụng BĐT Cauchy-Schwarz:
\((\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq (a+b+c)(2c+a+b+2a+b+c+2b+c+a)\)
\(\Leftrightarrow (\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq 4(a+b+c)^2\)
\(\Rightarrow \sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}\leq 2(a+b+c)(**)\)
Từ \((*); (**)\Rightarrow P\geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\)
Dấu "=" xảy ra khi $a=b=c=1$
Ta có : A+B+C= 180
=>sin(A+B)/2 = sin(180/2 - C/2) = cosC/2
ttcó: sinC/2 = cos(A+B)/2
=> sA+sB+sC =2cosC/2*cos(A-B)/2 + 2cos(A+B)/2*cosC/2
=2cosC/2
=4cosA/2cosB/2cosC/2
điều kiện : \(\dfrac{\pi}{2}\) < α < \(\pi\) (1)
\(\sin^2\dfrac{\alpha}{2}+\cos^2\dfrac{\alpha}{2}=1\)
⇔ \(\left(\dfrac{2}{\sqrt{5}}\right)^2+\cos^2\dfrac{\alpha}{2}=1\)
⇒ \(\cos\dfrac{\alpha}{2}=\pm\dfrac{1}{\sqrt{5}}\)
Do (1) nên ta có \(\dfrac{\pi}{4}< \dfrac{\alpha}{2}< \dfrac{\pi}{2}\): \(\cos\dfrac{\alpha}{2}>0\) ⇒ \(\cos\dfrac{\alpha}{2}=\dfrac{1}{\sqrt{5}}\) ⇒ \(\tan\dfrac{\alpha}{2}=\dfrac{\sin\dfrac{\alpha}{2}}{\cos\dfrac{\alpha}{2}}=\dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{1}{\sqrt{5}}}=2\)
Khi đó ta có:
A = \(\dfrac{\tan\dfrac{\alpha}{2}-\tan\dfrac{\pi}{4}}{1+\tan\dfrac{\alpha}{2}.\tan\dfrac{\pi}{4}}\) = \(\dfrac{2-1}{1+2.1}\) =\(\dfrac{1}{3}\)
VẬY..............................
Do \(a\le\left|a\right|,b\le\left|b\right|\) nên ta chỉ cần chứng minh
\(\dfrac{\left|a\right|}{\sqrt{a^2+b^2}}+\dfrac{\left|b\right|}{\sqrt{9a^2+b^2}}+\dfrac{2\left|a\right|\left|b\right|}{\sqrt{a^2+b^2}.\sqrt{9a^2+b^2}}\le\dfrac{3}{2}\)
Đặt \(a^2=x,b^2=3y^2\)
\(P=2\sqrt{\dfrac{x}{x+3y}}+2\sqrt{\dfrac{y}{y+3x}}+4\sqrt{\dfrac{xy}{\left(x+3y\right)\left(y+3x\right)}}\le3\)
Sử dụng BĐT AM-GM, ta có
\(2\sqrt{\dfrac{x}{x+3y}}\le\dfrac{x}{x+y}+\dfrac{x+y}{3x+y},2\sqrt{\dfrac{y}{y+3x}}\le\dfrac{y}{x+y}+\dfrac{x+y}{y+3x}\)\(4\sqrt{\dfrac{xy}{\left(x+3y\right)\left(y+3x\right)}}\le\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}+\dfrac{1}{2}\)
Cộng ba bất đẳng thức trên vế theo vế
\(P\le\dfrac{3}{2}+\dfrac{x+y}{x+3y}+\dfrac{x+y}{y+3x}+\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}\)
Và do đó chứng minh sẽ hoàn tất nếu ta chỉ ra được rằng:
\(\dfrac{x+y}{x+3y}+\dfrac{x+y}{y+3x}+\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}\le\dfrac{3}{2}\)
Ta có: \(\dfrac{3}{2}-\dfrac{x+y}{x+3y}-\dfrac{x+y}{y+3x}-\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}=\dfrac{3}{2}-\dfrac{4\left(x+y\right)^2+8xy}{\left(x+3y\right)\left(y+3x\right)}=\dfrac{\left(x-y\right)^2}{2\left(x+3y\right)\left(y+3x\right)}\ge0\)Bài toán được chứng minh xong. Đẳng thức xảy ra khi \(b=\sqrt{3}a>0\)
\(sin\dfrac{B}{2}=\dfrac{b}{2\sqrt{ac}}\Rightarrow sin^2\dfrac{B}{2}=\dfrac{b^2}{4ac}\Rightarrow\dfrac{1-cosB}{2}=\dfrac{b^2}{4ac}\)
\(\Rightarrow1-\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{b^2}{2ac}\Rightarrow2ac-a^2-c^2+b^2=b^2\)
\(\Rightarrow-\left(a-c\right)^2=0\Rightarrow a=c\)
\(\Rightarrow\Delta ABC\) cân tại B