Cho 3 góc \(\alpha,\beta,\gamma\) tạo thành một cấp số cộng theo thứ tự đó với công sai \(d=\dfrac{\pi}{3}\). Chứng minh :
a) \(\tan\alpha.\tan\beta+\tan\beta\tan\gamma+\tan\gamma.\tan\alpha=-3\)
b) \(4\cos\alpha.\cos\beta\cos\gamma=\cos3\beta\)
Đố: Cho \(\Delta ABC\), biết \(BC=a,AC=b,AB=c,\widehat{A}=\alpha,\widehat{B}=\beta,\widehat{C}=\gamma\) chứng minh:
a)\(\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}\) b) \(a^2=b^2+c^2-2bc\cos\alpha\)
c) \(\frac{a-b}{a+b}=\frac{\tan\left[\frac{1}{2}\left(\alpha-\beta\right)\right]}{\tan\left[\frac{1}{2}\left(\alpha+\beta\right)\right]}\)
d) Biết \(s=\frac{a+b+c}{2}\). Chứng minh \(\frac{\cot\frac{\alpha}{2}}{s-a}=\frac{\cot\frac{\beta}{2}}{s-b}=\frac{\cot\frac{\gamma}{2}}{s-c}\)
Chứng minh đẳng thức:
\(\dfrac{sin\left(\alpha-\beta\right)}{sin\alpha sin\beta}+\dfrac{sin\left(\beta-\gamma\right)}{sin\beta sin\gamma}+\dfrac{sin\left(\gamma-\alpha\right)}{sin\gamma sin\alpha}=0\)
\(\dfrac{sin\left(a-b\right)}{sina.sinb}+\dfrac{sin\left(b-c\right)}{sinb.sinc}+\dfrac{sin\left(c-a\right)}{sinc.sina}\)
\(=\dfrac{sina.cosb-cosa.sinb}{sina.sinb}+\dfrac{sinb.cosc-cosb.sinc}{sinb.sinc}+\dfrac{sinc.cosa-cosc.sina}{sina.sinc}\)
\(=\dfrac{cosb}{sinb}-\dfrac{cosa}{sina}+\dfrac{cosc}{sincc}-\dfrac{cosb}{sinb}+\dfrac{cosa}{sina}-\dfrac{cosc}{sincc}\)
\(=0\)
Chứng minh rằng các biểu thức sau là những hằng số không phụ thuộc \(\alpha,\beta\) :
a) \(\sin6\alpha\cot3\alpha-\cos6\alpha\)
b) \(\left[\tan\left(90^0-\alpha\right)-\cot\left(90^0+\alpha\right)\right]^2-\left[\cot\left(180^0+\alpha\right)+\cot\left(270^0+\alpha\right)\right]^2\)
c) \(\left(\tan\alpha-\tan\beta\right)\cot\left(\alpha-\beta\right)-\tan\alpha\tan\beta\)
d) \(\left(\cot\dfrac{\alpha}{3}-\tan\dfrac{\alpha}{3}\right)\tan\dfrac{2\alpha}{3}\)
a) \(sin6\alpha cot3\alpha cos6\alpha=2.sin3\alpha.cos3\alpha\dfrac{cos3\alpha}{sin3\alpha}-cos6\alpha\)
\(=2cos^23\alpha-\left(2cos^23\alpha-1\right)=1\) (Không phụ thuộc vào x).
b) \(\left[tan\left(90^o-\alpha\right)-cot\left(90^o+\alpha\right)\right]^2\)\(-\left[cot\left(180^o+\alpha\right)+cot\left(270^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+cot\left(90^o-\alpha\right)\right]^2\)\(-\left[cot\alpha+cot\left(90^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+tan\alpha\right]^2-\left[cot\alpha-tan\alpha\right]^2\)
\(=4tan\alpha cot\alpha=4\). (Không phụ thuộc vào \(\alpha\)).
c) \(\left(tan\alpha-tan\beta\right)cot\left(\alpha-\beta\right)-tan\alpha tan\beta\)
\(=\left(\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}\right).\dfrac{cos\left(\alpha-\beta\right)}{sin\left(\alpha-\beta\right)}-tan\alpha tan\beta\)
\(=\left(\dfrac{sin\alpha cos\beta-cos\alpha sin\beta}{cos\alpha cos\beta}\right).\dfrac{cos\left(\alpha-\beta\right)}{sin\left(\alpha-\beta\right)}\)\(-\dfrac{sin\alpha sin\beta}{cos\alpha cos\beta}\)
\(=\dfrac{sin\left(\alpha-\beta\right)}{cos\alpha cos\beta}.\dfrac{cos\left(\alpha-\beta\right)}{sin\left(\alpha-\beta\right)}-\dfrac{sin\alpha sin\beta}{cos\alpha cos\beta}\)
\(=\dfrac{cos\left(\alpha-\beta\right)}{cos\alpha cos\beta}-\dfrac{sin\alpha sin\beta}{cos\alpha cos\beta}\)
\(=\dfrac{cos\alpha cos\beta+sin\alpha sin\beta-sin\alpha sin\beta}{cos\alpha cos\beta}=\dfrac{cos\alpha cos\beta}{cos\alpha cos\beta}=1\).
Cho 0°< α<β< 90°. Chứng minh:
a) sin α < tan α
b) cos α < cotan α
c) sin α < sin β
d) cos α > cos β
e) tan α < tan β
f) cotan α > cotan β
Với \(\alpha\ge\beta\ge\gamma>0\) , \(a\ge\alpha\) , \(ab\ge\alpha\beta\) , \(abc\ge\alpha\beta\gamma\)
Chứng minh rằng \(a+b+c\ge\alpha+\beta+\gamma\)
\(VT=a+b+c=\alpha.\frac{a}{\alpha}+\beta.\frac{b}{\beta}+\gamma.\frac{c}{\gamma}\)
Áp dụng phương pháp nhóm ABEL
\(\Rightarrow VT=\left(\alpha-\beta\right)\frac{a}{\alpha}+\left(\beta-\gamma\right)\left(\frac{a}{\alpha}+\frac{b}{\beta}\right)+\gamma\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\left\{\begin{matrix}\frac{a}{\alpha}+\frac{b}{\beta}\ge2\sqrt{\frac{ab}{\alpha\beta}}\left(1\right)\\\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\ge3\sqrt[3]{\frac{abc}{\alpha\beta\gamma}}\left(3\right)\end{matrix}\right.\)
Ta có \(ab\ge\alpha\beta\Rightarrow\frac{ab}{\alpha\beta}\ge1\) \(\Rightarrow2\sqrt{\frac{ab}{\alpha\beta}}\ge2\left(2\right)\)
Ta có \(abc\ge\alpha\beta\gamma\Rightarrow\frac{abc}{\alpha\beta\gamma}\ge1\Rightarrow3\sqrt[3]{\frac{abc}{\alpha\beta\gamma}}\ge3\left(4\right)\)
Từ ( 1 ) và ( 2 )
\(\Rightarrow\frac{a}{\alpha}+\frac{b}{\beta}\ge2\)
\(\Rightarrow\left(\beta-\gamma\right)\left(\frac{a}{\alpha}+\frac{b}{\beta}\right)\ge2\left(\beta-\gamma\right)\) ( 5 )
Từ ( 3 ) và ( 4 )
\(\Rightarrow\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\ge3\)
\(\Rightarrow\gamma\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)\ge3\gamma\) ( 6 )
Theo đề bài ta có \(a\ge\alpha\Rightarrow\frac{a}{\alpha}\ge1\)\(\Rightarrow\left(\alpha-\beta\right)\frac{a}{\alpha}\ge\alpha-\beta\) ( 7 )
Từ ( 5 ) , ( 6 ) , ( 7 ) cộng theo từng vế
\(\Rightarrow VT=\left(\alpha-\beta\right)\frac{a}{\alpha}+\left(\beta-\gamma\right)\left(\frac{a}{\alpha}+\frac{b}{\beta}\right)+\gamma\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)\ge2\left(\beta-\gamma\right)+3\gamma+\alpha-\beta\)
\(\Rightarrow VT\ge2\beta-2\gamma+3\gamma+\alpha-\beta\)
\(\Rightarrow VT\ge\alpha+\beta+\gamma\)
\(\Leftrightarrow a+b+c\ge\alpha+\beta+\gamma\) ( đpcm )
Cho \(0< \alpha\); \(\beta< \frac{\pi}{2}\); \(\alpha+\beta=\frac{\pi}{4}\) và \(tan\alpha.tan\beta=3-2\sqrt{2}\)
a) Tính gtri của \(A=tan\left(\alpha+\beta\right)\)
b) Tính gtri của \(B=tan\alpha+tan\beta\)
c) TÍnh \(tan\alpha\) và \(tan\beta\). Suy ra \(\alpha\) và \(\beta\)
\(A=tan\left(a+b\right)=tan\frac{\pi}{4}=1\)
Ta có: \(tan\left(a+b\right)=\frac{tana+tanb}{1-tana.tanb}\)
\(\Rightarrow B=tana+tanb=tan\left(a+b\right)\left(1-tana.tanb\right)=1.\left(1-3+2\sqrt{2}\right)=2\sqrt{2}-2\)
\(\left\{{}\begin{matrix}tana+tanb=2\sqrt{2}-2\\tana.tanb=3-2\sqrt{2}\end{matrix}\right.\)
Theo Viet đảo, \(tana;tanb\) là nghiệm của:
\(x^2-\left(2\sqrt{2}-2\right)x+3-2\sqrt{2}=0\)
\(\Leftrightarrow\left(x-\sqrt{2}+1\right)^2=0\Rightarrow x=\sqrt{2}-1\)
\(\Rightarrow tana=tanb=\sqrt{2}-1\Rightarrow a=b=\frac{\pi}{8}\)
Chứng minh các đẳng thức :
a) \(\dfrac{\tan\alpha-\tan\beta}{\cot\beta-\cot\alpha}=\tan\alpha\tan\beta\)
b) \(\tan100^0+\dfrac{\sin530^0}{1+\sin640^0}=\dfrac{1}{\sin10^0}\)
c) \(2\left(\sin^6\alpha+\cos^6\alpha\right)+1=3\left(\sin^4\alpha+\cos^4\alpha\right)\)
a) \(\dfrac{tan\alpha-tan\beta}{cot\beta-cot\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}}{\dfrac{cos\beta}{sin\beta}-\dfrac{cos\alpha}{sin\alpha}}\)
\(=\dfrac{\dfrac{sin\alpha cos\beta-cos\alpha sin\beta}{cos\alpha cos\beta}}{\dfrac{cos\beta sin\alpha-cos\alpha sin\beta}{sin\beta sin\alpha}}\)
\(=\dfrac{sin\beta sin\alpha}{cos\beta cos\alpha}=tan\alpha tan\beta\).
b) \(tan100^o+\dfrac{sin530^o}{1+sin640^o}=tan100^o+\dfrac{sin170^o}{1+sin280^o}\)
\(=-cot10^o+\dfrac{sin10^o}{1-sin80^o}\)\(=\dfrac{-cos10^o}{sin10^o}+\dfrac{sin10^o}{1-cos10^o}\)
\(=\dfrac{-cos10^o+cos^210^o+sin^210^o}{sin10^o\left(1-cos10^o\right)}\) \(=\dfrac{1-cos10^o}{sin10^o\left(1-cos10^o\right)}=\dfrac{1}{sin10^o}\) .
c) \(2\left(sin^6\alpha+cos^6\alpha\right)+1=2\left(sin^2\alpha+cos^2\alpha\right)\)\(\left(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha\right)+1\)
\(=2\left(sin^4\alpha+cos^4\alpha-sin^2\alpha cos^2\alpha\right)+1\)
\(=2\left(sin^4\alpha+cos^4\alpha\right)+sin^2\alpha-sin^2\alpha cos^2\alpha+\)\(cos^2\alpha-sin^2\alpha cos^2\alpha\)
\(=2\left(sin^4\alpha+cos^4\alpha\right)+sin^2\alpha\left(1-cos^2\alpha\right)+\)\(cos^2\alpha\left(1-sin^2\alpha\right)\)
\(=2\left(sin^4\alpha+cos^4\alpha\right)+sin^2\alpha.sin^2\alpha+cos^2\alpha.cos^2\alpha\)
\(=3\left(sin^4\alpha+cos^4\alpha\right)\).
1.Cho \(\alpha,\beta\left(\alpha\ne\beta\right)\in\left(0;\dfrac{\pi}{2}\right)\)và thỏa mãn điều kiện \(\dfrac{cosx-cos\alpha}{cosx-cos\beta}=\dfrac{sin^2\alpha cos\beta}{sin^2\beta cos\alpha}\)
(giả sử \(x\) xác định). Chứng minh\(tan^2\dfrac{x}{2}=tan^2\dfrac{\alpha}{2}tan^2\dfrac{\beta}{2}\)
2. Giải hệ phương trình \(\left\{{}\begin{matrix}xy-2y-3=\sqrt{y-x-1}+\sqrt{y-3x+5}\\\left(1-y\right)\sqrt{2x-y}+2\left(x-1\right)=\left(2x-y-1\right)\sqrt{y}\end{matrix}\right.\)
3. Cho ba số thực dương a, b, c thỏa mãn \(\dfrac{1}{a+2}+\dfrac{1}{b+3}+\dfrac{1}{c+4}=1\). Tìm Min của biểu thức \(P=a+b+c+\dfrac{4}{\sqrt[3]{a\left(b+1\right)\left(c+2\right)}}+3\)
4. Tìm m để hệ bất phương trình \(\left\{{}\begin{matrix}x^2-5x+9\le\left|x-6\right|\\x^2+2x-3m^2+4\left|m\right|-4\le0\end{matrix}\right.\)
2.
ĐK: \(2x-y\ge0;y\ge0;y-x-1\ge0;y-3x+5\ge0\)
\(\left\{{}\begin{matrix}xy-2y-3=\sqrt{y-x-1}+\sqrt{y-3x+5}\left(1\right)\\\left(1-y\right)\sqrt{2x-y}+2\left(x-1\right)=\left(2x-y-1\right)\sqrt{y}\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(1-y\right)\sqrt{2x-y}+y-1+2x-y-1-\left(2x-y-1\right)\sqrt{y}=0\)
\(\Leftrightarrow\left(1-y\right)\left(\sqrt{2x-y}-1\right)+\left(2x-y-1\right)\left(1-\sqrt{y}\right)=0\)
\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(1+\sqrt{y}\right)+\left(\sqrt{2x-y}-1\right)\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}+1\right)=0\)
\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(\sqrt{y}+\sqrt{2x-y}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2x-1\end{matrix}\right.\) (Vì \(\sqrt{y}+\sqrt{2x-y}+2>0\))
Nếu \(y=1\), khi đó:
\(\left(1\right)\Leftrightarrow x-5=\sqrt{-x}+\sqrt{-3x+6}\)
Phương trình này vô nghiệm
Nếu \(y=2x-1\), khi đó:
\(\left(1\right)\Leftrightarrow2x^2-5x-1=\sqrt{x-2}+\sqrt{4-x}\) (Điều kiện: \(2\le x\le4\))
\(\Leftrightarrow2x\left(x-3\right)+x-3+1-\sqrt{x-2}+1-\sqrt{4-x}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1\right)=0\)
Ta thấy: \(1+\sqrt{x-2}\ge1\Rightarrow-\dfrac{1}{1+\sqrt{x-2}}\ge-1\Rightarrow1-\dfrac{1}{1+\sqrt{x-2}}\ge0\)
Lại có: \(\dfrac{1}{1+\sqrt{4-x}}>0\); \(2x>0\)
\(\Rightarrow\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1>0\)
Nên phương trình \(\left(1\right)\) tương đương \(x-3=0\Leftrightarrow x=3\Rightarrow y=5\)
Ta thấy \(\left(x;y\right)=\left(3;5\right)\) thỏa mãn điều kiện ban đầu.
Vậy hệ phương trình đã cho có nghiệm \(\left(x;y\right)=\left(3;5\right)\)
Cho \(\left\{{}\begin{matrix}\text{x, y, z > 0}\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{4}\end{matrix}\right.\). Tìm \(\min\limits_P=\dfrac{1}{\alpha\text{a}+\beta b+\gamma c}+\dfrac{1}{\beta\text{a}+\gamma b+\alpha c}+\dfrac{1}{\gamma\text{a}+\alpha b+\beta c} v\text{ới} \alpha; \beta;\text{ \gamma}\in\) \(\mathbb{N}^*\)