a.Ta có:$S=\dfrac{1}{2}ab\sin C=pr\Rightarrow r=\dfrac{ab\sin C}{a+b+c}=\dfrac{4R^2\sin A\sin B\sin C}{2R(\sin A+\sin B+\sin C)}$Suy ra: $\dfrac{r}{4R}=\dfrac{\sin A\sin B\sin C}{\sin A+\sin B+\sin C}$Mà ta có:$\sin A+\sin B+\sin C= 2\sin\dfrac{A}{2}\cos\dfrac{A}{2}+2\sin\dfrac{B+C}{2}\cos\dfrac{B-C}{2}$ $=2\cos\dfrac{A}{2}(\cos\dfrac{B+C}{2}+\cos\dfrac{B-C}{2})$ $=4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$$\Rightarrow \dfrac{r}{4R}=\tan\dfrac{A}{2}\tan\dfrac{B}{2}\tan\dfrac{C}{2}$

a.Ta có:$S=\dfrac{1}{2}ab\sin C=pr\Rightarrow r=\dfrac{ab\sin C}{a+b+c}=\dfrac{4R^2\sin A\sin B\sin C}{2R(\sin A+\sin B+\sin C)}$Suy ra: $\dfrac{r}{4R}=\dfrac{\sin A\sin B\sin C}{2(\sin A+\sin B+\sin C)}$Mà ta có:$\sin A+\sin B+\sin C= 2\sin\dfrac{A}{2}\cos\dfrac{A}{2}+2\sin\dfrac{B+C}{2}\cos\dfrac{B-C}{2}=2\cos\dfrac{A}{2}(\cos\dfrac{B+C}{2}+\cos\dfrac{B-C}{2})=4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$$\Rightarrow \dfrac{r}{4R}=\dfrac{8\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}}{8\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}}=\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}$

Ta có: $a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$$\Leftrightarrow a+b+c=ab+bc+ca$$\Leftrightarrow abc-ab-bc-ca+a+b+c-1=0$$\Leftrightarrow (a-1)(b-1)(c-1)=0$$\Leftrightarrow \left[\begin{array}{l}a=1\\b=1\\c=1\end{array}\right.\Rightarrow M=1$

Ta có: $a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$$\Leftrightarrow a+b+c=ab+bc+ca$$\Leftrightarrow abc-ab-bc-ca+a+b+c-1=0$$\Leftrightarrow (a-1)(b-1)(c-1)=0$$\Leftrightarrow \left[\begin{array}{l}a=1\\b=1\\c=1\end{array}\right.\Rightarrow M=0$

có bài tích phân ai giải giúp với

I=[tex]\int\limits_{-2}^{2}\frac{2x-5}{\sqrt{x^{2}+4x+13}}[tex]

Tích phân hàm chứa căn thức

có bài tích phân ai giải giúp với

$I=\int\limits_{-2}^{2}\frac{2x-5}{\sqrt{x^{2}+4x+13}}$

Tích phân hàm chứa căn thức

Áp dụng BĐT Cauchy ta có:$\sqrt{\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}\right)\left(\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)}\le\dfrac{1}{2}\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)$Tương tự suy ra:$\sum_{A,B,C}\sqrt{\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}\right)\left(\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)}\le2(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2})$Ta chỉ cần chứng minh: $\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\le\cot A+\cot B+\cot C$Ta có:$\cot A+\cot B=\dfrac{\sin(A+B)}{\sin A\sin B}=\dfrac{2\sin C}{\cos(A-B)-\cos(A+B)}\ge\dfrac{2\sin C}{1+\cos C}=\tan\dfrac{C}{2}$Tương tự: $\cot B+\cot C\ge\tan\dfrac{A}{2},\cot C+\cot A\ge\tan\dfrac{B}{2}$Cộng 3 BĐT trên ta có đpcm.

Áp dụng BĐT Cauchy ta có:$\sqrt{\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}\right)\left(\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)}\le\dfrac{1}{2}\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)$Tương tự suy ra:$\sum_{A,B,C}\sqrt{\left(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}\right)\left(\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\right)}\le2(\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2})$Ta chỉ cần chứng minh: $\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\le\cot A+\cot B+\cot C$Ta có:$\cot A+\cot B=\dfrac{\sin(A+B)}{\sin A\sin B}=\dfrac{2\sin C}{\cos(A-B)-\cos(A+B)}\ge\dfrac{2\sin C}{1+\cos C}=2\tan\dfrac{C}{2}$Tương tự:$\cot B+\cot C\ge2\tan\dfrac{A}{2},\cot C+\cot A\ge2\tan\dfrac{B}{2}$Cộng 3 BĐT trên ta có đpcm.

Điều kiện $\begin{cases}x>3\\ x^2 \ge 16 \end{cases}\Leftrightarrow \begin{cases}x>3 \\ \left[ {\begin{matrix} x\ge4\\ x\le-4 \end{matrix}} \right. \end{cases}\Leftrightarrow x>4$BPT$\Leftrightarrow \dfrac{2\sqrt{x^{2}-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}} > \dfrac{7-x}{\sqrt{x-3}} $$\Leftrightarrow 2\sqrt{x^{2}-16}+x-3 > 7-x$$ \Leftrightarrow 2\sqrt{x^{2}-16} > 10-2x $$\Leftrightarrow \sqrt{x^{2}-16} > 5-x$$ \Leftrightarrow \begin{cases}x^{2}-16>(5-x)^2 \\ 4\le x \le 5 \end{cases}$$\Leftrightarrow \begin{cases}x^{2}-16>x^2-10x+25 \\ 4\le x \le 5 \end{cases}$$\Leftrightarrow \begin{cases}10x&gt;41 \\ 4\le x \le 5 \end{cases}$$\Leftrightarrow \dfrac{41}{10}<x \le 5$

Điều kiện $\begin{cases}x>3\\ x^2 \ge 16 \end{cases}\Leftrightarrow \begin{cases}x>3 \\ \left[ {\begin{matrix} x\ge4\\ x\le-4 \end{matrix}} \right. \end{cases}\Leftrightarrow x>4$BPT $\Leftrightarrow \dfrac{2\sqrt{x^{2}-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}} > \dfrac{7-x}{\sqrt{x-3}}$$\Leftrightarrow 2\sqrt{x^{2}-16}+x-3 > 7-x$$\Leftrightarrow 2\sqrt{x^{2}-16} > 10-2x$$\Leftrightarrow \sqrt{x^{2}-16} > 5-x$$\Leftrightarrow \left[\begin{array}{l}5-x\le0\\\begin{cases}5-x>0\\x^{2}-16>(5-x)^2\end{cases}\end{array}\right.$$\Leftrightarrow \left[\begin{array}{l}x\ge5\\\begin{cases}x<5\\x^2-16>x^2-10x+25\end{cases}\end{array}\right.$$\Leftrightarrow \left[\begin{array}{l}x\ge5\\\begin{cases}x&lt;5\\x>\dfrac{41}{10}\end{cases}\end{array}\right. \Leftrightarrow x>\dfrac{41}{10}$

a.Ta có:$S(D)=\int\limits_{\pi/4}^{\pi/4}|\tan^3x|dx$ $=2\int\limits_0^{\pi/4}\tan^3xdx$ $=2\int\limits_0^{\pi/4}(\tan^3x+\tan x-\tan x)dx$ $=2\int\limits_0^{\pi/4}\tan x(\tan^2x+1)dx-2\int\limits_0^{\pi/4}\dfrac{\sin x}{\cos x}dx$ $=2\int\limits_0^{\pi/4}\tan xd(\tan x)+2\int\limits_0^{\pi/4}\dfrac{d(\cos x)}{\cos x}$ $=\tan^2x\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.+2\ln(\cos x)\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.=1-\ln2$

a.Ta có:$S(D)=\int\limits_{-\pi/4}^{\pi/4}|\tan^3x|dx=2\int\limits_0^{\pi/4}\tan^3xdx=2\int\limits_0^{\pi/4}(\tan^3x+\tan x-\tan x)dx=2\int\limits_0^{\pi/4}\tan x(\tan^2x+1)dx-2\int\limits_0^{\pi/4}\dfrac{\sin x}{\cos x}dx=2\int\limits_0^{\pi/4}\tan xd(\tan x)+2\int\limits_0^{\pi/4}\dfrac{d(\cos x)}{\cos x}=\tan^2x\left| $$\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}$$ \right.+2\ln(\cos x)\left| $$\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}$$ \right.=1-\ln2$

Chọn $a=\dfrac{-1+\sqrt{8041}}{2}\Rightarrow a^2=a+2010$Chọn $P(x)=(x-a)^{2010}$Ta có:$P(x^2-2010)=(x^2-a-2010)^{2010}$ $=(x^2-a^2)^{2010}$ $=(x-a)^{2010}(x+a)^{2010}=P(x)(x+a)^{2010}$Suy ra: $P(x^2-2010)\,\vdots\,P(x)$, thỏa mãn.

Choose $a=\dfrac{-1+\sqrt{8041}}{2}\Rightarrow a^2=a+2010$Choose $P(x)=(x-a)^{2010}$We have:$P(x^2-2010)=(x^2-a-2010)^{2010}$ $=(x^2-a^2)^{2010}$ $=(x-a)^{2010}(x+a)^{2010}=P(x)(x+a)^{2010}$It follows that $P(x^2-2010)\,\vdots\,P(x)$, satisfied.

Phương trình đã cho tương đương với: $3x^4-4x^3=(1-\sqrt{1+x^2})(2+x^2+\sqrt{1+x^2})$$\Leftrightarrow x^2(3x^2-4x)=\dfrac{-x^2(2+x^2+\sqrt{1+x^2})}{1+\sqrt{1+x^2}}$$\Leftrightarrow x^2\left(3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}\right)=0$Mà ta có:$3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}=3(x-\dfrac{2}{3})^2+\dfrac{2+5x^2+(\sqrt{1+x^2}-1)^2}{6(1+\sqrt{1+x^2})}>0$nên $x=0$ là nghiệm duy nhất.

The equation is equivalent to: $3x^4-4x^3=(1-\sqrt{1+x^2})(2+x^2+\sqrt{1+x^2})$or $x^2(3x^2-4x)=\dfrac{-x^2(2+x^2+\sqrt{1+x^2})}{1+\sqrt{1+x^2}}$or $x^2\left(3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}\right)=0$On the other hand, we have:$3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}=3(x-\dfrac{2}{3})^2+\dfrac{2+5x^2+(\sqrt{1+x^2}-1)^2}{6(1+\sqrt{1+x^2})}>0$then $x=0$.

c.Ta có:$\cos\frac{A}{2}+\cos\frac{B}{2}=2\cos(\frac{A}{4}+\frac{B}{4})\cos(\frac{A}{4}-\frac{B}{4})\le2\cos(\frac{A}{4}+\frac{B}{4})$$\cos\frac{C}{2}+\cos\frac{\pi}{6}=2\cos(\frac{C}{4}+\frac{\pi}{12})\cos(\frac{C}{4}-\frac{\pi}{12})\le2\cos(\frac{C}{4}+\frac{\pi}{12})$ $\cos(\frac{A}{4}+\frac{B}{4})+\cos(\frac{C}{4}+\frac{\pi}{12})=2\cos\frac{\pi}{6}\cos(\frac{A}{8}+\frac{B}{8}-\frac{C}{8}-\frac{\pi}{24})\le2\cos\frac{\pi}{6}$Suy ra: $\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}\le3\cos\frac{\pi}{6}=\frac{3\sqrt3}{2}$Ta có: $\frac{\displaystyle\cos\frac{A}{2}}{1+\cos A}+\frac{\displaystyle\cos\frac{B}{2}}{1+\cos B}+\frac{\displaystyle\cos\frac{C}{2}}{1+\cos C}$$=\frac{1}{\displaystyle 2\cos\frac{A}{2}}+\frac{1}{\displaystyle 2\cos\frac{B}{2}}+\frac{1}{\displaystyle 2\cos\frac{C}{2}}$$\ge\frac{9}{\displaystyle 2\cos\frac{A}{2}+2\cos\frac{B}{2}+2\cos\frac{C}{2}}\ge\sqrt3$Dấu bằng xảy ra khi:$\Delta ABC$ đều.

c.Ta có:$\cos\dfrac{A}{2}+\cos\dfrac{B}{2}=2\cos(\dfrac{A}{4}+\dfrac{B}{4})\cos(\dfrac{A}{4}-\dfrac{B}{4})\le2\cos(\dfrac{A}{4}+\dfrac{B}{4})$$\cos\dfrac{C}{2}+\cos\dfrac{\pi}{6}=2\cos(\dfrac{C}{4}+\dfrac{\pi}{12})\cos(\dfrac{C}{4}-\dfrac{\pi}{12})\le2\cos(\dfrac{C}{4}+\dfrac{\pi}{12})$$\cos(\dfrac{A}{4}+\dfrac{B}{4})+\cos(\dfrac{C}{4}+\dfrac{\pi}{12})=2\cos\dfrac{\pi}{6}\cos(\dfrac{A}{8}+\dfrac{B}{8}-\dfrac{C}{8}-\dfrac{\pi}{24})\le2\cos\dfrac{\pi}{6}$Suy ra:$\cos\dfrac{A}{2}+\cos\dfrac{B}{2}+\cos\dfrac{C}{2}\le3\cos\dfrac{\pi}{6}=\dfrac{3\sqrt3}{2}$Ta có:$\dfrac{\cos\dfrac{A}{2}}{1+\cos A}+\dfrac{\cos\dfrac{B}{2}}{1+\cos B}+\dfrac{\cos\dfrac{C}{2}}{1+\cos C}$$=\dfrac{1}{2\cos\dfrac{A}{2}}+\dfrac{1}{2\cos\dfrac{B}{2}}+\dfrac{1}{2\cos\dfrac{C}{2}}$$\ge\dfrac{9}{2\cos\dfrac{A}{2}+2\cos\dfrac{B}{2}+2\cos\dfrac{C}{2}}\ge\sqrt3$Dấu bằng xảy ra khi:$\Delta ABC$ đều.

c.Ta có:$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}=2\sin(\dfrac{A}{4}+\dfrac{B}{4})\cos(\dfrac{A}{4}-\dfrac{B}{4})\le2\sin(\dfrac{A}{4}+\dfrac{B}{4})$$\sin\dfrac{C}{2}+\sin\dfrac{\pi}{6}=2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})\cos(\dfrac{C}{4}-\dfrac{\pi}{12})\le2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})$$\sin(\dfrac{A}{4}+\dfrac{B}{4})+\sin(\dfrac{C}{4}+\dfrac{\pi}{12})=2\sin\dfrac{\pi}{6}\cos(\dfrac{A}{8}+\dfrac{B}{8}-\dfrac{C}{8}-\dfrac{\pi}{24})\le2\sin\dfrac{\pi}{6}$Suy ra: $\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\le3\sin\dfrac{\pi}{6}=\dfrac{3}{2}$Suy ra: $\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\le\left(\dfrac{\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}}{3}\right)^3\le\dfrac{1}{8}$Ta có: $\left(1+\dfrac{1}{\sin\dfrac{A}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{B}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{C}{2}}\right)$$=1+\left(\dfrac{1}{\sin\dfrac{A}{2}}+\dfrac{1}{\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{C}{2}}\right)+\left(\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{B}{2}\sin\dfrac{C}{2}}+\dfrac{1}{\sin\dfrac{C}{2}\sin\dfrac{A}{2}}\right)+\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}$$\ge1+\dfrac{3}{\sqrt[3]{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}}+\dfrac{3}{\sqrt[3]{\sin^2\dfrac{A}{2}\sin^2\dfrac{B}{2}\sin^2\dfrac{C}{2}}}+8\ge27$Dấu bằng xảy ra khi: $\Delta ABC$ đều.

e.Ta có:$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}=2\sin(\dfrac{A}{4}+\dfrac{B}{4})\cos(\dfrac{A}{4}-\dfrac{B}{4})\le2\sin(\dfrac{A}{4}+\dfrac{B}{4})$$\sin\dfrac{C}{2}+\sin\dfrac{\pi}{6}=2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})\cos(\dfrac{C}{4}-\dfrac{\pi}{12})\le2\sin(\dfrac{C}{4}+\dfrac{\pi}{12})$$\sin(\dfrac{A}{4}+\dfrac{B}{4})+\sin(\dfrac{C}{4}+\dfrac{\pi}{12})=2\sin\dfrac{\pi}{6}\cos(\dfrac{A}{8}+\dfrac{B}{8}-\dfrac{C}{8}-\dfrac{\pi}{24})\le2\sin\dfrac{\pi}{6}$Suy ra: $\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\le3\sin\dfrac{\pi}{6}=\dfrac{3}{2}$Suy ra: $\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\le\left(\dfrac{\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}}{3}\right)^3\le\dfrac{1}{8}$Ta có: $\left(1+\dfrac{1}{\sin\dfrac{A}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{B}{2}}\right)\left(1+\dfrac{1}{\sin\dfrac{C}{2}}\right)$$=1+\left(\dfrac{1}{\sin\dfrac{A}{2}}+\dfrac{1}{\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{C}{2}}\right)+\left(\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}}+\dfrac{1}{\sin\dfrac{B}{2}\sin\dfrac{C}{2}}+\dfrac{1}{\sin\dfrac{C}{2}\sin\dfrac{A}{2}}\right)+\dfrac{1}{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}$$\ge1+\dfrac{3}{\sqrt[3]{\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}}}+\dfrac{3}{\sqrt[3]{\sin^2\dfrac{A}{2}\sin^2\dfrac{B}{2}\sin^2\dfrac{C}{2}}}+8\ge27$Dấu bằng xảy ra khi: $\Delta ABC$ đều.

phương trình 23

$ biện luân số nghiệm thực của PT : \sqrt{x+1} +\sqrt{1-x} = m$

Giải và biện luận phương...

phương trình 23

biện luân số nghiệm thực của PT : $\sqrt{x+1} +\sqrt{1-x} = m$

Giải và biện luận phương...

Giới hạn cơ bản: $\mathop {\lim }\limits_{x \to 0}\frac{\sin x}{x}=\mathop {\lim }\limits_{x \to 0}\frac{x}{\sin x}=1$Ta có: $\mathop {\lim }\limits_{x \to 0}\frac{x}{\sqrt{1-\cos x}}$$=\mathop {\lim }\limits_{x \to 0}\frac{x}{\sqrt2\sin\frac{x}{2}}$$=\mathop {\lim }\limits_{x \to 0}\frac{\sqrt2.\frac{x}{2}}{\sin\frac{x}{2}}=\sqrt2$

Giới hạn cơ bản: $\mathop {\lim }\limits_{x \to 0}\frac{\sin x}{x}=\mathop {\lim }\limits_{x \to 0}\frac{x}{\sin x}=1$Ta có: $\mathop {\lim }\limits_{x \to 0^+}\frac{x}{\sqrt{1-\cos x}}$$=\mathop {\lim }\limits_{x \to 0^+}\frac{x}{\sqrt2\sin\frac{x}{2}}$$=\mathop {\lim }\limits_{x \to 0^+}\frac{\sqrt2.\frac{x}{2}}{\sin\frac{x}{2}}=\sqrt2$Mặt khác: $\mathop {\lim }\limits_{x \to 0^-}\frac{x}{\sqrt{1-\cos x}}$$=\mathop {\lim }\limits_{x \to 0^-}\frac{x}{-\sqrt2\sin\frac{x}{2}}$$=\mathop {\lim }\limits_{x \to 0^-}\frac{-\sqrt2.\frac{x}{2}}{\sin\frac{x}{2}}=-\sqrt2$Suy ra không tồn tại $\mathop {\lim }\limits_{x \to 0}\frac{x}{\sqrt{1-\cos x}}$

Áp dụng BĐT $a^2+b^2+c^2\ge 3(ab+bc+ca)$ta có:$(\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y})^2\geq 3(x^2+y^2+z^2)=9$$\Rightarrow \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\geq 3$Dấu bằng xảy ra khi: $x=y=z=1$

Áp dụng BĐT $(a+b+c)^2\ge 3(ab+bc+ca)$ta có:$(\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y})^2\geq 3(x^2+y^2+z^2)=9$$\Rightarrow \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\geq 3$Dấu bằng xảy ra khi: $x=y=z=1$

a.Ta có:$(x+1)^{2n+1}=C_{2n+1}^{2n+1}x^{2n+1}+C_{2n+1}^{2n}x^{2n}+\ldots+C_{2n+1}^1x+C_{2n+1}^0$Lấy nguyên hàm 2 vế ta được:$\frac{1}{2n+2}(x+1)^{2n+2}=\frac{C_{2n+1}^{2n+1}}{2n+2}x^{2n+2}+\frac{C_{2n+1}^{2n}}{2n+1}x^{2n+1}+\ldots+\frac{C_{2n+1}^1}{2}x^2+\frac{C_{2n+1}^0}{1}x$Lấy nguyên hàm 2 vế lần nữa ta có:$\frac{1}{(2n+2)(2n+3)}(x+1)^{2n+3}=\frac{C_{2n+1}^{2n+1}}{(2n+2)(2n+3)}x^{2n+3}+\frac{C_{2n+1}^{2n}}{(2n+1)(2n+2)}x^{2n+2}+\ldots+\frac{C_{2n+1}^1}{2.3}x^3+\frac{C_{2n+1}^0}{1.2}x^2$Cho $x=1$ ta được: $A=\frac{2^{2n+3}}{(2n+2)(2n+3)}$

a.Ta có:$(x+1)^{2n+1}=C_{2n+1}^{2n+1}x^{2n+1}+C_{2n+1}^{2n}x^{2n}+\ldots+C_{2n+1}^1x+C_{2n+1}^0$$\Rightarrow \int\limits_0^t(x+1)^{2n+1}dx=\int\limits_0^t\left(C_{2n+1}^{2n+1}x^{2n+1}+C_{2n+1}^{2n}x^{2n}+\ldots+C_{2n+1}^1x+C_{2n+1}^0\right)dx$$\Rightarrow \frac{1}{2n+2}(t+1)^{2n+2}=\frac{C_{2n+1}^{2n+1}}{2n+2}t^{2n+2}+\frac{C_{2n+1}^{2n}}{2n+1}t^{2n+1}+\ldots+\frac{C_{2n+1}^1}{2}t^2+\frac{C_{2n+1}^0}{1}t$$\Rightarrow \int\limits_0^x\frac{1}{2n+2}(t+1)^{2n+2}dt=\int\limits_0^x\left(\frac{C_{2n+1}^{2n+1}}{2n+2}t^{2n+2}+\frac{C_{2n+1}^{2n}}{2n+1}t^{2n+1}+\ldots+\frac{C_{2n+1}^1}{2}t^2+\frac{C_{2n+1}^0}{1}t\right)dt$$\Rightarrow \frac{1}{(2n+2)(2n+3)}(x+1)^{2n+3}=\frac{C_{2n+1}^{2n+1}}{(2n+2)(2n+3)}x^{2n+3}+\frac{C_{2n+1}^{2n}}{(2n+1)(2n+2)}x^{2n+2}+\ldots+\frac{C_{2n+1}^1}{2.3}x^3+\frac{C_{2n+1}^0}{1.2}x^2$Cho$x=1$ta được:$A=\frac{2^{2n+3}}{(2n+2)(2n+3)}$

Ta có: $\forall \alpha$ thì:$\cot \alpha-\cot2\alpha=\frac{\cos\alpha\sin2\alpha-\cos2\alpha\sin\alpha}{\sin\alpha\sin2\alpha}=\frac{\sin\alpha}{\sin\alpha\sin2\alpha}=\frac{1}{\sin2\alpha}$Suy ra:$\sum_{k=1}^n\frac{1}{\sin2^nx}=\sum_{k=1}^n(\cot2^{n-1}x-\cot2^nx)=\cot x-\cot2^nx$

Ta có: $\forall \alpha$ thì:$\cot \alpha-\cot2\alpha=\frac{\cos\alpha\sin2\alpha-\cos2\alpha\sin\alpha}{\sin\alpha\sin2\alpha}=\frac{\sin\alpha}{\sin\alpha\sin2\alpha}=\frac{1}{\sin2\alpha}$Suy ra:$\sum_{k=1}^n\frac{1}{\sin2^kx}=\sum_{k=1}^n(\cot2^{k-1}x-\cot2^kx)=\cot x-\cot2^nx$