\(=\dfrac{3-3x+4+3y+3x-7}{6x^4y}=\dfrac{3y}{6x^4y}=\dfrac{1}{2x^4}\)

\(23-y^2=7\left(x-2004\right)^2\ge0\\ \Leftrightarrow y^2\le23\)

Mà \(y\in N\Leftrightarrow y\in\left\{0;1;2;3;4\right\}\)

Với \(y=0\Leftrightarrow7\left(x-2004\right)^2=23\left(loại\right)\)

Với \(y=1\Leftrightarrow7\left(x-2004\right)^2=22\Leftrightarrow\left(x-2004\right)^2=\dfrac{22}{7}\left(loại\right)\)

Với \(y=2\Leftrightarrow7\left(x-2004\right)^2=19\Leftrightarrow\left(x-2004\right)^2=\dfrac{19}{7}\left(loại\right)\)

Với \(y=3\Leftrightarrow7\left(x-2004\right)^2=14\Leftrightarrow\left(x-2004\right)^2=2\left(loại\right)\)

Với \(y=4\Leftrightarrow7\left(x-2004\right)^2=7\Leftrightarrow\left[{}\begin{matrix}x-2004=1\\x-2004=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2005\\x=2003\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2005;4\right);\left(2003;4\right)\)

\(\Rightarrow\dfrac{z+y+1}{x}=\dfrac{x+z+1}{y}=\dfrac{x+y-2}{z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2=x+y+z\\ \Rightarrow\left\{{}\begin{matrix}z+y+1=2x\\x+z+1=2y\\x+y-2=2z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z=2x-1\\x+z=2y-1\\x+y=2z+2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x-1=2-x\\2y-1=2-y\\2z+2=2-z\end{matrix}\right.\Rightarrow\left(x,y,z\right)=\left(1;1;0\right)\)

Áp dụng PTG ta có: \(c^2=a^2+b^2\) với \(n=1\)

Giả sử đúng với \(n=k\)

\(\Rightarrow A_k=a^{2k}+b^{2k}\le c^{2k}\)

Cần cm nó cũng đúng với \(n=k+1\)

\(\Rightarrow A_{k+1}=a^{2k+2}+b^{2k+2}=c^{2k+2}\\ \Rightarrow\left(a^{2k}+b^{2k}\right)\left(a^2+b^2\right)-a^2b^{2k}-a^{2k}b^2\le c^{2k}\cdot c^2=c^{2k+2}\)

Vậy BĐT đúng với \(n=k+1\)

\(\RightarrowĐpcm\)

:O đc 2 vk ck tick lun

ai lại học kiểu việt hóa đấy :v

Xét tam giác ABC: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Rightarrow\widehat{B}+\widehat{C}=180^0-\widehat{A}\)

Mặt khác: \(\widehat{B}-\widehat{C}=18^0\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\dfrac{180^0-\widehat{A}+18^0}{2}=99^0-\dfrac{\widehat{A}}{2}\\\widehat{C}=99^0+\dfrac{\widehat{A}}{2}-18^0=81^0-\dfrac{\widehat{A}}{2}\end{matrix}\right.\)

Xét tam giác ABD: \(\widehat{ADC}=\widehat{BAD}+\widehat{B}=\dfrac{\widehat{A}}{2}+99^0-\dfrac{\widehat{A}}{2}=99^0\)

\(\widehat{ABD}=180^0-\widehat{ADC}=81^0\)

Sửa đề \("="\rightarrow"+"\)

Áp dụng BĐT cauchy, ta có:\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)

\(\Leftrightarrow\sum\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\right)\\ \Leftrightarrow\sum\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{ab^2c+abc+ab}+\dfrac{b}{abc+ab+b}\right)=\dfrac{1}{2}\cdot1=\dfrac{1}{2}\)

Dấu \("="\Leftrightarrow a=b=c=1\)

\(4,a^2+b^2=4a+2b+540\\ \Leftrightarrow\left(a-2\right)^2+\left(b-1\right)^2=545\)

\(P=23a+4b+2013=23\left(a-2\right)+4\left(b-1\right)2063\\ \Leftrightarrow P\le\sqrt{\left(23^2+4^2\right)\left[\left(a-2\right)^2+\left(b-1\right)^2\right]}+2063\\ \Leftrightarrow P\le\sqrt{545\cdot545}+2063=2608\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a-2}{23}=\dfrac{b-1}{4}\\\left(a-2\right)^2+\left(b-1\right)^2=545\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=25\\b=5\end{matrix}\right.\)