Hu hu tui cũng thế !

từ C vẽ  Ct // DE

ta co goc tCD+goc CDE=180 ( 2 goc trong cung phia va Ct//DE)

==> goc tCD+100=180

---> goc tCD =180-100=80

ta co goc tCD+ goc tCA=goc ACD

--->80+goc tCA=140

--> goc tCA=140-80=60

ta co : goc CAB + goc tCA=120+60=180

ma goc CAB va goc tCA nam o vi tri trong cung phia nen AB//Ct

ma DE// Ct (cmt)

nen AB// DE

áp dụng BĐT bnyacovsky :\(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow2a^2+2b^2\ge\left(a+b\right)^2\)

\(\left(4+4\right)\left(a^4+b^4\right)\ge\left(2a^2+2b^2\right)^2\ge\left(a+b\right)^4\)

\(\Leftrightarrow a^4+b^4\ge\frac{\left(a+b\right)^4}{8}\)

dấu = xảy ra khi a=b

\(x^2+2xy+6x+6y+2y^2+8=0\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+y^2=-8\)

\(y^2\ge0\Rightarrow\left(x+y\right)^2+6\left(x+y\right)\le-8\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9\le1\)

\(\Leftrightarrow\left(x+y+3\right)^2\le1\rightarrow\left|x+y+3\right|\le1\)

\(\Rightarrow-1\le x+y+3\le1\Leftrightarrow2012\le B\le2014\)

dấu = xảy ra: #_MIn: \(\left\{\begin{matrix}x+y+2016=2012\\y=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-4\\y=0\end{matrix}\right.\)

#_MAX:\(\left\{\begin{matrix}x+y+2016=2014\\y=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-2\\y=0\end{matrix}\right.\)

x+y+z=a\(\rightarrow\frac{1}{x+y+z}=\frac{1}{a}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\frac{xy+yz+xz+z^2}{xyz\left(x+y+z\right)}=0\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left[\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}a=z\\a=x\\a=y\end{matrix}\right.\)

thay vào ta đều tính được S=0