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A B C D E

Kẻ tia Cx sao cho \(\widehat{ABD}=\widehat{ACx}\) . Tia Cx cắt BD tại E

+ \(\Delta ABD~\Delta ECD\left(g.g\right)\)

\(\Rightarrow\hept{\begin{cases}\frac{AD}{BD}=\frac{ED}{CD}\\\widehat{BAD}=\widehat{CEB}\end{cases}}\)

\(\Rightarrow AD.CD=BD.ED\left(1\right)\)

+ \(\Delta ABD~\Delta EBC\left(g.g\right)\)

\(\Rightarrow\frac{AB}{BD}=\frac{EB}{BC}\Rightarrow AB.BC=EB.BD\left(2\right)\)

Từ (1) và (2) suy ra 

\(AB.BC-AD.DC=BD.EB-BD.ED=BD^2\)

PT (1) \(\Leftrightarrow\left(x-1\right)\left(x+y-2\right)=0\)

Rút: \(c=-\left(a+b\right)\) ta cần chứng minh:

\(a^2+b^2+\left(a+b\right)^2< 2\) với \(-1< a\le b\le-\left(a+b\right)< 1\)

Từ \(-1< a\le b\le-\left(a+b\right)< 1\Rightarrow-1< a+b< 1\)

Xét hiệu: \(\left(a+b\right)^2-1=\left(a+b-1\right)\left(a+b+1\right)< 0\).Vậy \(\left(a+b\right)^2< 1\)

Ta có: \(VT=a^2+b^2+\left(a+b\right)^2=2\left(a+b\right)^2-2ab< 2\left(a+b\right)^2< 2.1=2\)

Ta có đpcm.

_{Is that true?}

\(Q=\frac{1}{x^2+y^2}+\frac{2}{xy}+4xy+2016=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{4xy}+4xy+\frac{5}{4xy}+2016\)

Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\). Dấu "=" khi a=b (bạn tự chứng minh)

\(\frac{1}{x^2+y^2}+\frac{1}{2xy}\ge\frac{4}{\left(x+y\right)^2}=4\)

Vì x>0, y>0 nên xy>0

Áp dụng bất đẳng thức Cô si cho 2 số dương

\(\frac{1}{4xy}+4xy\ge2\sqrt{\frac{1}{4xy}.4xy}=2\)

Ta có: \(1=x+y\ge2\sqrt{xy}\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\Rightarrow\frac{5}{4xy}\ge5\)

Dấu "=" khi \(\hept{\begin{cases}x^2+y^2=2xy\\\frac{1}{4xy}=4xy\\x=y\end{cases}\Rightarrow x=y=\frac{1}{2}}\)

\(\Rightarrow Q\ge4+2+5+2016=2027\)

Vậy \(minQ=2027\)khi \(x=y=\frac{1}{2}\)

\(\left(a-\sqrt{2011}\right)\left(b+\sqrt{2011}\right)=14\)

\(\Leftrightarrow ab+\sqrt{2011}\left(a-b\right)=2025\)

Có: a,b nguyên => a-b nguyên

=> VP=VT <=> \(\sqrt{2011}\left(a-b\right)\)nguyên

=> a-b=0 <=> a=b

=> pt <=> a^2=2025

Làm nốt nhé.