\(P\left(x\right)=\frac{2012x+2013\sqrt{1-x^2}+2014}{\sqrt{1-x^2}}=\frac{2012x+2014}{\sqrt{1-x^2}}+\frac{2013\sqrt{1-x^2}}{\sqrt{1-x^2}}\)

\(=\frac{2012x+2014}{\sqrt{1-x^2}}+2013=2012+\frac{2012\left(1+x\right)+1-x}{\sqrt{1-x^2}}\)

Áp dụng BĐT AM-GM ta có: 

\(P\left(x\right)\ge2012+\frac{2\sqrt{2012\left(1+x\right)\left(1-x\right)}}{\sqrt{1-x^2}}=2012+2\sqrt{2012}\)

=\(2013\) \(+\frac{2014+2012x}{\sqrt{1-x^2}}\) =\(\frac{2013\left(1+x\right)+1-x}{\sqrt{1-x^2}}\) \(\ge2013+\frac{2\sqrt{2013\left(1+x\right)\left(1-x\right)}}{\sqrt{1-x^2}}=2013+2\sqrt{2013}\)

dau = xay ra khi \(2013\left(1+x\right)=1-x\)

               \(\Leftrightarrow x=-\frac{1001}{1002}\)

min p(x) =\(2013+2\sqrt{2013}\Leftrightarrow x=-\frac{1001}{1002}\)

c/ ĐKXĐ: \(x\ge3\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x-3}-\sqrt{x-2}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\left(\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}\right)-\left(\sqrt{\left(x-1\right)\left(x+3\right)}-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}-\sqrt{x+3}=0\\\sqrt{x-1}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2< 3\left(ktm\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

a/ ĐKXĐ: \(\left\{{}\begin{matrix}x>2013\\y>2014\\z>2015\end{matrix}\right.\)

\(\Leftrightarrow\frac{1}{4}-\frac{\sqrt{x-2013}-1}{x-2013}+\frac{1}{4}-\frac{\sqrt{y-2014}-1}{y-2014}+\frac{1}{4}-\frac{\sqrt{z-2015}-1}{z-2015}=0\)

\(\Leftrightarrow\frac{x-2013-4\sqrt{x-2013}+4}{4\left(x-2013\right)}+\frac{y-2014-4\sqrt{y-2014}+4}{4\left(y-2014\right)}+\frac{z-2015-4\sqrt{z-2015}+4}{4\left(z-2015\right)}=0\)

\(\Leftrightarrow\left(\frac{\sqrt{x-2013}-2}{2\sqrt{x-2013}}\right)^2+\left(\frac{\sqrt{y-2014}-2}{2\sqrt{y-2014}}\right)^2+\left(\frac{\sqrt{z-2015}-2}{2\sqrt{z-2015}}\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2013}-2=0\\\sqrt{y-2014}-2=0\\\sqrt{z-2015}-2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)

b/ Trừ vế cho vế 2 pt ta được:

\(x^3-y^3=2\left(y-x\right)\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+y^2-xy\right)+2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+y^2-xy+2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}+2\right]=0\)

\(\Leftrightarrow x-y=0\Leftrightarrow x=y\)

Thay vào pt đầu:

\(x^3+1=2x\Leftrightarrow x^3-2x+1=0\)

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

\(\Leftrightarrow...\)

Ai trả lời nhanh và chính xác mình k

Ta có: \(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\)

\(\Leftrightarrow\left(x-\sqrt{x^2+2013}\right)\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)

\(\Leftrightarrow-2013\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)

\(\Leftrightarrow-y-\sqrt{y^2+2013}=x-\sqrt{x^2+2013}\)

⇔\(x+y=\sqrt{x^2+2013}-\sqrt{y^2+2013}\)(1)

Nhân liên hợp tương tự nhân \(y-\sqrt{y^2+2013}\)vào hai về rút được

\(x+y=\sqrt{y^2+2013}-\sqrt{x^2+2013}\)(2)

Cộng vế theo vế (1)(2) ta được \(x+y=0\Rightarrow x=-y\)

Thay vào \(A=\left(-y\right)^{2014}-y^{2014}+1=1\)