c: Ta có: \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}\)

\(=4+\sqrt{10}-4+\sqrt{10}\)

\(=2\sqrt{10}\)

d: Ta có: \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}\)

\(=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1\)

\(=2\sqrt{2}\)

a) \(=\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2=12-18=-6\)

b) \(=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}-\sqrt{2015}=-\sqrt{2013}-\sqrt{2015}\)

c) \(=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)

d) \(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)

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

\(\sqrt{\left(x-2013\right)^{10}}+\sqrt{\left(x-2014\right)^{14}}=1\)

Mà \(\sqrt{\left(x-2013\right)^{10}};\sqrt{\left(x-2014\right)^{14}}\ge0\)

=> \(\sqrt{\left(x-2013\right)^{10}}=0;\sqrt{\left(x-2014\right)^{14}}=1\)

HOặc \(\sqrt{\left(x-2013\right)^{10}}=1;\sqrt{\left(x-2013\right)^{14}}=0\)

\(\sqrt{\left(x-2013\right)^{10}}=1\rightarrow x-2013\in\left\{-1;1\right\};x\in\left\{2014;2012\right\}\)

\(\sqrt{\left(x-2014\right)^{14}}=0;x-2014=0;x=2014\)

=> x = 2014 (thích hợp)

\(\sqrt{\left(x-2013\right)^{10}}=0;x-2013=0;x=2013\)

\(\sqrt{\left(x-2014\right)^{14}}=1;x-2014\in\left\{-1;1\right\};x\in\left\{2013;2015\right\}\)

=> x = 2013 (thích hợp)

Vậy x = 2013 hoặc x = 2014          

Đặt \(x-2003=t\)

Ta có: \(\sqrt{t^{10}}+\sqrt{\left(1-t\right)^{14}}=1\Leftrightarrow\left|t\right|^5+\left|1-t\right|^7=1\text{(*)}\)

\(\left(\text{*}\right)\Rightarrow\left|t\right|;\left|1-t\right|\le1\)

\(+t1\text{ (loại)}\)

\(+t=0\) thì \(\left(\text{*}\right)\) thỏa

\(+0

Ta có:

\(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)=2013\\ \Leftrightarrow\left(x^2-x^2-2013\right)\left(y+\sqrt{y^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\\ \Leftrightarrow y+\sqrt{y^2+2013}=\sqrt{x^2+2013}-x\left(1\right)\)

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

Do đó: 2x=-2y

Suy ra: x=-y

Do đó:

\(x^{2013}+y^{2013}=\left(-y\right)^{2013}+y^{2013}=0\left(ĐPCM\right)\)

=3-3+3:1/9+1=3-3+1/27+1=1/27+1=28/7

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

Theo đề bài  và (1) => dpcm

b) theo a có \(y+\sqrt{y^2+\sqrt{2013}}=-x+\sqrt{x^2+\sqrt{2013}}\)(2)

tương tự ta có \(x+\sqrt{x^2+\sqrt{2013}}=-y+\sqrt{y^2+\sqrt{2013}}\)(3)

Cộng 2 vế (2)  với (3) => x+y = -x -y

hay 2(x+y) =0 =>S= x+y =0