Đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a\\\sqrt{x+2=b}\end{matrix}\right.\)

Thì được:

\(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(a-b\right)=0\)

Làm tiếp

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

Gọi chân hải đăng là A thì ta để ý thấy. Lần đầu quan sát thì tam giác DAB là nửa tam giác đều
\(\Rightarrow DA^2+AB^2=DB^2\)

\(\Leftrightarrow DA^2+AB^2=4AB^2\)

\(\Leftrightarrow DA^2=3AB^2=3.75^2=11250\)

\(\Leftrightarrow DA=106,066\)
Lần thứ 2 quan sát thì tam giác CAB là tam giác vuông cân
\(\Rightarrow CA=AB=75\)

Vậy quãng đường thuyền đi được là:
\(DC=DA-CA=106,066-75=31,066\)

\(A=\sqrt{7-\sqrt{48}}+\sqrt{13+\sqrt{48}}\)

\(=\sqrt{7-2.2\sqrt{3}}+\sqrt{13+2.2\sqrt{3}}\)

\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}+1\right)^2}\)

\(=2-\sqrt{3}+2\sqrt{3}+1=\sqrt{3}+3\)

Ta có:

\(\left(\sqrt{5-3\sqrt{2}}+\sqrt{3\sqrt{2}-4}\right)^2=5-3\sqrt{2}+3\sqrt{2}-4+2\sqrt{5-3\sqrt{2}}\sqrt{3\sqrt{2}-4}\)

\(=1+2\sqrt{27\sqrt{2}-38}\)

Áp dụng vào bài toán t được

\(\dfrac{\sqrt{1+2\sqrt{27\sqrt{2}-38}}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}-4}}\)

\(=\dfrac{\sqrt{\left(\sqrt{5-3\sqrt{2}}+\sqrt{3\sqrt{2}-4}\right)^2}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}-4}}\)

\(=\dfrac{\sqrt{5-3\sqrt{2}}+\sqrt{3\sqrt{2}-4}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}-4}}=1\)

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

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

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

(1) thì tự làm nốt
\(\left(2\right)\Leftrightarrow x^2-x\left(y+1\right)+y^2-y=0\)

Xem phương trình ẩn x. Để phương trình có nghiệm thì:
\(\Delta_x=\left(y+1\right)^2-4\left(y^2-y\right)\ge0\)

\(\Leftrightarrow0\le y\le2\)

Làm nốt

b/ 
\(N=x^3+y^3+6x^2y^2\left(x+y\right)+3xy\left(x^2+y^2\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+6x^2y^2+3xy\left(\left(x+y\right)^2-2xy\right)\)

\(=\left(\left(x+y\right)^2-3xy\right)+6x^2y^2+3xy\left(1-2xy\right)\)

\(=\left(1-3xy\right)+6x^2y^2+3xy\left(1-2xy\right)=1\)

6/ Ta có:

\(a+b=-c-d\)

\(\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)

\(\Leftrightarrow a^3+3ab^2+3a^{2b}+b^3=c^3+3c^2d+3cd^2+d^3\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3\left(a^2b+ab^2+c^2d+cd^2\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3\left[ab\left(a+b\right)+cd\left(c+d\right)\right]\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3\left[-ab\left(c+d\right)+cd\left(c+d\right)\right]\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3\left(c+d\right)\left(cd-ab\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\)

\(C=\dfrac{2014\left(2015^2+2016\right)-2016\left(2015^2-2014\right)}{2014\left(2013^2-2012\right)-2012\left(2013^2+2014\right)}\)

\(=\dfrac{2.2014.2016+2014.2015^2-2016.2015^2}{2014.2013^2-2012.2013^2-2.2012.2014}\)

\(=\dfrac{2.\left(2015+1\right)\left(2015-1\right)-2.2015^2}{2.2013^2-2.\left(2013+1\right)\left(2013-1\right)}\)

\(=\dfrac{2.\left(2015^2-1\right)-2.2015^2}{2.2013^2-2.\left(2013^2-1\right)}=\dfrac{-2}{2}=-1\)

\(\left(x^3-y^3\right)-\left(x^2-y^2\right)=0\)

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

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

Làm nốt