Lời giải:
Trừ theo vế 2 pt trên ta có:
$x^3-y^3=5y-5x$

$\Leftrightarrow (x-y)(x^2+xy+y^2)+5(x-y)=0$

$\Leftrightarrow (x-y)(x^2+xy+y^2+5)=0$

Ta thấy: $x^2+xy+y^2+5=(x+\frac{y}{2})^2+\frac{3y^2}{4}+5\geq 5>0$ với mọi $x,y$

$\Rightarrow x-y=0$

$\Leftrightarrow x=y$.

Thay vào pt (1): $x^3=3x+8x=11x$

$\Leftrightarrow x(x^2-11)=0$

$\Leftrightarrow x\in\left\{0; \pm \sqrt{11}\right\}$

Vậy........

Yêu cầu đề bài là gì bạn nên ghi đầy đủ để được hỗ trợ tốt hơn.

\(\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}}=\left(\dfrac{x-1}{\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}.\dfrac{\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\)

Ta có \(\widehat{EDF}=\widehat{ECF}\) (chắn hai cung bằng nhau AI và BI của đường tròn (O))

\(\Rightarrow\) Tứ giác CDEF nội tiếp

\(\Rightarrow\widehat{DEF}+\widehat{DCF}=180^0\)

Mà \(\widehat{DCF}+\widehat{DAB}=180^0\) (tứ giác ABCD nội tiếp)

\(\Rightarrow\widehat{DEF}=\widehat{DAB}\)

\(\Rightarrow EF||AB\) (hai góc đồng vị bằng nhau)

Ta có:

\(x^2+3y^2=x^2+y^2+2y^2\ge2\sqrt{\left(x^2+y^2\right).2y^2}=2y\sqrt{2\left(x^2+y^2\right)}\)

\(\Rightarrow P=\sum\dfrac{xy}{x^2+3y^2}\le\sum\dfrac{x}{2\sqrt{2\left(x^2+y^2\right)}}=\dfrac{1}{2\sqrt{2}}\sum\sqrt{\dfrac{x^2}{x^2+y^2}}=\dfrac{1}{2\sqrt{2}}\sum\sqrt{\dfrac{a}{a+b}}\)

\(P\le\dfrac{1}{2\sqrt{2}}\sum\sqrt{a+c}.\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}\)

\(\Rightarrow P^2\le\dfrac{1}{8}\left(\sum\sqrt{a+c}.\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}\right)^2\le\dfrac{1}{8}\left(2a+2b+2c\right)\left(\dfrac{a}{\left(a+b\right)\left(a+c\right)}+\dfrac{b}{\left(a+b\right)\left(b+c\right)}+\dfrac{c}{\left(a+c\right)\left(b+c\right)}\right)\)

\(\Rightarrow P^2\le\dfrac{1}{2}.\dfrac{\left(a+b+c\right)\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Lại có \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Rightarrow P^2\le\dfrac{1}{2}.\dfrac{\left(a+b+c\right)\left(ab+bc+ca\right)}{\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)}=\dfrac{9}{16}\)

\(\Rightarrow P\le\dfrac{3}{4}\)

Dấu "=" xảy ra khi \(a=b=c\) hay \(x=y=z\)

iu