Ta có: $\left|x_1-x_2\right|=17\iff \sqrt{{\left(x_1+x_2\right)}^2-4x_1{x}_2}=17\mathrm{làm\; cách\; nào\; để\; ra\; đc\; vậy??}$ a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}{b}^2-...+a^2{b}^{n-3}-a{b}^{n-2}+b^{n-1})$$ a,b,c,d>0$$ \frac{a-d}{b+d}+\frac{d-b}{b+c}+\frac{b-c}{c+a}+\frac{c-a}{a+d}\ge 0$\{\begin{array}{c}{x}^3+x-2=y^3+3y^2+4y\\ x^5+y^3+1=0\end{array}$ x^3+y^3-3xy\left(x^2+y^2\right)+4x^2{y}^2\left(x+y\right)-4x^3{y}^3=0$$ x,y,z$$ x+y+z=0$$ x^2+y^2+z^2=a^2$$ x^4+y^4+z^4$$ a$$ \frac{2\sqrt{x}}{x+1}$$ \sqrt{p}$$ \sqrt{3x-1}-\sqrt{x+1}=3x^2-2x-1$$ (x+1)\sqrt{x^2-5x+3}=x^2+1$$ x^2+4x+5=2\sqrt{2x+3}$$ a,b>0$$ \frac{1}{a^3}+\frac{a^3}{b^3}+b^3\ge \frac{1}{a}+\frac{a}{b}+b$$ \begin{array}{c}\sqrt{x+10}+\sqrt{y-1}=11\\ \sqrt{y+10}+\sqrt{x-1}=11\end{array}$$ \sqrt{x-2}+\sqrt{10-x}=x^2-12x+40$$ x^2+y^2=1$$ 1$$ \frac{1}{\sqrt{2}}\le x^3+y^3\le 1$$