BiBi Chi

cho $\left\{\begin{matrix} a,b,c >0 & & \\ a+b=c\leq 1& & \end{matrix}\right.$

tìm GTNN

$a, P=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}$

$b, \frac{1}{a^{2}+b^{2}}+\frac{1}{b^{c}+c^{2}}+\frac{1}{a^{2}+c^{2}}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}$

$c, \frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ac}+\frac{1}{c^{2}+ab}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}$

$\sqrt{3x+1}-\sqrt{6-x}=14x-3x^{2}+8$

a, $\left\{\begin{matrix} x^{3}+x+2-\frac{4}{y}=0 & & \\ 1+y^{2}-y^{3}(4x-2)=0& & \end{matrix}\right.$

b, $\left\{\begin{matrix} 3x^{3}y^{3}=2x^{3}+y^{3} & & \\ x+y=2xy^{3}& & \end{matrix}\right.$

c, $\left\{\begin{matrix} x+y+\sqrt{x^{2}-y^{2}}=12 & & \\ y\sqrt{x^{2}-y^{2}}=12 & & \end{matrix}\right.$

1, $4x^{3}+18x^{2}+27x+14=\sqrt[3]{4x+5}$

2, $\left\{\begin{matrix} x^{2}-3x=(y-2)(y+1) & & \\ (x+y)\sqrt{x^{2}-4x+5}+(x-2)\sqrt{x^{2}+y^{2}+2xy+1}=0& & \end{matrix}\right.$

3, $\left\{\begin{matrix} x^{2}+\frac{1}{x^{2}}+y^{2}+\frac{1}{y^{2}}=5 & & \\ (x^{2}-1)(y^{2}+1)\left [ (x+y)xy+x-y \right ]=x^{2}& & \end{matrix}\right.$

4, $\left\{\begin{matrix} 8(x+y)-3xy=2y^{2}+x^{2} & & \\ 4\sqrt{2-x}+\sqrt{3-y2}=2x^{2}-y^{2}+5& & \end{matrix}\right.$

5, $\left\{\begin{matrix} x^{2}+y^{2}-y=(2x+1)(y-1) & & \\ \sqrt{3x+8}-y=\frac{5}{x+y-12}& & \end{matrix}\right.$

1, $\left\{\begin{matrix} (x+\sqrt{x^{2}+4})(y+\sqrt{y^{2}+1})=2 & & \\ 12y^{2}-10y+2=2\sqrt[3]{x^{3}+1}& & \end{matrix}\right.$

2, $\left\{\begin{matrix} x^{2}-3y+2+2\sqrt{x^{2}y+2y}=0 & & \\ \sqrt{x^{2}+4x-y+1}+\sqrt[3]{2x-1}=1& & \end{matrix}\right.$

3, $\left\{\begin{matrix} 2+6y=\frac{x}{y}-\sqrt{x-2y} & & \\ \sqrt{x+\sqrt{x-2y}}=x+3y-2& & \end{matrix}\right.$