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1 $\left\{\begin{matrix} y^{3}+3xy^{2}=-28 & & \\ x^{2}-6xy+y^{2}=6x-10y & & \end{matrix}\right.$

2$\left\{\begin{matrix} 6x^{2}y+2y^{3}+35=0 & & \\ 5x^{2}+5y^{2}+2xy+5x+13y=0 & & \end{matrix}\right.$

3 $\left\{\begin{matrix} x^{3}+5x^{2}y=-35 & & \\ 2x^{2}-5xy-5y^{2}+x+10y-35=0 & & \end{matrix}\right.$

4$\left\{\begin{matrix} x^{3}+3x^{2}y=6xy-3x-49 & & \\ x^{2}-8xy+y^{2}=10y-25x-9 & & \end{matrix}\right.$

$4x^{2}+2xy+y^{2}=3$

tìm min và max P =$x^{2}+2xy-y^{2}$

x,y>0.tìm GTLN:GTNN P=$x^{3}+y^{3}+x^{2}y+xy^{2}-5xy$

2 ĐK ,pt2$\Leftrightarrow (x-y-1)(\frac{x-y-2}{\sqrt{2(x-y)^{2}+6y-2x+4}}+\frac{1}{\sqrt{y+1}+\sqrt{x}})=0$

$\Leftrightarrow x=y+1$ cái còn lại cm vô nghiệm 

x= y+1 thay vào 1 ta dc;

$\sqrt{3-y}+\sqrt{y-8}=y^{2}+7y+6$ ai giải giúp 

a,b,c>0:$a^{2}+b^{2}+c^{2}=3$

cmr: $\frac{1}{4-\sqrt{ab}}+\frac{1}{4-\sqrt{ca}}+\frac{1}{4-\sqrt{ca}}\leq 1$

$4x^{3}-2x+3=0$

1 $a^{2}+b^{2}+c^{2}=3$

CMR $\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c}\geqslant 3$

2.a,b.c>0 

CMR $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} \leq \frac{3\sqrt{2}}{2}$

$a^{2}+b^{2}+c^{2}+abc=4$

CMR a+b+c$\leq$3

áp dụng $\frac{a^{2}}{c}+\frac{b^{2}}{d}\geq \frac{\left ( a+b \right )^{2}}{c+d}$ 

E=$\sum \frac{x^{2}y^{2}z^{2}}{x^{3}\left ( y+z \right )}\geq \frac{\left ( yz+zx+xy \right )^{2}}{2\left ( xy+xz+yz \right )}\geq\frac{3\sqrt[3]{x^{2}y^{2}z^{2}}}{2} =\frac{3}{2}$

dấu '=' xãy ra khi x=y=z=1

CMR $\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}\leq \frac{2}{\sqrt{1+ab}}$