x căn(x^2+y) + y = căn(x^4+x^3) +x (1)
x+căn y + căn(x-1) + căn(y(x-1)) =9/2 (2)
ĐK : [tex]x\geq 1;y\geq 0[/tex]
[tex]pt (1)\iff \sqrt{x^4+x^3}-x\sqrt{x^2+y} +x-y =0 \iff x\frac{x-y}{\sqrt{x^4+x^3}+\sqrt{x^2+y} }+(x-y)=0|\iff x=y[/tex]
pt(2) [tex]\iff 2x +2(\sqrt{x}+\sqrt{x-1}+\sqrt{x(x-1)})=9 \iff x-1 +x +2(\sqrt{x}+\sqrt{x-1}+\sqrt{x(x-1)})=8[/tex]
Đặt [tex](\sqrt{x};\sqrt{x-1})=(a;b)(a,b\geq 0)[/tex]
pt trở thành : [tex]\left\{\begin{matrix} a^2+b^2+2(ab+a+b)=8 & & \\ a^2-b^2=1 & & \end{matrix}\right.[/tex]
Nhận thấy pt $(1) \iff (a+b)^2+2(a+b)+1=9 \iff a+b+1=3...$
sau đó rút thế^^