ĐK:...Đặt u=$\sqrt[3]{x^{2}-xy+1}$ ; v=$\sqrt[3]{y^{2}-xy+1}$$\Rightarrow$$u^{3}$+$v^{3}$=$(x-y)^{2}$+2$\geq$2pt(1) $\Leftrightarrow $u+v-2=2($u^{3}$+$v^{3}$-2)(*)$\Leftrightarrow$$\frac{u^{3}+v^{3}}{2}$=$\frac{u+v+2}{4}$$\geq$1(do $u^{3}$+$v^{3}$$\geq$2)$\Rightarrow$u+v$\geq$2(1)Ta cm đc:$\frac{u^{3}+v^{3}}{2}$$\geq$$(\frac{u+v}{2})^{3}$$\Leftrightarrow$(u+v-2)$\left[ {(u+v)^{2}+2(u+v)+2} \right]$$\leq$0$\Leftrightarrow$u+v$\leq$2(2)Từ(1)&(2)$\Rightarrow$u+v=2. Từ(*)$\Rightarrow$$u^{3}$+$v^{3}$=2$\Leftrightarrow$x=yThế vào pt(2) of hệ$\Rightarrow$32$x^{2}$$\sqrt{x}$-10$\sqrt{x}$+4=0Đặt t=$\sqrt{x}$(t$\geq$0)$\Rightarrow$32$t^{5}$-10t+4=0$\Leftrightarrow$$(t-\frac{1}{2})^{2}$(32$t^{3}$+32$t^{2}$+24t+16)=0$\Rightarrow$t=$\frac{1}{2}$$\Rightarrow$x=y=$\frac{1}{4}$(t/m đk)
ĐK:...Đặt
$u=\sqrt[3]{x^{2}-xy+1} ; v=\sqrt[3]{y^{2}-xy+1}$$\Rightarrow
u^{3}+v^{3}=(x-y)^{2}+2
\geq
2
$pt(1) $\Leftrightarrow u+v-2=2(u^{3}+v^{3}-2)
$ (*)$\Leftrightarrow
\frac{u^{3}+v^{3}}{2}=\frac{u+v+2}{4}
\geq
1
$(do $u^{3}+v^{3}
\geq
2)\Rightarrow
u+v
\geq
2$(1)Ta cm đc:$\frac{u^{3}+v^{3}}{2}
\geq
(\frac{u+v}{2})^{3}
\Leftrightarrow
(u+v-2)
\left[ {(u+v)^{2}+2(u+v)+2} \right]
\leq
0$$\Leftrightarrow
u+v
\leq
2$(2)Từ(1)&(2)$\Rightarrow
u+v=2
$. Từ
(*)
$\Rightarrow
u^{3}
+
v^{3}
=2
\Leftrightarrow
x=y
$Thế vào pt(2)
của hệ$\Rightarrow
32
x^{2}
\sqrt{x}
-10\sqrt{x}
+4=0
$Đặt
$t=\sqrt{x}$
($
t \geq
0
$)$\Rightarrow
32
t^{5}-10t+4=0$
$\Leftrightarrow
(t-\frac{1}{2})^{2}
(32t^{3}+32t^{2}+24t+16)=0
$$\Rightarrow
t=
\frac{1}{2}
\Rightarrow
x=y=\frac{1}{4}$(t/m đk)