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$\begin{cases}x+y=m (1)\\ x^{4}+y^{4}=m^{4} (2) \end{cases}$
từ $(1): y=m-x$ thế vào $(2)$:
$x^{4}+\left(m-x\right)^{4}=m^{4}$
đặt $y=x-\frac{m}{2}\Rightarrow x=y+\frac{m}{2}, x-m=y+\frac{m}{2}-m=y-\frac{m}{2}$
được phương trình: $\left(y+\frac{m}{2}\right)^{4}+\left(y-\frac{m}{2}\right)^{4}=m^{4}$
$\Leftrightarrow
y^{4}+4.y^{3}.\frac{m}{2}+6y^{2}.\frac{m^{2}}{4}.m+4y\left(\frac{m}{2}\right)^{3}+\frac{m^{4}}{16}+y^{4}-4.y^{3}.\frac{m}{2}+6y^{2}.\frac{m^{2}}{4}-4y\left(\frac{m}{2}\right)^{3}+\frac{m^{4}}{16}$
$=m^{4}$
$\Leftrightarrow 2y^{4}+3m^{2}y^{2}-7\frac{m^{4}}{8}=0(*)$
+/ $m=0: y=0\Rightarrow x=0$
+/ $m\neq 0: (*)$ có $\Delta=9m^{4}+7m^{4}=16m^{4}$
$\Rightarrow y^{2}=\frac{-3m^{2}+4m^{2}}{4}=\frac{m^{2}}{4}\Leftrightarrow y=\pm\frac{m}{2}$
nếun $y=\frac{m}{2}\Rightarrow x=m, y=-\frac{m}{2}\Rightarrow x=0$
hệ luôn có hai nghiệm $(0;m), (m; 0)$
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