Giải phương trình trùng phương. Giup mình với ạ!!! Bài 4: Giải phương trình:,$a,~x^4-20x^2+4=0.$ $b,~3x^4-x^2-10=0.$ $c,~5x^4+13x^2-6=0$,Bài 5: Giải phương trình:,$a,~x^4-3x^2-4=0.$ $b,~x^4-2x^2+1=0.$ $c,~x^4+2x^2-15=0$

Question

Accepted Answer

Bài 4)

a) $\displaystyle x^{4} -20x^{2} +4=0$

Đặt $\displaystyle x^{2} =t\ ( t\geqslant 0)$

⟹ $\displaystyle t^{2} -20t+4=0$

$\displaystyle \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow \left[ \begin{array}{l l}
t=10+4\sqrt{6} & \
t=10-4\sqrt{6} &
\end{array} ight. \Leftrightarrow \left[ \begin{array}{l l}
x^{2} =10+4\sqrt{6} & \
x^{2} =10-4\sqrt{6} &
\end{array} ight.\
\Leftrightarrow \left[ \begin{array}{l l}
x=\sqrt{10+4\sqrt{6}} & \
x=\sqrt{10-4\sqrt{6}} &
\end{array} ight. \Leftrightarrow \left[ \begin{array}{l l}
x=\sqrt{\left( 2+\sqrt{6} ight)^{2}} & \
x=\sqrt{\left( 2-\sqrt{6} ight)^{2}} &
\end{array} ight.\
\Leftrightarrow \left[ \begin{array}{l l}
x=2+\sqrt{6} & \
x=\sqrt{6} -2 &
\end{array} ight.
\end{array}$

b) $\displaystyle 3x^{4} -x^{2} -10=0$

Đặt $\displaystyle x^{2} =t\ ( t\geqslant 0)$

⟹ $\displaystyle 3t^{2} -t-10=0$

$\displaystyle \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow 3t^{2} -6t+5t-10=0\
\Leftrightarrow 3t( t-2) +5( t-2) =0\
\Leftrightarrow ( t-2)( 3t+5) =0\
\Leftrightarrow \left[ \begin{array}{l l}
t=2 & \
t=\frac{-5}{3} &
\end{array} ight. \Longrightarrow x^{2} =2\Leftrightarrow x=\pm \sqrt{2}
\end{array}$

c) $\displaystyle 5x^{4} +13x^{2} -6=0$

Đặt $\displaystyle x^{2} =t\ ( t\geqslant 0)$

⟹ $\displaystyle 5t^{2} +13t-6=0$

$\displaystyle \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow 5t^{2} +15t-2t-6=0\
\Leftrightarrow 5t( t+3) -2( t+3) =0\
\Leftrightarrow ( t+3)( 5t-2) =0\
\Leftrightarrow \left[ \begin{array}{l l}
t=-3 & \
t=\frac{2}{5} &
\end{array} ight. \Longrightarrow t=\frac{2}{5} \Longrightarrow x^{2} =\frac{2}{5}\
\Leftrightarrow x=\pm \sqrt{\frac{2}{5}}
\end{array}$