Đáp án:
a)$2x^{3}$ -5$x^{2}$ +8x-3=0
(2$x^{3}$ -$x^{2}$ )-(4$x^{2}$ -2x)+(6x-3)=0
$x^{2}$ (2x-1)-2x(2x-1)+3(2x-1)=0 (2x-1)($x^{2}$ -2x+3)=0
=> \(\left[ \begin{array}{l}2x-1=0\\x^2-2x+3=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}2x=1\\(x-3)(x-1)=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=0,5\\x=3 hoặc x=1\end{array} \right.\)
b)$(x+3)^{4}$ + $(x+5)^{4}$ =2
Đặt x+4=a
Do đó: $(a+1)^{4}$ +$(a-1)^{4}$=2
=> ($a^{4}$+ 4$a^{3}$+6 $a^{2}$ +4a+1)+($a^{4}$ -4$x^{3}$ +6$a^{2}$ -4a+1)=2
=>2$a^{4}$ +2+12$a^{2}$ =2
=>$a^{2}$( $a^{2}$ +12)=0
mà$a^{2}$ +12>0 ∀a
Vậy a=0
