Đáp án:
Giải thích các bước giải:
$\text{36,}$
$\text{a/ x²+4x+3= x²+3x+x+3}$
$\text{=x(x+3)+(x+3)=(x+1)(x+3)}$
$\text{b/ 2x²+3x-5= 2x²-2x+5x-5}$
$\text{= 2x(x-1)+5(x-1)= (2x+5)(x-1)}$
$\text{c, 16x-5x²-3=-5x²+15x+x-3}$
$\text{=-5x(x-3)+(x-3)= (1-5x)(x-3)}$
$\text{37.}$
$\text{a, 5x(x-1)=x-1}$
$\text{⇔5x(x-1)-(x-1)=0}$
$\text{⇔(5x-1)(x-1)=0}$
$\text{⇔\(\left[ \begin{array}{l}5x-1=0\\x-1=0\end{array} \right.\) }$
$\text{⇔\(\left[ \begin{array}{l}x=\frac{1}{5} \\x=1\end{array} \right.\) }$
$\text{Vậy S={$\frac{1}{5}$ ; 1} }$
$\text{b, 2(x+5)-x²-5x=0}$
$\text{⇔2x+10-x²-5x=0}$
$\text{⇔10-x²-3x=0}$
$\text{⇔10-5x+2x-x²=0}$
$\text{⇔5(2-x)+x(2-x)=0}$
$\text{⇔(5+x)(2-x)=0}$
$\text{⇔\(\left[ \begin{array}{l}5+x=0\\2-x=0\end{array} \right.\) }$
$\text{⇔\(\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.\) }$
$\text{Vậy S= {-5;2} }$
$\text{38}$
$\text{Có a+b+c=0⇔ a+b=-c⇔(a+b)³=-c³}$
$\text{⇔a³+3a²b+3ab²+b²=-c³}$
$\text{⇔a³+3ab(a+b)+b³=-c³}$
$\text{Vì a+b=-c(cmt)}$
$\text{⇒a³+3ab(-c)+b³=-c³}$
$\text{⇔a²+b³-3abc=-c³}$
$\text{⇔a³+b³+c³-3abc=0}$
$\text{⇔a³+b³+c³=3abc(đpcm)}$
Xin hay nhất ạ!
