Toán Lớp 8: Tìm x c)3x(x-1)+x-1=0 d)8x^3-50x=0

Question

Toán Lớp 8:  Tìm x c)3x(x-1)+x-1=0 d)8x^3-50x=0

maingocquynhnhu2k1 · Answer

c) 3x(x-1)+x-1=0   &rArr;3x(x-1)+(x-1)=0   &rArr;(3x+1).(x-1)=0   &rArr; ( left[  begin{array}{l}3x+1=0  x-1=0 end{array}  right. )   &rArr; ( left[  begin{array}{l}x=- frac{1}{3}  x=1 end{array}  right. )     d) 8$x^{3}$ -50x=0     &rArr;2x(4$x^{2}$ -25)=0      ( left[  begin{array}{l}2x=0  4x^{2} -25=0 end{array}  right. )      ( left[  begin{array}{l}x=0  4x^{2}=25&rArr;x^{2}= frac{25}{4}  end{array}  right. ) &rArr; ( left[  begin{array}{l}x= frac{5}{2}  x=- frac{5}{2} end{array}  right. )

ankim · Answer

$c)$ $3x$. $( x$ $-$ $1)$ $+$ $x$ $-$ $1$ $=$ $0$           $3x$. $( x$ $-$ $1)$ $+$ $( x$ $-$ $1)$ $=$ $0$              $( x$ $-$ $1)$ . $( 3x$ $+$ $1)$ $=$ $0$   $ Rightarrow$ $ left[ begin{matrix} x- 1=0   3x + 1=0 end{matrix} right.$   $ Rightarrow$ $ left[ begin{matrix} x = 1   3x =- 1 end{matrix} right.$   $ Rightarrow$ $ left[ begin{matrix} x =1   x =-1/3  end{matrix} right.$   Vậy $x$ $ in$ { $1$; $ frac{-1}{3}$}         $d)$ $8x&sup3;$ $-$ $50x$ $=$ $0$       $2x.$ $( 4x&sup2;$ $-$ $25x)$ $=$ $0$   $ Rightarrow$ $ left[ begin{matrix} 2x =0   4x&sup2; - 25x =0 end{matrix} right.$   $ Rightarrow$ $ left[ begin{matrix} x = 0   4x&sup2; = 25x end{matrix} right.$   $ Rightarrow$ $ left[ begin{matrix} x = 0   x&sup2; = 25/4  end{matrix} right.$   $ Rightarrow$ $ left[ begin{matrix} x = 0  end{matrix} right.$         $ left[ begin{matrix} x = 5/2   x = -5/2  end{matrix} right.$   $Vậy$ $x$ $ in$ {$0$; $ frac{5}{2}$; $ frac{-5}{2}$}