Toán Lớp 8: Bài 3: Tìm x, biết: a) 2x + 50 = 0 b) (x – 2 )(x + 1) = 0 c) 5x(x – 1) = 1 – x d) (3x – 4)2 – (x + 1)2 = 0 e) x2 + 7x + 12 =0

Question

Toán Lớp 8:  Bài 3: Tìm x, biết: a) 2x + 50 = 0  b) (x – 2 )(x + 1) = 0  c) 5x(x – 1) = 1 – x d) (3x – 4)2 – (x + 1)2 = 0  e) x2 + 7x + 12 =0

minhtamnguyet88 · Answer

Giải đáp:       Lời giải và giải thích chi tiết:    a) 2x+50=0   2x=-50   x = -50:2   x = -25   b) (x-2)(x+1)=0    &rArr; x-2 =0 &hArr; x= 0+2 &hArr; x=2   hoặc x+1 =0 &hArr; x= 0-1 &hArr;x= -1    vậy S={ 2 ; -1}   c) 5x(x-1)=1-x   5x^2-5x -1+x=0   5x^2 -4x-1= 0   5x^2 -5x +x - 1= 0   5x(x-1)+(x-1)=0   (x-1)(5x+1)=0   &rArr; x-1 =0 &hArr; x= 1   hoặc 5x+1 =0 &hArr;5x= -1 &hArr;x=-1/5   vậy S= { 1 ; -1/5}   d)   (3x-4)^2 - (x+1)^2 =0    (3x-4 +x +1)(3x-4 -x-1)=0   (4x -3)(2x-5)=0    &rArr; 4x-3= 0 &hArr; 4x= 3 &hArr; x= 3/4    hoặc 2x-5 =0 &hArr;2x=5 &hArr; x= 5/2   vậy S= { 3/4 ; 5/2}   e)   x^2+7x+12 =0    x^2 +3x+4x+12=0   x(x+3)+ 4(x+3)=0   (x+3)(x+4)=0   &rArr; x+3 =0 &hArr; x= -3    hoặc x+4 =0 &hArr; x=-4    vậy S={ -3 ; -4}

hiennhi98 · Answer

$ a) 2x + 50 = 0 $     $&rArr;2x = -50$     $&rArr;x = -25$     $b)(x-2)(x+1)=0$     $&rArr;$ ( left[  begin{array}{l}x-2=0  x+1=0 end{array}  right. )      $&rArr;$ ( left[  begin{array}{l}x=2  x=-1 end{array}  right. )      $c)5x(x &ndash; 1) = 1 &ndash; x$     $&rArr;5x(x &ndash; 1) - (1 - x)=0$     $&rArr;5x(x &ndash; 1) + (x - 1)=0$     $&rArr;(x - 1)(5x + 1)=0$     $&rArr;$ ( left[  begin{array}{l}x - 1=0  5x + 1=0 end{array}  right. )      $&rArr;$ ( left[  begin{array}{l}x=1  5x=-1 end{array}  right. )      $&rArr;$ ( left[  begin{array}{l}x=1  x= frac{-1}{5} end{array}  right. )      $d)(3x &ndash; 4)&sup2; &ndash; (x + 1)&sup2; = 0$    $&rArr;(3x-4-x-1)(3x-4+x+1)=0$   $&rArr;(2x-5)(4x-3)=0$    $&rArr;$ ( left[  begin{array}{l}2x-5=0  4x-3=0 end{array}  right. )      $&rArr;$ ( left[  begin{array}{l}2x=5  4x=3 end{array}  right. )      $&rArr;$ ( left[  begin{array}{l}x= frac{5}{2}  x= frac{3}{4} end{array}  right. )     $e)x&sup2; + 7x + 12 =0$     $&rArr;x&sup2; + 3x + 4x + 12 =0$     $&rArr;(x&sup2; + 3x)+(4x + 12) = 0$     $&rArr;x(x+3)+4(x+3)=0$     $&rArr;(x+4)(x+3)=0$     $&rArr;$ ( left[  begin{array}{l}x+4=0  x+3=0 end{array}  right. )      $&rArr;$ ( left[  begin{array}{l}x=-4  x=-3 end{array}  right. )