Toán Lớp 8: a) x^2+3x-18=0 b) x^2-5x-24=0 c) 8x^2+30x+7=0

Question

Toán Lớp 8:  a) x^2+3x-18=0 b) x^2-5x-24=0 c) 8x^2+30x+7=0

buianhtuan · Answer

Giải đáp + Lời giải và giải thích chi tiết:     a) $x^{2}$ + 3x - 18 = 0     &rArr; ($x^{2}$ - 3x) + (6x - 18) = 0     &rArr; x. (x - 3) + 6.(x - 3) = 0     &rArr; (x - 3). (x + 6) = 0     &rArr;  ( left[  begin{array}{l}x-3=0  x+6=0 end{array}  right. )      &rArr;  ( left[  begin{array}{l}x=3  x=-6 end{array}  right. )             Vậy x = 3 hoặc x = -6     b) $x^{2}$ - 5x - 24 = 0     &rArr; ($x^{2}$ + 3x) + (-8x - 24)     &rArr; x. (x + 3) - 8. (x + 3) = 0     &rArr; (x + 3). (x - 8) = 0     &rArr;  ( left[  begin{array}{l}x+3=0  x-8=0 end{array}  right. )      &rArr;  ( left[  begin{array}{l}x=-3  x=8 end{array}  right. )              Vậy x = - 3 hoặc x = 8     c) $8x^{2}$ + 30x + 7 = 0     &rArr; ($8x^{2}$ + 2x) + (28x + 7) = 0     &rArr; 2x. (4x + 1) + 7. (4x + 1)= 0     &rArr; (4x + 1). (2x + 7) = 0     &rArr;  ( left[  begin{array}{l}4x+1=0  2x+7=0 end{array}  right. )      &rArr;  ( left[  begin{array}{l}4x=-1  2x=-7 end{array}  right. )      &rArr;  ( left[  begin{array}{l}x= frac{-1}{4}  x= frac{-7}{2} end{array}  right. )               Vậy x = $ frac{-1}{4}$ oặc x = $ frac{-7}{2}$        Ch&uacute;c bạn hc tốt         Mong bạn cho mk ctlhn ạ

quynhthanhnghi · Answer

a) x&sup2; + 3x - 18 = 0 x&sup2; -3x + 6x - 18 = 0 x(x-3) + 6(x-3) = 0 (x-3)(x+6) = 0 x - 3 = 0 hay x + 6 = 0 x = 3 hay x = -6 b) x&sup2; - 5x - 24 = 0 x&sup2; - 8x  + 3x - 24 = 0 x(x-8) + 3(x-8) = 0 (x-8)(x+3) = 0 x - 8 = 0 hay x + 3 = 0 x = 8 hay x = -3 c) 8x&sup2; + 30x + 7 = 0 8x&sup2; + 28x + 2x + 7 = 0 4x(2x+7) + (2x+7) = 0 (2x+7)(4x+1) = 0 2x + 7 = 0 hay 4x + 1 = 0 2x = -7 hay 4x = -1 x = $ frac{-7}{2}$ hay x = $ frac{-1}{4}$