Toán Lớp 8: Tìm x a) 4x²+3x-7=0 b) 5x²-x-4=0 c) 6x²-5x-14=0 Giả hộ mik ạ

Question

Toán Lớp 8:  Tìm x  a) 4x²+3x-7=0 b) 5x²-x-4=0 c) 6x²-5x-14=0  Giả hộ mik ạ

minhtuquynh · Answer

Giải đáp + Lời giải và giải thích chi tiết ! to Tìm x: a) 4x^2+3x-7 = 0 <=> 4x^2+7x-4x-7 = 0 <=> (4x^2-4x)+(7x-7) = 0 <=> 4x(x-1)+7(x-1) = 0 <=> (4x+7)(x-1) = 0 <=> ( left[ begin{array}{l}4x+7=0 x-1=0 end{array} right. ) <=> ( left[ begin{array}{l}4x=-7 x=1 end{array} right. ) <=> ( left[ begin{array}{l}x=- dfrac{7}{4} x=1 end{array} right. ) Vậy S= {-7/4; 1} b) 5x^2-x-4 = 0 <=> 5x^2-5x+4x-4 = 0 <=> (5x^2-5x)+(4x-4) = 0 <=> 5x(x-1)+4(x-1) = 0 <=> (5x+4)(x-1) = 0 <=> ( left[ begin{array}{l}5x+4=0 x-1=0 end{array} right. ) <=> ( left[ begin{array}{l}5x=-4 x=1 end{array} right. ) <=> ( left[ begin{array}{l}x=- dfrac{4}{5} x=1 end{array} right. ) Vậy S= {-4/5; 1} c) 6x^2-5x-14 = 0 <=> 6x^2-12x+7x-14 = 0 <=> (6x^2-12x)+(7x-14) = 0 <=> 6x(x-2)+7(x-2) = 0 <=> (6x+7)(x-2) = 0 <=> ( left[ begin{array}{l}6x+7=0 x-2=0 end{array} right. ) <=> ( left[ begin{array}{l}6x=-7 x=2 end{array} right. ) <=> ( left[ begin{array}{l}x=- dfrac{7}{6} x=2 end{array} right. ) Vậy S= {-7/6; 2}

tuanh · Answer

Giải đáp:   $a) 4x&sup2;+3x-7=0$  $4x^2-4x+7x-7=0$  $4x(x-1)+7(x-1)=0$  $(x-1)(4x+7)=0$   ( left[  begin{array}{l}x=1  x=-7/4 end{array}  right. )   $b) 5x&sup2;-x-4=0$  $5x^2-5x+4x-4=0$  $5x(x-1)+4(x-1)=0$  $(x-1)(5x+4)=0$   ( left[  begin{array}{l}x=1  x=-4/5 end{array}  right. )  $c) 6x&sup2;-5x-14=0 $  $6x^2-12x+7x-14=0$  $6x(x-2)+7(x-2)=0$  $(x-2)(6x+7)=0$   ( left[  begin{array}{l}x=2  x=-7/6 end{array}  right. )