Toán Lớp 8: tìm x a)(x+3)(x-3)-(x-2)(x+5)=-2 b)x^3-5x^2+6x=0 c)2x^2-2x=(x-1)^2

Question

Toán Lớp 8:  tìm x a)(x+3)(x-3)-(x-2)(x+5)=-2 b)x^3-5x^2+6x=0 c)2x^2-2x=(x-1)^2

thudan · Answer

a) (x + 3)(x - 3) - (x - 2)(x + 5) = -2 <=> (x + 3)(x - 3) - (x - 2)(x + 5) = -(3x - 1) <=> -(3x - 1) = -2 <=>1 - 3x = -2 <=> -3x = -3 <=> x = 1 b) x^3 - 5x^2 + 6x = 0 <=> x^3 - 5x^2 + 6x = (x - 3)(x - 2)x <=> x(x - 3)(x - 2) = 0 <=> ( left[ begin{array}{l}x=0 x=2 x = 3 end{array} right. ) c) 2x^2 - 2x = (x - 1)^2 <=> 2x^2 - 2x = x^2 - 2x + 1 <=> x^2 = 1 (Rút gọn 2x^2 - 2x cả hai vế) <=> x = 1

thanhthuythu · Answer

text{a)} (x + 3)(x - 3) - (x - 2)(x + 5) = - 2 <=> x^2 - 3^2 - (x^2 + 5x - 2x - 10) = - 2 <=> x^2 - 9 - x^2 - 5x + 2x + 10 = - 2 <=> - 3x + 1 = - 2 <=> - 3x = - 3 <=> x = 1 text{b)} x^3 - 5x^2 + 6x = 0 <=> x (x^2 - 5x + 6) = 0 <=> x (x^2 - 2x - 3x + 6) = 0 <=> x [x (x - 2) - 3 (x - 2)] = 0 <=> x (x - 2)(x - 3) = 0 <=> $ left[ begin{array}{ccc}x - 0 = 0 x - 2 = 0 x - 3 = 0 end{array} right.$ <=> $ left[ begin{array}{ccc}x = 0 x = 2 x = 3 end{array} right.$ text{c)} 2x^2 - 2x = (x - 1)^2 <=> 2x (x - 1) = (x - 1)^2 <=> 2x = x - 1 <=> 2x - x = - 1 <=> x = - 1