Toán Lớp 8: Tìm x biết : `a)` `(x-3)^2-x^2=-15` `b)` `2(x-3)=3x(x-3)`

Question

Toán Lớp 8:  Tìm x biết : a) (x-3)^2-x^2=-15 b) 2(x-3)=3x(x-3)

tuyetanhthu · Answer

Giải đáp + giải thích các bước giải: a, (x - 3)^2 - x^2 = -15 <=> x^2 - 6x + 9 - x^2 = -15 <=> -6x + 9 = -15 <=> -6x = -24 <=> x = 4 Vậy x = 4 b, 2(x - 3) = 3x(x - 3) <=> 2(x - 3) - 3x(x -3) = 0 <=> (2 - 3x)(x - 3) = 0 <=> ( left[ begin{array}{l}2 - 3x = 0 x - 3 = 0 end{array} right. ) <=> ( left[ begin{array}{l}x = dfrac{2}{3} x = 3 end{array} right. ) Vậy x in {2/3 ; 3}

diepbich93 · Answer

Answer    a, (x - 3)^2 - x^2 = -15   => x^2 - 6x + 9 - x^2 = -15   => (x^2 - x^2) - 6x + 9 = -15   => -6x + 9 = -15   => -6x = -15 - 9   => -6x = -24   => x = (-24) : (-6)   => x = 4   $ text{Vậy}$ x = 4   ______________________   b, 2 . (x - 3) = 3x . (x - 3)   => 2 . (x - 3) - 3x . (x - 3) = 0   => (3 - x) . (3x - 2) = 0   => $ left[ begin{matrix} 3 - x = 0   3x - 2 = 0 end{matrix} right.$   => $ left[ begin{matrix} x = 3 - 0   3x = 0 + 2 end{matrix} right.$   => $ left[ begin{matrix} x = 3   3x = 2 end{matrix} right.$   => $ left[ begin{matrix} x = 3   x = 2 : 3 end{matrix} right.$   => $ left[ begin{matrix} x = 3   x =  dfrac{2}{3} end{matrix} right.$   $ text{Vậy}$ x  in {3 ; 2/3}