Tìm x: (5x-1)(x+1)-2(x-3)^2=(x+2)(3x-1)-(x+4)^2+(x^2-x) 12/08/2021 Bởi Rose Tìm x: (5x-1)(x+1)-2(x-3)^2=(x+2)(3x-1)-(x+4)^2+(x^2-x)
Đáp án: $(5x-1)(x+1)-2(x-3)^2 =(x+2)(3x-1)-(x+4)^2+(x^2-x)$ $ ⇔( 5x^2+5x-x-1) -2(x^2-6x+9) = (3x^2-x+6x-2)- (x^2+8x+16) +(x^2-x)$ $ ⇔ 5x^2+4x-1 -2x^2 +12x -18 = 3x^2 +5x -2 – x^2-8x-16 +x^2-x$ $ ⇔ 5x^2 -3x^2-x^2+x^2 -2x^2 +4x +12x -5x+8x +x = -2 -16 +1 +18$ $ ⇔19x = 1$ $⇔ x = \dfrac{1}{19}$ Vậy $x =\dfrac{1}{19}$ Bình luận
Đáp án:
$(5x-1)(x+1)-2(x-3)^2 =(x+2)(3x-1)-(x+4)^2+(x^2-x)$
$ ⇔( 5x^2+5x-x-1) -2(x^2-6x+9) = (3x^2-x+6x-2)- (x^2+8x+16) +(x^2-x)$
$ ⇔ 5x^2+4x-1 -2x^2 +12x -18 = 3x^2 +5x -2 – x^2-8x-16 +x^2-x$
$ ⇔ 5x^2 -3x^2-x^2+x^2 -2x^2 +4x +12x -5x+8x +x = -2 -16 +1 +18$
$ ⇔19x = 1$
$⇔ x = \dfrac{1}{19}$
Vậy $x =\dfrac{1}{19}$