Tìm x biết: 3*(2*x-1)*(x+2)-2*(3*x+2)*(x-4)=5 mong mn giúp ạ 09/07/2021 Bởi Quinn Tìm x biết: 3*(2*x-1)*(x+2)-2*(3*x+2)*(x-4)=5 mong mn giúp ạ
Đáp án: `3 (2x – 1) (x + 2) – 2 (3x + 2) (x – 4) = 5` `⇔ 3 [(2x – 1) (x + 2)] – 2 [(3x + 2) (x-4)] = 5` `⇔ 3 [2x (x + 2) – 1 (x + 2)] – 2 [3x (x-4) + 2 (x-4)] = 5` `⇔ 3 [2x^2 + 4x – x – 2] – 2 [3x^2 – 12x + 2x – 8] = 5` `⇔ 3 [2x^2 + 3x – 2] – 2 [3x^2 – 10x – 8] = 5` `⇔ 6x^2 + 9x – 6 – 6x^2 + 20x + 16 = 5` `⇔ (6x^2 – 6x^2) + (9x + 20x) + (-6 + 16) = 5` `⇔ 29x + 10 = 5` `⇔ 29x=5-10` `⇔29x =-5` `⇔x=-5 ÷ 29` `⇔x=(-5)/29` Vậy `x=(-5)/29` Bình luận
\( 3(2x-1)(x+2)-2(3x+2)(x-4)=5\\↔3(2x^2+3x-2)-2(3x^2-10x-8)=5\\↔6x^2+9x-6-6x^2+20x+16=5\\↔29x+10=5\\↔29x=-5\\↔x=-\dfrac{5}{29}\) Vậy \(x=-\dfrac{5}{29}\) Bình luận
Đáp án:
`3 (2x – 1) (x + 2) – 2 (3x + 2) (x – 4) = 5`
`⇔ 3 [(2x – 1) (x + 2)] – 2 [(3x + 2) (x-4)] = 5`
`⇔ 3 [2x (x + 2) – 1 (x + 2)] – 2 [3x (x-4) + 2 (x-4)] = 5`
`⇔ 3 [2x^2 + 4x – x – 2] – 2 [3x^2 – 12x + 2x – 8] = 5`
`⇔ 3 [2x^2 + 3x – 2] – 2 [3x^2 – 10x – 8] = 5`
`⇔ 6x^2 + 9x – 6 – 6x^2 + 20x + 16 = 5`
`⇔ (6x^2 – 6x^2) + (9x + 20x) + (-6 + 16) = 5`
`⇔ 29x + 10 = 5`
`⇔ 29x=5-10`
`⇔29x =-5`
`⇔x=-5 ÷ 29`
`⇔x=(-5)/29`
Vậy `x=(-5)/29`
\( 3(2x-1)(x+2)-2(3x+2)(x-4)=5\\↔3(2x^2+3x-2)-2(3x^2-10x-8)=5\\↔6x^2+9x-6-6x^2+20x+16=5\\↔29x+10=5\\↔29x=-5\\↔x=-\dfrac{5}{29}\)
Vậy \(x=-\dfrac{5}{29}\)