Giúp mình với ạ. (2x+3)×(x-4)+(x-5)×(x-2)=(3x-5)×(x-4) Đây là bài tìm x nha mn???????? 09/07/2021 Bởi Delilah Giúp mình với ạ. (2x+3)×(x-4)+(x-5)×(x-2)=(3x-5)×(x-4) Đây là bài tìm x nha mn????????
`(2x+3)×(x-4)+(x-5)×(x-2)=(3x-5)×(x-4)` `⇔(2x+3)×(x-4)+(x-5)×(x-2)-(3x-5)×(x-4)=0` `⇔(x-4)(2x+3-3x+5)+(x-5)×(x-2)=0` `⇔(x-4)(8-x)+(x-5)×(x-2)=0` `⇔8x-x^2-32+4x+x^2-5x-2x+10=0` `⇔5x-22=0` `⇔5x=22` `⇔x=(22)/5` Bình luận
`(2x+3)(x-4) + (x-5)(x-2) = (3x-5)(x-4)` `=> 2x^2 – 8x + 3x – 12 + x^2 – 2x – 5x + 10 = 3x^2 – 12x – 5x + 20` `=> (2x^2 + x^2) + (-8x + 3x – 2x-5x) + (10-12) = 3x^2 – 17x + 20` `=> 3x^2 -12x – 2 = 3x^2 – 17x + 20` `=> 3x^2 – 12x – 2 – 3x^2 + 17x – 20 = 0` `=> (3x^2 – 3x^2) + (17x – 12x) – (2 + 20) = 0` `=> 5x – 22 = 0` `=> 5x=22` `=> x = 22/5` Vậy `x=22/5` Bình luận
`(2x+3)×(x-4)+(x-5)×(x-2)=(3x-5)×(x-4)`
`⇔(2x+3)×(x-4)+(x-5)×(x-2)-(3x-5)×(x-4)=0`
`⇔(x-4)(2x+3-3x+5)+(x-5)×(x-2)=0`
`⇔(x-4)(8-x)+(x-5)×(x-2)=0`
`⇔8x-x^2-32+4x+x^2-5x-2x+10=0`
`⇔5x-22=0`
`⇔5x=22`
`⇔x=(22)/5`
`(2x+3)(x-4) + (x-5)(x-2) = (3x-5)(x-4)`
`=> 2x^2 – 8x + 3x – 12 + x^2 – 2x – 5x + 10 = 3x^2 – 12x – 5x + 20`
`=> (2x^2 + x^2) + (-8x + 3x – 2x-5x) + (10-12) = 3x^2 – 17x + 20`
`=> 3x^2 -12x – 2 = 3x^2 – 17x + 20`
`=> 3x^2 – 12x – 2 – 3x^2 + 17x – 20 = 0`
`=> (3x^2 – 3x^2) + (17x – 12x) – (2 + 20) = 0`
`=> 5x – 22 = 0`
`=> 5x=22`
`=> x = 22/5`
Vậy `x=22/5`