tìm x: a. (5x+1)^2-(5x+3)(5x-3)=30 b. ( 6x-2)^2+(5x-2)^2-4(3x-1)(5x-2)=0 tlhn, cn, 5vt 04/07/2021 Bởi Adalyn tìm x: a. (5x+1)^2-(5x+3)(5x-3)=30 b. ( 6x-2)^2+(5x-2)^2-4(3x-1)(5x-2)=0 tlhn, cn, 5vt
`a)` `(5x+1)^2 – (5x+3)(5x-3) = 30` `=> [ (5x)^2 + 2 . 5x . 1 + 1^2] – [ (5x)^2 – 3^2] = 30` `=> (25x^2 + 10x + 1) – (25x^2 – 9) = 30` `=> 25x^2 + 10x + 1 – 25x^2 + 9 =30` `=> 10x +10 = 30` `=> 10x = 20` `=> x = 2` Vậy `x=2` `b)` `(6x-2)^2 + (5x-2)^2 – 4 . (3x-1)(5x-2) = 0` `=> [ (6x)^2 – 2 . 6x .2 + 2^2] + [ (5x)^2 – 2 .5x . 2 + 2^2] – (12x -4)(5x-2)=0` `=> (36x^2 – 24x + 4) + (25x^2 – 20x + 4) – (60x^2 – 24x – 20x + 8) = 0` `=> 36x^2 – 24x + 4 + 25x^2 – 20x + 4 – 60x^2 + 24x + 20x – 8 =0` `=> x^2 =0` `=> x=0` Vậy `x=0` Bình luận
`a,` `(5x+1)^2-(5x+3)(5x-3)=30` `⇔25x^2+10x+1-25x^2+9-30=0` `⇔(25x^2-25x^2)+10x+(1+9-30)=0` `⇔10x-20=0` `⇔10x=20` `⇔(25x^2-25x^2)+10x+(1+9-30)=0` `⇔10x-20=0` `⇔x=2` `b,` `(6x-2)^2+(5x-2)^2-4(3x-1)(5x-2)=0` `⇔(6x-2)^2-2(6x-2)(5x-2)+(5x-2)^2=0` `⇔[(6x-2)-(5x-2)]^2=0` `⇔6x-2-5x+2=0` `⇔x=0` Bình luận
`a)`
`(5x+1)^2 – (5x+3)(5x-3) = 30`
`=> [ (5x)^2 + 2 . 5x . 1 + 1^2] – [ (5x)^2 – 3^2] = 30`
`=> (25x^2 + 10x + 1) – (25x^2 – 9) = 30`
`=> 25x^2 + 10x + 1 – 25x^2 + 9 =30`
`=> 10x +10 = 30`
`=> 10x = 20`
`=> x = 2`
Vậy `x=2`
`b)`
`(6x-2)^2 + (5x-2)^2 – 4 . (3x-1)(5x-2) = 0`
`=> [ (6x)^2 – 2 . 6x .2 + 2^2] + [ (5x)^2 – 2 .5x . 2 + 2^2] – (12x -4)(5x-2)=0`
`=> (36x^2 – 24x + 4) + (25x^2 – 20x + 4) – (60x^2 – 24x – 20x + 8) = 0`
`=> 36x^2 – 24x + 4 + 25x^2 – 20x + 4 – 60x^2 + 24x + 20x – 8 =0`
`=> x^2 =0`
`=> x=0`
Vậy `x=0`
`a,` `(5x+1)^2-(5x+3)(5x-3)=30`
`⇔25x^2+10x+1-25x^2+9-30=0`
`⇔(25x^2-25x^2)+10x+(1+9-30)=0`
`⇔10x-20=0`
`⇔10x=20`
`⇔(25x^2-25x^2)+10x+(1+9-30)=0`
`⇔10x-20=0`
`⇔x=2`
`b,` `(6x-2)^2+(5x-2)^2-4(3x-1)(5x-2)=0`
`⇔(6x-2)^2-2(6x-2)(5x-2)+(5x-2)^2=0`
`⇔[(6x-2)-(5x-2)]^2=0`
`⇔6x-2-5x+2=0`
`⇔x=0`