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Toán Lớp 8: 1 (3x-1)^2-(3x+2)(3x-2)=0 2 (2x+5)^2-(x-1)(x+1)-3x^2=0 (Giup em voi )

Home/ Toán Lớp 8: 1 (3x-1)^2-(3x+2)(3x-2)=0 2 (2x+5)^2-(x-1)(x+1)-3x^2=0 (Giup em voi )

Toán Lớp 8: 1 (3x-1)^2-(3x+2)(3x-2)=0 2 (2x+5)^2-(x-1)(x+1)-3x^2=0 (Giup em voi )

Khánh Giao Tháng Tám 13, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: 1 (3x-1)^2-(3x+2)(3x-2)=0
2 (2x+5)^2-(x-1)(x+1)-3x^2=0 (Giup em voi )

Comments ( 2 )

quynhthanhnhu
13 Tháng Tám, 2022 at 11:21 sáng

Reply

#tnvt

(3x-1)^2-(3x+2)(3x-2)=0

<=>(3x-1)^2-(3x)^2+2^2=0

<=>(3x-1-3x)(3x-1+3x)=-4

<=>-(6x-1)=-4

<=>-6x+1=-4

<=>-6x=-5

<=>x=5/6

Vậy x=5/6

(2x+5)^2-(x-1)(x+1)-3x^2=0

<=>4x^2+20x+25-x^2+1-3x^2=0

<=>20=-26

<=>x=-13/10

Vậy x=-13/10
diepbich93
13 Tháng Tám, 2022 at 11:21 sáng

Reply

1)

$(=) 9x^2 – 6x + 1 -(9x^2 – 4) =0$

$(=) 9x^2 – 6x + 1 – 9x^2 + 4 =0$

$(=) -6x=-5$

$=) x=5/6$

2)

$(=) 4x^2 + 20x + 25 – (x^2 – 1) – 3x^2 = 0$

$(=) 4x^2 + 20x + 25 – x^2 + 1 – 3x^2 = 0$

$(=) 20x = -26$

$=) x=-13/10$

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