Đã cố gắng hết sức:(

3.3 a) x²

A)x²-(x-1)²=15

=>(x+x-1)(x-x+1)=15

=>(2x-1)1=15

=>2x-1=15

=>2x=16

=>x=8

B)16x-(4x-5)=15

=>(4x)-(4x-5)=15

=>(4x+4x-5)(4x-4x+5)=15

=>(8x-5)5=15

=>8x-5=3

=>8x=8

=>x=1

Chất adiponitrile cthh: C6H8N2:)

Bạn cho mik xin file nghe dứ= ))

bài ni làm sao vậy bạn hè..

1. True

2. True

3. False

4. False

1 T

2 F

3 F

4 F

P = 1+ \(\dfrac{x+3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x^2-4\right)}-\dfrac{1}{x+2}\right)\)

    = 1+ \(\dfrac{1}{x+2}:\left(\dfrac{2}{x-2}-\dfrac{x}{\left(x+2\right)\left(x-2\right)}-\dfrac{1}{x+2}\right)\)

    =1+\(\dfrac{1}{x+2}:\left(\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x}{\left(x+2\right)\left(x-2\right)}-\dfrac{1.\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right)\)

   = 1 + \(\dfrac{1}{x+2}:\left(\dfrac{2x+4-x-x+2}{\left(x+2\right)\left(x-2\right)}\right)\)

   =1 + \(\dfrac{1}{x+2}\): (\(\dfrac{6}{\left(x+2\right)\left(x-2\right)}\))

   =1 + \(\dfrac{1}{x+2}\).\(\dfrac{\left(x+2\right)\left(x-2\right)}{6}\)

   = 1 + \(\dfrac{x-2}{6}\)

  = \(\dfrac{6+x-2}{6}\)

= \(\dfrac{x+4}{6}\)

1. 5.(x+2)-2x.(x+2)=0

(x+2).(5-2x) = 0

+) TH1: x+2=0

=> x=-2

+) TH2: 5-2x=0

=>2x=5

=>x=5/2.

2. (x-1)^2 - 25=0

=> (x-1)^2 = 25

=> (x-1)^2 = 5^2

=> x-1 = 5 

=> x=6.

mn giúp mik vs , pls 

Bạn tham khảo nhé.

Đặt x - 2 = t
=> t^4 + (t - 1)^4 = 1
<=> t^4 + t^4 - 4t^3 + 6t² - 4t + 1 - 1 = 0
<=> 2t^4 - 4t^3 + 6t² - 4t = 0
<=> t(2t^3 - 4t² + 6t - 4) = 0
<=> t( 2t^3 - 2t² + 4t - 2t² + 2t - 4 ) = 0
<=> t[t(2t² - 2t + 4) - 1(2t² - 2t + 4)] = 0
<=> t(t - 1)(2t² - 2t + 4) = 0
=> t = 0
=> t - 1 = 0
=> 2t² - 2t + 4 = 0

=> t = 0
=> t = 1
=> Không có nghiệm

=> x - 2 = 0
=> x - 2 = 1

=> x = 2
=> x = 3

xl bạn mình dốt vật lí lắm