A

Giả sử cả quãng đường ab là S

*Người 1:

  Thời gian đi của người 1: 

   \(t_1=\dfrac{S}{\dfrac{2}{30}}+\dfrac{S}{\dfrac{2}{20}}=\dfrac{S}{24}\)

Thời gian đi của người 2:

   t₂₌\(\dfrac{S}{30}+\dfrac{S}{20}=\dfrac{S}{12}\)

Cùng quãng đường S\(\Rightarrow\)\(\dfrac{S}{24}>\dfrac{S}{12}\)

Vậy người 1 đến đích trước.

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1. did you go

3. Has lost

Pt tương đương :

a^2+b^2-2ab=(a-b) ^2

(a-b) ^2= (a-b) ^2

Đpcm

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Đổi: 18km/h=5m/s

 Gia tốc vật:  v²-v₀²=2aS \(\Leftrightarrow\) a=\(\dfrac{3^2-5^2}{2\cdot50}=-0,16\)

thời gian lên dốc: v=v₀+at\(\Leftrightarrow\) t=\(\dfrac{v-v_0}{a}=\dfrac{3-5}{-0,16}=12,5\)s

Zn, Al₂O₃, FeO, NaOH

B