tìm x 1/2x^2-x=0 (x+1)(4x-20)=0 1/3+2/3x=5/6 (x-1)^2021=(x-1)^2019

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1. Đáp án + giải thích bước giải :

   $1/$ `1/2x^2 – x = 0`

   `-> 1/2x . x – x = 0`

   `-> x (1/2x – 1) = 0`

   `-> x = 0` hoặc `1/2x – 1 = 0`

   `-> x=  0` hoặc `x = 2`

   $2/$ `(x + 1) (4x – 20) = 0`

   `-> x + 1 = 0` hoặc `4x – 20 = 0`

   `-> x = -1` hoặc `x = 5`

   $3/$ `1/3 + 2/3x = 5/6`

   `-> 2/3x = 1/2`

   `-> x = 3/4`

   $4/$ `(x – 1)^{2021} = (x – 1)^{2019}`

   `-> (x – 1)^{2021} – (x – 1)^{2019} = 0`

   `-> (x – 1)^{2019} (x – 1 – 1) (x – 1+ 1) = 0`

   `-> x (x – 1) (x – 2) = 0`

   `-> x=  0` hoặc `x – 1 = 0` hoặc `x – 2 = 0`

   `-> x=  0` hoặc `x = 1` hoặc `x = 2`

    

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2. $#Dino$

   `2x²-x=0`

   `⇒x(2x-1)=0`

   `1) x=0`

   `2) 2x-1=0⇒x=1/2`

   Vậy…………………..

   `(x+1)(4x-20)=0`

   `1) x+1=0⇒x=-1`

   `2) 4x-20=0⇒x=5`

   Vậy…………………

   `1/3+2/3x=5/6`

   `⇒2/3x=5/6-1/3=1/2`

   `⇒x=1/2:2/3`

   `⇒x=3/4`

   Vậy……………..

   `(x-1)^2021=(x-1)^2019`

   `⇒(x-1)^2021-(x-1)^2019=0`

   `⇒(x-1)^2019*[(x-1)²-1)=0`

   `⇔(x-1)^2019*(x-1-1)(x-1+1)=0`

   `⇔x(x-1)(x-2)=0`

   `1) x=0`

   `2) x-1=0⇒x=1`

   `3) x-2=0⇒x=2`

   Vậy………….

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