Bài 2 tìm x a, 3/7-1/7 : |2x-1/3|=3/14 b, x+7/2018 + x+6/2019 + x+5/2020 + x+4/2021

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1. Đáp án:

   a, `3/7 – 1/7 : |2x – 1/3| = 3/14`

   `<=> 1/7 : |2x – 1/3| = 3/7 – 3/14`

   `<=> 1/7 : |2x – 1/3| = 3/14`

   `<=> |2x – 1/3| = 1/7 : 3/14`

   `<=> |2x – 1/3| = 2/3`

   <=>  \(\left[ \begin{array}{l}2x – 1/3 = 2/3\\2x – 1/3 = -2/3\end{array} \right.\) 

   <=> \(\left[ \begin{array}{l}x= 1/2\\x=-1/6\end{array} \right.\) 

   b, Ta có : 

   `(x + 7)/2018 + (x + 6)/2019 = (x + 5)/2020 + (x + 4)/2021`

   `<=> ((x + 7)/2018 + 1) + ((x + 6)/2019  + 1) = ((x + 5)/2020  + 1) + ((x + 4)/2021 + 1)`

   `<=> (x + 2025)/2018 + (x + 2025)/2019 = (x + 2025)/2020 + (x + 2025)/2021`

   `<=> (x + 2025)/2018 + (x + 2025)/2019 –  (x + 2025)/2020 –  (x + 2025)/2021 = 0`

   `<=> (x + 2025) . (1/2018 + 1/2019 – 1/2020 – 1/2021) = 0`

   Do `1/2018 + 1/2019 – 1/2020 – 1/2021 \ne 0`

   `<=> x + 2025 = 0`

   `<=> x = -2025`

   Giải thích các bước giải:

    

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2. Đáp án:

    

   Giải thích các bước giải:

    a) `3/7-1/7:|2x-1/3|=3/14`

   `⇔1/7:|2x-1/3|=3/7-3/14`

   `⇔1/7:|2x-1/3|=3/14`

   `⇔|2x-1/3|=1/7.(14/3)`

   `⇔|2x-1/3|=2/3`

   `⇔2x-1/3=2/3 hoặc 2x-1/3=-2/3`

   `⇔2x=1 hoặc 2x=-1/3`

   `⇔x=1/2 hoặc x=-1/6`

   vậy `S={1/2;-1/6}`

   b) `(x+7)/2018+(x+6)/2019=(x+5)/2020+(x+4)/2021`

   `⇔[(x+7)/2018+1]+[(x+6)/2019+1]=[(x+5)/2020+1]+[(x+4)/2021+1]`

   `⇔(x+7+2018)/2018+(x+6+2019)/2019-(x+5+2020)/2020-(x+4+2021)/2021=0`

   `⇔(x+2025)/2018+(x+2025)/2019-(x+2025)/2020-(x+2025)/2021=0`

   `⇔(x+2025)(1/2018+1/2019-1/2020-1/2021)=0`

   do `1/2018+1/2019-1/2020-1/2021` khác 0

   `⇒x+2025=0⇔x=-2025`

   vậy `S={-2025}`

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