Giải các phương trình sau: a) (x+1)^3+(x-2)^3=(2x-1)^3 b) x+1/2019+x+2/2018=x+3/2017+x+4/2016

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1. Đáp án+Giải thích các bước giải:

   `a)(x+1)^3+(x-2)^3=(2x-1)^3` `(1)`

    Đặt:`x+1=a;x-2=b`

   `=>a+b=(x+1)+(x-2)=(x+x)+(1-2)=2x-1`

   `=>` Phương trình `(1)` trở thành:

   `a^3+b^3=(a+b)^3`

   `<=>a^3+b^3=a^3+b^3+3ab(a+b)`

   `<=>3ab(a+b)=a^3+b^3-a^3-b^3`

   `<=>3ab(a+b)=0`

    Với `a=x+1;b=x-2;a+b=2x-1`

   `=>(x+1)(x-2)(2x-1)=0`

   `TH1:x+1=0<=>x=-1`

   `TH2:x-2=0<=>x=2`

   `TH3:2x-1=0<=>x=1/2`

    Vậy `S=\{-1;2;1/2\}`

   `b)(x+1)/2019+(x+2)/2018=(x+3)/2017+(x+4)/2016`

   `<=>(x+1)/2019+(x+2)/2018+2=(x+3)/2017+(x+4)/2016+2`

   `<=>((x+1)/2019+1)+((x+2)/2018+1)=((x+3)/2017+1)+((x+4)/2016+1)`

   `<=>(x+2020)/2019+(x+2020)/2018=(x+2020)/2017+(x+2020)/2016`

   `<=>(x+2020)/2019+(x+2020)/2018-(x+2020)/2017-(x+2020)/2016=0`  

   `<=>(x+2020)(1/2019+1/2018-1/2017-1/2016)=0`

   `<=>x+2020=0(` vì `1/2019+1/2018-1/2017-1/2016\ne 0“)`

   `<=>x=-2020` 

    Vậy `S=\{-2020\}`

    

   Trả lời

2. Giải thích các bước giải:

   a.Ta có:

   $(x+1)^3+(x-2)^3=(2x-1)^3$ 

   $\to (x+1+x-2)^3-3(x+1)(x-2)(x+1+x-2)=(2x-1)^3$ 

   $\to (2x-1)^3-3(x+1)(x-2)(2x-1)=(2x-1)^3$

   $\to3(x+1)(x-2)(2x-1)=0$

   $\to(x+1)(x-2)(2x-1)=0$

   $\to x\in\{-1, 2, \dfrac12\}$

   b.Ta có:

   $\dfrac{x+1}{2019}+\dfrac{x+2}{2018}=\dfrac{x+3}{2017}+\dfrac{x+4}{2016}$

   $\to (\dfrac{x+1}{2019}+1)+(\dfrac{x+2}{2018}+1)=(\dfrac{x+3}{2017}+1)+(\dfrac{x+4}{2016}+1)$

   $\to \dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}=\dfrac{x+2020}{2017}+\dfrac{x+2020}{2016}$

   $\to \dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{2017}-\dfrac{x+2020}{2016}=0$

   $\to (x+2020)(\dfrac1{2019}+\dfrac1{2018}-\dfrac1{2017}-\dfrac1{2016})=0$

   $\to x+2020=0$

   $\to x=-2020$

   Trả lời