d) 2-x/2016-1=1-x/2017-x/2018 e) x²+6x+9=144 f) x³-3x²+4=0

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1. `d. 2-x/2016-1=1-x/2017-x/2018`

   `⇔2-x/2016-1+2=1-x/2017+1-x/2018+1`

   `⇔2018-x/2016=2018-x/2017-x+2018/2018`

   `⇔(2018-x)(1/2016-1/2017+1/2018)=0`

   `⇔2018-x=0 (vì 1/2016-1/2017+1/2018` $\neq$ `0)`

   `⇔x=2018`

   `e.x²+6x+9=144`

   `<=> (x-3)^2 =144`

   `<=> x-3=±12`

   `+) x-3=12`

   `=> x=15`

   `+) x-3=-12`

   `=>x=-9`

   `<=> x=15` hoặc `x=-9`

   `f. x^3 – 3x^2 + 4 =0`

   `<=> (x^3 + 1) – (3x^2 – 3 ) =0`

   `<=> (x+1)(x^2 – x +1 ) – 3 (x-1)(x +1) =0`

   `<=> (x +1)(x^2 – 4x +4) =0`

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

   `<=> x =-1` hoặc `x =2`

   $army×blink$

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2. d)2-x/2016-1=1-x/2017-x/2018

   ⇔2-x/2016-1+2=1-x/2017+1-x/2018+1

   ⇔2018-x/2016=2018-x/2017-x+2018/2018

   ⇔(2018-x)(1/2016-1/2017+1/2018)=0

   ⇔2018-x=0 (vì 1/2016-1/2017+1/2018 khác 0)

   ⇔x=2018

   Vậy S={2018}

   e)x²+6x+9=144

   ⇔(x+3)²=144

   ⇔x+3=12 hoặc x+3=-12

   ⇔x=9                 x=-15

   Vậy S={-15;9}

   f)x³-3x²+4=0

   ⇔x³-2x²-x²+4=0

   ⇔x²(x-2)-(x-2)(x+2)=0

   ⇔(x-2)(x²-x-2)=0

   ⇔(x-2)(x²+x-2x-2)=0

   ⇔(x-2)[x(x+1)-2(x+1)]=0

   ⇔(x-2)²(x+1)=0

   ⇔x-2=0 hoặc x+1=0

   ⇔x=2             x=-1

   Vậy S={-1; 2}

   Chúc bn hc tốt, cho mk 5 sao và ctlhn nha^^

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