Giải giúp mình với ! Mình cần gấp !!! Tìm x, biết: x-15/21 + x-17/23 + x-19/25 + x-21/27 = -4

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1. Giải thích các bước giải:

   $\dfrac{x-15}{21}+\dfrac{x-17}{23}+\dfrac{x-19}{25}+\dfrac{x-21}{27}=-4$

   ⇔ $\dfrac{x-15}{21}+1+\dfrac{x-17}{23}+1+\dfrac{x-19}{25}+1+\dfrac{x-21}{27}+1=0$

   ⇔ $\dfrac{x-15+21}{21}+\dfrac{x-17+23}{23}+\dfrac{x-19+25}{25}+\dfrac{x-21+27}{27}=0$

   ⇔ $\dfrac{x+6}{21}+\dfrac{x+6}{23}+\dfrac{x+6}{25}+\dfrac{x+6}{27}=0$

   ⇔ $(x+6)(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27})=0$

   Vì $\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}) > 0$

   nên $x+6=0$

   ⇔ $x=-6$

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2. $\dfrac{x-15}{21}+\dfrac{x-17}{23}+\dfrac{x-19}{25}+\dfrac{x-21}{27}=-4$

   $\dfrac{x-15}{21}+\dfrac{x-17}{21}+\dfrac{x-19}{25}+\dfrac{x-21}{27}+4=0$

   $(\dfrac{x-15}{21}+1)+(\dfrac{x-17}{21}+1)+(\dfrac{x-19}{25}+1)+(\dfrac{x-21}{27}+1)=0$

   $\dfrac{x+6}{21}+\dfrac{x+6}{23}+\dfrac{x+6}{25}+\dfrac{x+6}{27}=0$

   $(x+6).(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27})=0$

   $⇒x+6=0$ (vì $\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}>0$)

   $⇒x=-6$

   Vậy $x=-6$ thỏa mãn đề bài

    

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