tìm x lớp 6 nha $\frac{3.x}{2.5}$+$\frac{3.x}{8.11}$+$\frac{3.x}{11.14}$=$\frac{1}{21}$

0 bình luận về “tìm x lớp 6 nha $\frac{3.x}{2.5}$+$\frac{3.x}{8.11}$+$\frac{3.x}{11.14}$=$\frac{1}{21}$”

1. $\dfrac{3x}{2.5}+\dfrac{3}{8.11}+\dfrac{3.x}{11.14}=\dfrac{1}{21}$

   $ ⇔ x.(\dfrac{1}{2}-\dfrac{1}{14}) = \dfrac{1}{21}$

   $⇔ x . \dfrac{3}{7} = \dfrac{1}{21}$

   $⇔ x= \dfrac{1}{9}$

    

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2. `(3x)/2.5+(3x)/8.11+(3x)/11.14=1/21`

   `=> 3x.(1/2.5+1/8.11+1/11.14)=1/21`

   `=>3x.(1/10 + 1/88 + 1/154)=1/21“

   `=>3x.(308/3080+35/3080+20/3080)=1/21`

   `=>3x. 33/280=1/21`

   `⇒ x = 1/21 : 33/280 :3`

   `⇒ x = 1/21 . 280/33 . 1/3`

   `⇒ x = 40/297`

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