1.tìm x,y a,6 ×8^x-1+8^x+1=6 ×8^19+8^21(x ∈ N*) b,4^5+4^5+4^5/3^5+3^5+3^5 ×6^5+6^5+6^5+6^5+6^5+6^5/2^5+2^5=2^x

0 bình luận về “1.tìm x,y a,6 ×8^x-1+8^x+1=6 ×8^19+8^21(x ∈ N*) b,4^5+4^5+4^5/3^5+3^5+3^5 ×6^5+6^5+6^5+6^5+6^5+6^5/2^5+2^5=2^x”

1. Giải thích các bước giải:

    a) `6 ×8^x-1+8^x+1=6 ×8^19+8^21`

   `=>7×8^x=8^19×(6+8^2)`

   `=>7xx8^x=8^19xx70`

   `=>7xx8^x-8^19xx70=0`

   `=>8^x-8^19xx10=0`

   `=>8^x-8^x xx8^(19-x) xx10=0`

   `=>8^x xx(1-8^(19-x)xx10)=0`

   `=>`\(\left[ \begin{array}{l}8^x=0\\10×8^{19-x}=\text{1 (vì x∉N*)}\end{array} \right.\) `=>8^x=0=>x=1`

         Vậy `x=1.`

    b) `(4^5+4^5+4^5)/(3^5+3^5+3^5) ×(6^5+6^5+6^5+6^5+6^5+6^5)/(2^5+2^5)=2^x`

   `=>(3xx4^5)/(3xx3^5) ×(6xx6^5)/(2xx2^5)=2^x`

   `=>(4/3)^5×6^6/2^6=2^x`

   `=>(4/3)^5×6^6/2^6=2^x`

   `=>(4/3)^5×3^6=2^x`

   `=>(4/3)^5×3^5xx3=2^x`

   `=>4^5xx3=2^x`

   `=>1024=2^x`

   `=>x=10`

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