bài 3 :tìm x a,4^x+4^x+1=320 b,5^x.5^3=-(15-40) c,2x^2-15=1 d,1762-(4X+3)=-5 e,(x^2+1)(x^2-1)=0

0 bình luận về “bài 3 :tìm x a,4^x+4^x+1=320 b,5^x.5^3=-(15-40) c,2x^2-15=1 d,1762-(4X+3)=-5 e,(x^2+1)(x^2-1)=0”

1. a, $4^x+4^{x+1}=320$

   $=>4^x.(1+4)=320$

   $=>4^x.5=320$

   $=>4^x=64$

   $=>4^x=4^3$

   $=>x=3$

   b, $5^x.5^3=-(15-40)$

   $=>5^x.125=25$

   $=>5^x=$$\frac{1}{5}$ 

   $=>5^x=5^{-1}$

   $=>x=-1$

   c, $2x^2-15=1$

   $=>2x^2=16$

   $=>x^2=4$

   $=>x^2=2^2$

   $=>x=2$

   d, $1762-(4x+3)=-5$

   $=>4x+3=1767$

   $=>4x=1764$

   $=>x=441$

   e, $(x^1+1)(x^2-1=0$

   $=>x^2-1=0$

   $=>x^2=1$

   $=>$\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\) 

   Vậy x=1, x=-1

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