Question

tìm x bt (3+2 căn 2)^x+(3-2 căn 2)^x=2
Làm giúp mình nha 
mình cần gấp :)

Answer 1

Ta có:

\(\left(3+2\sqrt{2}\right)^x+\left(3-2\sqrt{2}\right)^x=\left(\sqrt{2}+1\right)^{2x}+\left(\sqrt{2}-1\right)^{2x}\)

   \(=\left(\left(\sqrt{2}+1\right)^x\right)^2+\left(\left(\sqrt{2}-1\right)^x\right)^2\)\(\ge2.\left(\sqrt{2}+1\right)^x.\left(\sqrt{2}-1\right)^x\)

     \(=2.\left(2-1\right)^x\)= 2

Dấu "=" xảy ra <=> \(\left(\sqrt{2}+1\right)^x=\left(\sqrt{2}-1\right)^x\)

\(\Rightarrow x=0\)