=>x^2-6x+9=\(\frac{4\left(4^{16}+16\right)}{4^{16}+16}\)
=>x^2-6x+9=4

=>x^2-6x+5=0

=>(x^2-x)-(5x-5)=0

=>x(x-1)-5(x-1)=0

=>(x-1)(x-5)=0

=>\(\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}}\)

=>\(\orbr{\begin{cases}x=1\\x=5\end{cases}}\)

vậy x=1;x=5

sao khó quá!!!!

khó quá tui ko biết lớp 7 à

64*16+ 81*4 + 17*16

= 256 . 4 + 81 . 4 + 68 . 4

= 4 ( 256 + 81 + 68 )

= 4 . 405 = 1620

    64 . 16 + 81 . 4 + 17 . 16

= 16 ( 64 + 17 ) + 81 . 4

= 16 . 81 + 81 . 4

= 81 ( 16 + 4 )

= 81 . 20

= 1620

Hok tốt

64x16+81x4+17x16

=(64x16+17x16)+81x4

=16(64+17)+81x4

=16x81+81x4

=81x(16+4)

=81x20

=1620

4^2 . 120 - 64 . 17 + 16 . 34

=16 . 120 - 64 . 17 + 16 . 34

=16.(120+34)-64.17

=16.154-64.17

=2464-1088

=1376

ta có : \(\frac{4^{17}+64}{4^{16}+16}=\frac{2^{34}+2^6}{2^{32}+2^4}=\frac{2^6\left(2^{28}+1\right)}{2^4\left(2^{28}+1\right)}=2^2=4\)

Ta có: \(\frac{4^{17}+64}{4^{16}+16}\) = \(\frac{2^{34}+2^6}{2^{32}+2^4}\) = \(\frac{2^6\left(2^{28}+1\right)}{2^4\left(2^{28}+1\right)}\) = \(\frac{2^6}{2^4}\) = \(\frac{64}{16}\) = 4

( 4¹⁷ + 64 ) : ( 4¹⁶ + 16 )

= ( 4¹⁷ + 4³ ) : (4¹⁶ + 4²)

= 4²⁰ : 4¹⁸

= 4²

đổi 4² = 16

 Đáp số :.........