(x-3)^2=(4^17+64)/(4^16+16)
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\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{256}-\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)
= \(\frac{1.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{4}\right)}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{3}{64}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{12}{256}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.3.\left(\frac{1}{4}+\frac{21}{256}\right)}{3.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}+\frac{17}{256}\right)}{\frac{1}{4}+\frac{1}{64}}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}\right)+3.\frac{17}{256}:\left(\frac{1}{4}+\frac{1}{64}\right)}{1.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(3+\frac{51}{256}:\frac{17}{64}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\left(3+\frac{3}{4}\right)+\frac{5}{8}\)
= \(\frac{1}{2}.\frac{15}{4}+\frac{5}{8}\)
= \(\frac{15}{8}+\frac{5}{8}\)
= \(\frac{5}{2}\)
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)
\(=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{111}{68}+\frac{5}{8}\)
\(=\frac{49}{34}\)
\(\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\)\(\frac{65}{64}-7\)
\(=\frac{96}{64}+\frac{80}{64}+\frac{72}{64}+\frac{66}{64}+\frac{65}{64}-7\)
\(=\frac{96+80+72+66+64}{64}-7\)
\(=\frac{378}{64}-\frac{7}{1}\)
\(=\frac{189}{32}-\frac{224}{32}\)
\(=\frac{-35}{32}\)
Bài 1:
a) =25,97+(6,54+103,46)
=25,97+110
=135,97
b)136x75+75x64
=75x(136+64)
=75x200
=15 000
c) (21/8+1/2):5/16
=(21/8+4/8)x16/5
=25/8x16/5
=10
d)3/17-4/5+14/17
=(3/17+14/17)-4/5
=1-4/5
=1/5
Bài 2:
a)720:\([41-(2x-5)]\)=120
41 - (2x-5) =720:120
41 - (2x-5) =6
2x-5 =41-6
2x-5 =35
2x =35+5
2x =40
x =40:2
x =20
b)2/3 x X +3/4=3
2/3 x X =3-3/4
2/3 x X =12/4-3/4
2/3 x X =9/4
x =9/4:2/3
x =9/4x3/2
x =27/8
c) x+0,34=1,19x1,02
x+0,34=1,2138
x =1,2138-0,34
x =0,8738
+) Ta có : \(2^{32}=\left(2^2\right)^{16}=4^{16}\)
=> 316 < 232
+) Ta có : \(64^2=\left(4^3\right)^2=4^6\\ 16^4=\left(4^2\right)^4=4^8\) \(\Rightarrow64^2< 16^4\left(4^6< 4^8\right)\)
=>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