\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{1+x+1-x}{\left(1+x\right)\left(1-x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{4+4x^4+4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)

\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)

\(=\frac{16+16x^{16}+16-16x^{16}}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)

\(=\frac{32}{1-x^{32}}\)

Bạn kiểm tra lại đề bài nhé!

À nhầm x^2-3x+1=0

a)Xét \(\Delta\) NAM và \(\Delta\)BAC có:

\(\frac{BA}{AC}=\frac{4}{5};\frac{NA}{AM}=\frac{4}{5}\)

^A_chung

Vậy\(\Delta\)NAM đồng dạng\(\Delta\) BAC (c.g.c) 

=> đpcm

 b, Xét \(\Delta\)NAB và \(\Delta\)MAC ta có :

 \(\frac{AM}{AC}=\frac{1}{3};\frac{AN}{AB}=\frac{1}{3}\)

\(\Rightarrow\frac{AM}{AC}=\frac{AN}{AB}\)

^A_chung

Vậy \(\Delta\)NAB đồng dạng với \(\Delta\)MAC (c.g.c)

=>  ^ANB = ^AMC

=> \(\Delta\)BOM đồng dạng với \(\Delta\)COM(gg)

Vì có ^ABN = ^ACM ; ^MOB = ^NOC (đđ)

=> \(\frac{OM}{OB}=\frac{ON}{OC}\Rightarrowđpcm\)          

câu a mình ko hiểu

vì x^2>=0;x^8>=0

=> x^2+x^8>=0

mà x^2+x^8=0

=>x=0

X kia mũ 3 mừ =w=

1/M+1/BD+1/CN>1/a+1b+1/c

2,4 cm đó

x^2+2x-8

=x^2+2x+1-9

=(x+1)^2-9

=(x+1-3)(x+1+3)

=(x-2)(x+4)