a: ĐKXĐ: x∉{2;-2;3}

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\right):\frac{16}{x^2-x-6}\)

\(=\left(\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\frac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right):\frac{16}{\left(x-3\right)\left(x+2\right)}\)

\(=\left(\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\frac{\left(x-3\right)\cdot\left(x+2\right)}{16}\)

\(=\frac{4x\left(x+2\right)-4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\cdot\frac{\left(x-3\right)\left(x+2\right)}{16}=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)}\cdot\frac{x-3}{16}\)

\(=\frac{-16\left(x-3\right)}{16\left(x+2\right)}=\frac{-x+3}{x+2}\)

b: A<0

=>\(\frac{-x+3}{x+2}<0\)

=>\(\frac{x-3}{x+2}>0\)

=>x>3 hoặc x<-2

c: A>=5

=>\(\frac{-x+3}{x+2}-5\ge0\)

=>\(\frac{-x+3-5x-10}{x+2}\ge0\)

=>\(\frac{-6x-7}{x+2}\ge0\)

=>\(\frac{6x+7}{x+2}\le0\)

=>-2<x<=-7/6

bạn ơi!! 2⁸ +1 chứ, bn xem lại đề cấy

A=\(2^{n-1}+2.2^n+3-8.2^{n-4}-16.2^n=\)\(\frac{2^n}{2}+2.2^n-8.\frac{2^n}{2^4}-16.2^n+3\)

=\(2^n\left(\frac{1}{2}+2-\frac{8}{16}-16\right)+3\)=\(-14.2^n+3\)

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

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

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

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

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

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

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

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

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

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

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

bạn có chơi tok không?

Không bít

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

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

Tưng tự 

= \(\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}=\frac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\frac{32}{1-x^{32}}\)