b) rút gọn còn :

\(\frac{10xy^5}{12xy}\)

\(a.=\frac{2x}{5y}\)

\(b.=\frac{5y^4}{6}\)

\(c.=\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}=\frac{x+1}{5x^2}\)

\(d.=\frac{2x\left(x+1\right)}{x+1}=2x\)

\(e.=\frac{-3\left(x-y\right)}{\left(x-y\right)}=-3\)

\(f.=\frac{-x\left(x-1\right)}{\left(x-1\right)}=-x\)

Ta có:

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

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

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

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

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

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

M = \(\frac{1}{1-x^{32}}\)

chỉ cần CM \(Q=2^{2^n}+4^n+1⋮3\) là ok 

Với n=1 thì \(Q⋮3\)

Giả sử Q vẫn chia hết cho 3 đến n=k, ta có: \(Q=2^{2^k}+4^k+1⋮3\)

Với n=k+1 thì \(Q=2^{2^k.2}+4^{k+1}+1=2^{2^k}.2^{2^k}+4^k.4+1\)

\(=\left(2^{2^k}.2^{2^k}+2^{2^k}.4^k+2^{2^k}\right)-\left(2^{2^k}.4^k+2^{2^k}-4^k.4-4\right)-3\)

\(=2^{2^k}\left(2^{2^k}+4^k+1\right)-\left(4^k+1\right)\left(2^{2^k}-4\right)-3\)

\(=2^{2^k}Q-\left(4^k+1\right)\left(4^{2^{k-1}}-1-3\right)-3⋮3\) do \(\left(4^{2^{k-1}}-1\right)⋮\left(4-1\right)=3\)

b)Ta có:  \(a^{2000}+b^{2000}=a^{2001}+b^{2001}\)

\(\Rightarrow a^{2001}+b^{2001}\)\(-a^{2000}-b^{2000}=0\)

\(\Rightarrow a^{2000}\left(a-1\right)+b^{2000}\left(b-1\right)=0\)(1)

và \(a^{2001}+b^{2001}=a^{2002}+b^{2002}\)

\(\Rightarrow a^{2002}+b^{2002}\)\(-a^{2001}-b^{2001}=0\)

\(\Rightarrow a^{2001}\left(a-1\right)+b^{2001}\left(b-1\right)=0\)(2)

Lấy (2) - (1), ta được: \(a^{2000}\left(a-1\right)^2+b^{2000}\left(b-1\right)^2=0\)(3)

Mà \(a^{2000}\left(a-1\right)^2\ge0\forall a\)và \(b^{2000}\left(b-1\right)^2\ge0\forall b\)

nên (3) xảy ra\(\Leftrightarrow\hept{\begin{cases}a^{2000}\left(a-1\right)^2=0\\b^{2000}\left(b-1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1hoaca=0\\b=1hoacb=0\end{cases}}\)

Mà a,b dương nên a = 1 và b = 1

a) Áp dụng BĐT Svac - xơ:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=9\)

(Dấu "="\(\Leftrightarrow a=b=c=\frac{1}{3}\))

\(\left(1-x\right)\frac{3}{x^3-1}\)

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

\(=\frac{-3}{x^2+x+1}\)

\(\Leftrightarrow\left(a+b+c\right).\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)

\(\Leftrightarrow\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}=a+b+c\)

\(\Leftrightarrow\frac{a^2+a.\left(b+c\right)}{b+c}+\frac{b^2+b.\left(c+a\right)}{c+a}+\frac{c^2+c.\left(a+b\right)}{a+b}=a+b+c\)

\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)

\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\left(dpcm\right)\)