1.rút gọn biểu thức
\(A=\dfrac{1^2}{2^2-1}.\dfrac{3^2}{4^2-1}.\dfrac{5^2}{6^2-1}......\dfrac{n^2}{\left(n+1\right)^2-1}\)
2. rút gọn biểu thức
\(B=\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)
3. rút gọn biểu thức
\(C=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+......+\dfrac{2n+1}{[n\left(n+1\right)]^2}\)
4. cho a + b + c = 0 và (a.b.c khác 0)
rút gọn : \(D=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}\)
giúp mk vs 
A= \(\dfrac{1^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(4-1\right)\left(4+1\right)}\cdot...\cdot\dfrac{n^2}{n\left(n+2\right)}\)
= \(\dfrac{1}{1\cdot3}\cdot\dfrac{3^2}{3\cdot5}\cdot\dfrac{5^2}{5\cdot7}\cdot...\cdot\dfrac{n^2}{n\left(n+2\right)}\)
=\(\dfrac{1}{n+2}\)
B = \(\dfrac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)
= \(\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)
= \(\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)
= \(\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}=\dfrac{16}{1-x^{16}}\)
C = \(\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot7}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)
= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)
= \(1-\dfrac{1}{\left(n+1\right)^2}=\dfrac{\left(n+1\right)^2-1}{\left(n+1\right)^2}=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)
D = \(\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}\)
Ta có a + b + c = 0
=> \(\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2=\left(-c\right)^2\\\left(b+c\right)^2=\left(-a\right)^2\\\left(c+a\right)^2=\left(-b\right)^2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a^2+2ab+b^2=c^2\\b^2+2bc+c^2=a^2\\c^2+2ac+a^2=b^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2+b^2-c^2=-2ab\\b^2+c^2-a^2=-2bc\\c^2+a^2-b^2=-2ac\end{matrix}\right.\)
=> D = \(\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ca}{-2ca}=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}\)
= \(-\dfrac{3}{2}\)
