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\(A=\frac{\left(a+2\right)^2\left(5a-15a^2\right)}{\left(a-3\right)\left(4a-a^3\right)}=\frac{\left(a+2\right)^2.5a.\left(1-3a\right)}{\left(a-3\right).a.\left(2-a\right)\left(a+2\right)}\)
\(=\frac{\left(a+2\right).5.\left(1-3a\right)}{\left(a-3\right).\left(2-a\right)}\)
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{n^2-1}{n^2}\right)\)
\(=\text{[}\frac{\left(2-1\right)\left(2+1\right)}{2^2}\text{]}.\text{[}\frac{\left(3-1\right)\left(3+1\right)}{3^2}\text{]}.\text{[}\frac{\left(4-1\right)\left(4+1\right)}{4^2}\text{]}...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\left(\frac{1.3}{2^2}\right).\left(\frac{2.4}{3^2}\right).\left(\frac{3.5}{4^2}\right)...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\frac{\text{[}1.2.3...\left(n-1\right)\text{]}.\text{[}3.4.5...\left(n+1\right)\text{]}}{\text{[}2.3.4...n\text{]}.\text{[}2.3.4...n\text{]}}\)
\(=\frac{1}{n}.\frac{n+1}{2}\)
\(=\frac{n+1}{2n}\)
\(\frac{\left(x^2+a\right)\left(1+a\right)a^2x^2+1}{\left(x^2-a\right)\left(1-a\right)+a^2x^2+1}=\frac{x^2+x^2a+a+a^2+a^2x^2+1}{x^2-x^2a-a+a^2+a^2x^2+1}\)\(=\frac{x^2+x^2a+a^2x^2+1+a+a^2}{x^2-x^2a+a^2x^2+1-a+a^2}=\frac{x^2\left(1+a+a^2\right)+\left(1+a+a^2\right)}{x^2\left(1-a+a^2\right)+\left(1-a+a^2\right)}\)
\(=\frac{\left(x^2+1\right)\left(1+a+a^2\right)}{\left(x^2+1\right)\left(1-a+a^2\right)}=\frac{1+a+a^2}{1-a+a^2}\)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{c-b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{a-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{b-a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{c-b+b-a+a-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4\left(x^2+1\right)}{\left(2x+1\right)\left(x^2+1\right)}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4x^2+4+4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\frac{\left(2x+1\right)^2}{\left(x^2+1\right)\left(2x+1\right)}\frac{x^2+1}{x^2+2}=\frac{2x+1}{x^2+2}\)
a: \(A=25a^2+50a+25+10\left(a^2-2a-3\right)+a^2-6a+9\)
\(=26a^2+46a+34+10a^2-20a-30\)
\(=36a^2+26a+4\)
b: \(B=\dfrac{1}{4}\left(x^2-2x+1\right)+x^2-1+x^2+2x+1\)
\(=\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}+2x^2+2x\)
\(=\dfrac{9}{4}x^2+\dfrac{3}{2}x+\dfrac{1}{4}\)
$\frac{5a^2-33a+60}{a\left(a-1\right)\left(a-4\right)\left(a-5\right)}$