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\(\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\sqrt{\frac{k^2\left(k+1\right)^2+\left(k+1\right)^2+k^2}{k^2\left(k+1\right)^2}}=\sqrt{\frac{k^2\left(k+1\right)^2+2k\left(k+1\right)+1}{k^2\left(k+1\right)^2}}\)
\(=\sqrt{\frac{\left[k\left(k+1\right)+1\right]^2}{k^2\left(k+1\right)^2}}=\frac{k\left(k+1\right)+1}{k\left(k+1\right)}=1+\frac{1}{k\left(k+1\right)}\)
\(\Rightarrow A=1+\frac{1}{2.3}+1+\frac{1}{3.4}+...+1+\frac{1}{k\left(k+1\right)}\)
\(=k-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{k}-\frac{1}{k+1}\)
\(=k-1+\frac{1}{2}-\frac{1}{k+1}=...\)
mẫu các phân số này có dạng a4 + 4 = a4 + 4a2 + 4 - 4a2 = (a2 - 2a + 2)(a2 + 2a + 2)
do đó các phân số sẽ biến đổi như sau:
\(\frac{a}{4+a^4}=\frac{a}{\left(a^2-2a+2\right)\left(a^2+2a+2\right)}=\frac{1}{4}\frac{4a}{\left(a^2-2a+2\right)\left(a^2+2a+2\right)}\)
\(=\frac{1}{4}\left(\frac{1}{a^2-2a+2}-\frac{1}{a^2+2a+2}\right)\)
do đó biểu thức M = \(\frac{1}{4}\left(\frac{1}{1}-\frac{1}{\left(2n-1\right)^2+2\left(2n-1\right)+2}\right)=\frac{n^2}{4n^2+1}\)
a) ĐK : \(x\ne1;x\ne2;x\ne3\)
\(K=\left(\frac{x^2}{x^2-5x+6}+\frac{x^2}{x^2-3x+2}\right).\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(\Leftrightarrow K=\left(\frac{x^2}{\left(x-3\right)\left(x-2\right)}+\frac{x^2}{\left(x-2\right)\left(x-1\right)}\right).\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(\Leftrightarrow K=\left(\frac{2x^2}{\left(x-1\right)\left(x-3\right)}\right).\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(\Leftrightarrow K=\frac{2x^2}{x^4+x^2+1}\)
a, \(K=\left(\frac{x^2}{x^2-5x+6}+\frac{x^2}{x^2-3x+2}\right).\frac{\left(x-1\right)\left(x-2\right)}{x^4+x^2+1}\)
\(=\left(\frac{x^2}{\left(x-3\right)\left(x-2\right)}+\frac{x^2}{\left(x-2\right)\left(x-1\right)}\right).\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(=\left(\frac{x^2\left(x-1\right)+x^2\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\right).\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(=\frac{x^3-x^2+x^3-3x^2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}.\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(=\frac{2x^3-4x^2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}.\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(=\frac{2x^3-4x^2}{\left(x-2\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2x^2\left(x-2\right)}{\left(x-2\right)\left(x^4+x^2+1\right)}\)
\(=\frac{2x^2}{x^4+x^2+1}\)
ở mẫu n4+n2+1=(n2+n+1)(n2-n+1)
\(\frac{2n}{n^4+n^2+1}=\frac{\left(n^2+n+1\right)-\left(n^2-2+1\right)}{\left(n^2-n+1\right)\left(n^2+n+1\right)}\)
\(\frac{4k}{4k^4+1}=\frac{4k}{4k^4+4k^2+1-4k^2}=\frac{4k}{\left(2k^2+1\right)^2-\left(2k\right)^2}=\frac{4k}{\left(2k^2+2k+1\right)\left(2k^2-2k+1\right)}=\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1}\)
\(=\frac{1}{2k\left(k-1\right)+1}-\frac{1}{2k\left(k+1\right)+1}\)
\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+...+\frac{1}{2k\left(k-1\right)+1}-\frac{1}{2k\left(k+1\right)+1}\)
\(=1-\frac{1}{2k\left(k+1\right)+1}=...\)