\(\left.\left(\right.x-1\right)^2=\frac{1}{4}\)
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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)
\(=\frac{1}{n+1}\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=115\)
Xét số hạng tổng quát:
\(k^4+\frac{1}{4}=\left(k^4+2\cdot\frac{1}{2}\cdot k^2+\frac{1}{4}\right)-k^2\)=\(\left(k^2+\frac{1}{2}\right)^2-k^2\)
= \(\left(k^2+\frac{1}{2}-k\right)\left(k^2+\frac{1}{2}+k\right)\)
Thay k từ 1 đến 12 ta được:
A=\(\frac{\frac{1}{2}\cdot\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(110+\frac{1}{2}\right)\left(132+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)...\left(132+\frac{1}{2}\right)\left(152+\frac{1}{2}\right)}\)=\(\frac{\frac{1}{2}}{152+\frac{1}{2}}=\frac{1}{305}\)
(\(x-1\))\(^2\) = \(\frac14\)
(\(x-1\))\(^2\) = (\(\frac12\))\(^2\)
\(\left[\begin{array}{l}x-1=\frac12\\ x-1=-\frac12\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12+1\\ x=-\frac12+1\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac32\\ x=\frac12\end{array}\right.\)
Vậy \(x\) ∈ {\(\frac12;\frac32\)}
\(\left(x-1\right)^2=\frac14\)
\(\left(x-1\right)^2=\) \((\frac12)^2\)
\(x-1=\frac12\)
\(x=\frac12+1\)
\(x=\frac32\)