Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{3^2-1}{5^2-1}:\frac{9^2-1}{7^2-1}:\frac{13^2-1}{11^2-1}:...:\frac{55^2-1}{53^2-1}\)
\(=\frac{\left(3-1\right)\left(3+1\right)}{\left(5-1\right)\left(5+1\right)}:\frac{\left(9-1\right)\left(9+1\right)}{\left(7-1\right)\left(7+1\right)}:\frac{\left(13-1\right)\left(13+1\right)}{\left(11-1\right)\left(11+1\right)}:...:\frac{\left(55-1\right)\left(55+1\right)}{\left(53-1\right)\left(53+1\right)}\)
\(=\frac{2.4}{4.6}:\frac{8.10}{6.8}:\frac{12.14}{10.12}:...:\frac{54.56}{52.54}\)
\(=\frac{2.4.6.8.10.12......52.54}{4.6.8.10.12.....54.56}\)
\(=\frac{2}{56}\)
\(=\frac{1}{28}\)
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\) ( có lẽ đề như này )
\(\Leftrightarrow\frac{x-1}{2013}-1+\frac{x-2}{2012}-1+\frac{x-3}{2011}-1=\frac{x-4}{2010}-1+\frac{x-5}{2009}-1+\frac{x-6}{2008}-1\)
\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2014}{2011}-\frac{x-2014}{2010}-\frac{x-2014}{2009}-\frac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Leftrightarrow x-2014=0\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\right)\)
\(\Leftrightarrow x=2014\)
...
Ta có : \(x^2+9x+20=x^2+4x+5x+20=\left(x+4\right)\left(x+5\right)\)
\(x^2+11x+30=x^2+5x+6x+30=\left(x+5\right)\left(x+6\right)\)
\(x^2+13x+42=x^2+6x+7x+42=\left(x+6\right)\left(x+7\right)\)
\(\Rightarrow Pt\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\) (*)\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)
(*) \(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow3.18=x^2+4x+7x+28\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)
\(\frac{1}{4}.\frac{2}{6}............\frac{31}{64}=2^x\)
\(\Rightarrow\frac{1.2........31}{2.2.2.3...........2.31.64}=2^x\)
\(\Rightarrow\frac{1}{2^{30}.2^4}=2^x\)
\(\Rightarrow\frac{1}{2^{34}}=2^x\)
\(\Rightarrow x=-34\)
S=\(3\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(S=3\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{5050}\right)\)
\(S=3.\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{10100}\right)\)
\(S=\frac{3}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)
\(S=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(S=\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(S=\frac{3}{2}.\frac{100}{101}=\frac{150}{101}\)
Lời giải:
\(B=\frac{1}{11}.2\frac{30}{31}+3\frac{1}{55}.\frac{1}{31}-\frac{3}{11}+\frac{6}{55.31}\)
\(=\frac{1}{11}(2+\frac{30}{31})+(3+\frac{1}{55}).\frac{1}{31}-\frac{3}{11}+\frac{6}{55.31}\)
\(=\frac{2}{11}+\frac{30}{11.31}+\frac{3}{31}+\frac{1}{55.31}-\frac{3}{11}+\frac{6}{55.31}\)
\(=(\frac{2}{11}-\frac{3}{11})+\frac{31-1}{11.31}+\frac{3}{31}+(\frac{1}{55.31}-\frac{6}{55.31})\)
\(=\frac{-1}{11}+\frac{1}{11}-\frac{1}{11.31}+\frac{3}{31}-\frac{1}{11.31}\)
\(=\frac{-2}{11.31}+\frac{3}{31}=\frac{-2}{11.31}+\frac{33}{11.31}=\frac{33-2}{11.31}=\frac{31}{11.31}=\frac{1}{11}\)
Thanks bạn nhiều nhé!