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1)
a. Để B là phân số thì:\(n-3\ne0\Leftrightarrow n\ne3\)
b. Có: \(B=\frac{n-8}{n-3}=\frac{n-3-5}{n-3}=1-\frac{5}{n-3}\)
Để B là số nguyên thì \(n-3\inƯ\left(5\right)\)
Mà: Ư(5)={1;-1;5;-5}
=> n-3={1;-1;5;-5}
Ta có bảng sau:
| n-3 | 1 | -1 | 5 | -5 |
| n | 4 | 2 | 8 | -2 |
Vậy n={-2;2;4;8} thì B nguyên
Ta có :
(x+1)/2009 + (x+2)/2008 = (x+3)/2007 + (x+4)/2006
<=> (x+1)/2009 + 1 + (x+2)/2008 + 1 = (x+3)/2007 +1 + (x+4)/2006 + 1
<=> (x+2010)/2009 + (x+2010)/2008 = (x+2010)/2007 + (x+2010)/2006
<=> (x + 2010).[ 1/2009 + 1/2008 - 1/2007 - 1/2006 ] = 0
<=> x = -2010
Ta có: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
\(Q=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+10}\)
\(Q=\frac{1}{\frac{2.\left(2+1\right)}{2}}+\frac{1}{\frac{3.\left(3+1\right)}{2}}+....+\frac{1}{\frac{10.\left(10+1\right)}{2}}\)
\(Q=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+....+\frac{1}{\frac{10.11}{2}}\)
\(Q=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)
\(\frac{1}{2}Q=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}\)
\(\frac{1}{2}Q=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
=>\(Q=\frac{9}{22}.2=\frac{9}{11}\)
\(Q=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{55}\\ \Rightarrow\frac{1}{2}Q=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}\)
Tiếp theo tự tính nhé
\(=\frac{\frac{5}{11.2}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{1}{11}+\frac{3}{2}}=\frac{5}{\frac{2}{4}}=\frac{5}{\frac{1}{2}}\)
a: \(=\dfrac{17}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{7}{2}\)
\(=\dfrac{35}{8}+\dfrac{8}{8}-\dfrac{37}{100}-\dfrac{128}{100}\)
\(=\dfrac{43}{8}-\dfrac{165}{100}=\dfrac{149}{40}\)
b: \(=\left(\dfrac{22\cdot26+3\cdot10-65}{130}\right):\left(\dfrac{4\cdot22-2\cdot26+3\cdot143}{286}\right)\)
\(=\dfrac{537}{130}\cdot\dfrac{286}{465}=\dfrac{1969}{775}\)
!Cám ơn các bạn nhiều...!
Theo đề bài ta có :
\(A=\frac{n+1}{n-1}=\frac{1}{2}\)
\(\Leftrightarrow2\left(n+1\right)=n-1\)
\(\Leftrightarrow2n+2=n-1\)
\(\Leftrightarrow2n-n=-1-2\)
\(\Rightarrow n=-3\)
Vậy với n = - 3 thì A = \(\frac{1}{2}\)
2. \(\frac{1995.1994-1}{1993.1995+1994}=\frac{1995.\left(1993+1\right)-1}{1993.1995+1994}=\frac{1995.1993+1995-1}{1993.1995+1994}=\frac{1995.1993+1994}{1993.1995+1994}\)
1. \(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}\)
\(=\frac{7-3}{3.7}+\frac{12-7}{7.12}+\frac{13-12}{12.13}+\frac{23-20}{20.23}\)
\(=\left[\frac{7}{3.7}-\frac{3}{3.7}\right]+\left[\frac{12}{7.12}-\frac{7}{7.12}\right]+\left[\frac{13}{12.13}-\frac{12}{12.13}\right]+\left[\frac{20}{13.20}-\frac{13}{13.20}\right]+\left[\frac{23}{20.23}-\frac{20}{20.23}\right]\) \(=\left[\frac{1}{3}-\frac{1}{7}\right]+\left[\frac{1}{7}-\frac{1}{12}\right]+\left[\frac{1}{12}-\frac{1}{13}\right]+\left[\frac{1}{13}-\frac{1}{20}\right]+\left[\frac{1}{20}-\frac{1}{23}\right]\) \(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{23}\) \(=\frac{1}{3}-\frac{1}{23}\\ =\frac{20}{69}\)
Có: \(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
=>a=1; b=2 ; c=3 ; d=4
\(\frac{30}{43}=\frac{1}{\frac{43}{30}}\)
\(=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}\)
\(=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
Vậy a = 1 ; b = 2 ; c = 3 ; d = 4