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LN
1
T
6 tháng 9 2015
\(y=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(y=\frac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(y=\frac{2^9.3^9\left(2^3.3+120\right)}{2^{11}.3^{11}\left(2.3+1\right)}\)
\(y=\frac{6^9\left(2^3.3+120\right)}{6^{11}.7}\)
\(y=\frac{2^3.3+120}{6^2.7}\)
\(y=\frac{144}{252}\)
\(y=\frac{4}{7}\)
PM
19 tháng 10 2016
B=\(\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2016}{2017}\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2016}{2017}\)
\(=\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\right):\frac{2016}{2017}\)
\(=\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right):\frac{2016}{2017}=\left(\frac{2}{7}-\frac{2}{7}\right):\frac{2016}{2017}=0\)
11 tháng 7 2017
Câu A
X + (X+1) + (X+3) +...+ (X+2003) = 2004
Số số hạng trong tổng 1 + 3 + ... + 2003 là
(2003 - 1) : 2 + 1 = 1002
Tổng dãy 1 + 3 + ... + 2003 là:
(1 + 2003) * 1002 : 2 = 1004004
=> (1003.X) + 1004004 = 2004
=> (1003.X)= 2004 - 1004004
=> 1003.X = - 1002000
X = - 1002000/1003
E chỉ giải đc đến đây thui!!!!!!!!!!!!!!! :)))
11 tháng 7 2017
x + ( x + 1) + (x + 3) ... + (x + 2003) = 2004
x + x + x + ... + x (có 1003 x) + 1 + 3 + 5 + ... + 2003 = 2004
x . 1003 + 1004004 = 2004
x . 1003 = 2004 - 1004004
x . 1003 = -1002000
x = -1002000 : 1003
x = -999,00299 = ~-999
14 tháng 12 2017
Ta có: \(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+.....+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+....+\frac{1}{x+9}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{x+11-x+1}{\left(x+1\right)\left(x+11\right)}=\frac{12}{\left(x+1\right)\left(x+11\right)}\)
VN
0
29 tháng 11 2016
\(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(=\frac{1+1+1+1+1}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+11\right)\left(x+3\right)\left(x+9\right)\left(x+5\right)\left(x+7\right)}\)
\(=\frac{5}{\left(x^2+11x+x+11\right)\left(x^2+9x+3x+27\right)\left(x^2+7x+5x+35\right)}\)
\(=\frac{5}{\left(x^2+12x+11\right)\left(x^2+12x+27\right)\left(x^2+12x+35\right)}\)
29 tháng 11 2016
A=\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+11}\)
Rút gọn hết đi ta có \(\frac{1}{x+1}-\frac{1}{x+11}\)=\(\frac{x+11}{\left(x+1\right).\left(x+11\right)}-\frac{x+1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{x+11-x-1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{10}{x^2+12x+11}\)
C0
2
1 tháng 4 2018
Ta có :
\(\frac{x-5}{2009}+\frac{x-7}{2007}=\frac{x-9}{2005}+\frac{x-11}{2003}\)
\(\Leftrightarrow\)\(\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)=\left(\frac{x-9}{2005}-1\right)+\left(\frac{x-11}{2003}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-2014}{2009}+\frac{x-2014}{2007}=\frac{x-2014}{2005}+\frac{x-2014}{2003}\)
\(\Leftrightarrow\)\(\frac{x-2014}{2009}+\frac{x-2014}{2007}-\frac{x-2014}{2005}-\frac{x-2014}{2003}=0\)
\(\Leftrightarrow\)\(\left(x-2014\right)\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\)
Nên \(x-2014=0\)
\(\Rightarrow\)\(x=2014\)
Vậy \(x=2014\)
Chúc bạn học tốt ~
1 tháng 4 2018
\(\frac{x-5}{2009}+\frac{x-7}{2007}=\frac{x-9}{2005}+\frac{x-11}{2003}\)
Trừ cả 2 vế cho 2 ta được :
\(\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)=\left(\frac{x-9}{2005}-1\right)+\left(\frac{x-11}{2003}-1\right)\)
\(\Leftrightarrow\frac{x-2014}{2009}+\frac{x-2014}{2007}=\frac{x-2014}{2005}+\frac{x-2014}{2003}\)
\(\Leftrightarrow\frac{x-2014}{2009}+\frac{x-2014}{2007}-\frac{x-2014}{2005}-\frac{x-2014}{2003}=0\)
\(\Leftrightarrow\left(x-2014\right)\times\left(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\right)=0\)
Mà : \(\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{2003}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
\(\frac{12.3^{11}+5.3^{12}}{9^7}\)
\(=\frac{3.4.3^{11}+5.3^{12}}{\left(3^2\right)^7}\)
\(=\frac{3^{12}.4+5.3^{12}}{3^{14}}\)
\(=\frac{3^{12}\left(4+5\right)}{3^{14}}=\frac{9}{9}=1\)
\(\frac{12.3^{11}+5.3^{12}}{9^7}\)
\(=\frac{4.3.3^{11}+5.3^{12}}{\left(3^2\right)^7}\)
\(=\frac{4.3^{12}+5.3^{12}}{3^{14}}\)
\(=\frac{\left(4+5\right).3^{12}}{3^{14}}\)
\(=\frac{9.1}{3^2}=1\)