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\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\) .Trừ 1 ở mỗi hạng tử của 2 vế ,ta có :
\(\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)\left(x-11\right)=\left(\frac{1}{14}+\frac{1}{15}\right)\left(x-11\right)\)
\(\Rightarrow\left[\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\right]\left(x-11\right)=0\)
\(\frac{1}{12}>\frac{1}{14};\frac{1}{13}>\frac{1}{15}\Rightarrow\frac{1}{12}+\frac{1}{13}>\frac{1}{14}+\frac{1}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\)
\(\Leftrightarrow\frac{x+1}{12}-1+\frac{x+2}{13}-1=\frac{x+3}{14}-1+\frac{x+4}{15}-1\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}-\frac{x-11}{14}-\frac{x-11}{15}=0\)
\(\Leftrightarrow\left(x-11\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
Mà: \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
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Đk:\(x\ne0;1;2;3;4\)
\(pt\Leftrightarrow\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}=2-\frac{1}{4-x}\)
\(\Leftrightarrow\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x}=2-\frac{1}{4-x}\)
\(\Leftrightarrow\frac{1}{x-4}-\frac{1}{x}=2-\frac{1}{4-x}\)\(\Leftrightarrow\frac{4}{x\left(x-4\right)}=\frac{2x-7}{x-4}\)
Dễ thấy \(x\ne4\) nên nhân 2 vế của pt vừa biến đổi với \(x-4\) ta dc:
\(\Leftrightarrow\frac{4}{x}=2x-7\Leftrightarrow x\left(2x-7\right)=4\)
\(\Leftrightarrow2x^2-7x=4\Leftrightarrow2x^2-7x-4=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x+1\right)=0\)\(\Leftrightarrow x=-\frac{1}{2}\left(x\ne4\right)\)
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a) \(\left(x+7\right)^2-x\left(x-3\right)=12\)
\(\Leftrightarrow x^2+14x+49-x^2+3x=12\)
\(\Leftrightarrow17x=-37\)
\(\Leftrightarrow x=-\frac{37}{17}\)
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c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)
Nên x + 2009 = 0 => x = -2009
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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!!!!!!!!!!!!!!! :)))
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
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\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}\)
\(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
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\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(-\frac{1}{x}+\frac{1}{x-4}=\frac{1}{x-4}\)
\(\Leftrightarrow\)\(\frac{-\left(x-4\right)+x}{x\left(x-4\right)}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(4-x+x=x\)
\(\Leftrightarrow x=4\)
lo nói mk làm cách lâu chứ m cx hỏi người khác!!!!!!!!!!!
Ta có : \(\frac{x}{x^2+1}=\frac{1}{12}\)
=> 12x = x2 + 1
=> x2 - 12x + 1 = 0
=> x2 - 12x + 36 - 35 = 0
=> (x - 6)2 = 35
\(\Leftrightarrow\orbr{\begin{cases}x-6=\sqrt{35}\\x-6=-\sqrt{35}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6+\sqrt{35}\\x=6-\sqrt{35}\end{cases}}\)
\(\frac{x}{x^2+1}=\frac{1}{12}\Leftrightarrow x^2+1=12x\Leftrightarrow x^2-12x+1=0.\)
Giải phương trình bậc hai, được hai giá trị : \(x_1=6-\sqrt{35}.\) \(x_2=6+\sqrt{35}\)