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a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\) ĐKXĐ : x #0, x#2, x#-2
<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)
<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)
=> 10 - 2x + 7x - 14 = 4x - 4 + x
<=>-2x + 7x - 4x + x = -4 - 10 + 14
<=>x=-14
\(A=\frac{1}{x.\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\)
\(\Leftrightarrow2A=\frac{2}{x.\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{2}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\)
\(=\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\)
\(=\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\right)\)
Ta có:\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}=\frac{1}{x}-\frac{1}{x+6}=\frac{x+6}{x\left(x+6\right)}-\frac{x}{x\left(x+6\right)}=\frac{6}{x\left(x+6\right)}\)k mik nha
ĐKXĐ : \(x\ne0;-1;-2;-3;-4;-5;-6\)
Giá trị của của tổng trên rất dễ
Giá trị của nó là:
\(\frac{1}{x}-\frac{1}{x+6}\)
Tiếp
\(=\left(\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right).\left(\frac{x^2+x+1}{2x+1}\right)=\left(\frac{x^2+x+1}{x^2-1}\right)=1+\frac{x+2}{x^2-1}\)
ta có
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}\)
Ta có: \(\frac{1}{n\left(n+1\right)}=\frac{\left(n+1\right)-n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\left(n\in N\right)\)
Như vậy,
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5}{x\left(x+5\right)}-\frac{x}{x\left(x+5\right)}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}=\frac{x+6}{x.\left(x+6\right)}-\frac{x}{x.\left(x+6\right)}=\frac{6}{x^2+6x}\)