1+1x2+3
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Phân tích các mẫu thành nhân tử sau đó nhân cả 2 vế của phương trình với 2 ta được:
Pt tương đương:
1 ( x + 1 ) ( x + 3 ) + 1 ( x + 3 ) ( x + 5 ) + 1 ( x + 5 ) ( x + 7 ) + 1 ( x + 7 ) ( x + 9 ) = 1 5
⇔ 2 ( x + 1 ) ( x + 3 ) + 2 ( x + 3 ) ( x + 5 ) + 2 ( x + 5 ) ( x + 7 ) + 2 ( x + 7 ) ( x + 9 ) = 2 5
ĐKXĐ: x ≠ -1; -3; -5; -7; -9
Khi đó:
<=> 1 x + 1 - 1 x + 3 + 1 x + 3 - 1 x + 5 + 1 x + 5 - 1 x + 7 + 1 x + 7 - 1 x + 9 = 2 5
<=> 1 x + 1 - 1 x + 9 = 2 5
<=> 1 ( x + 9 ) - 1 ( x + 1 ) ( x + 1 ) ( x + 9 ) = 2 ( x + 1 ) ( x + 9 ) 5 ( x + 1 ) ( x + 9 )
=> 5[x + 9 – (x + 1)] = 2(x + 1) (x + 9)
ó 5(x + 9 – x – 1) = 2 x 2 + 20x + 18
ó 2 x 2 + 20x – 22 = 0
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\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{996}{997}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{996}{997}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{996}{997}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{997}\)
\(\Rightarrow x+1=997\)
\(\Rightarrow x=996\)
\(\Leftrightarrow\)1-1/2+1/2-1/3+1/3-1/4+..+1/x-1/(x+1)=996/997
\(\Leftrightarrow\)1-1/(x+1)=996/997
\(\Leftrightarrow\)\(\frac{x}{x+1}\)\(=\frac{996}{997}\)
\(\Leftrightarrow x=996\)
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7
6