giải phương trình
1/x.(x-1) + 1/x^2-3x+2 +1/x^2-5x+6 + 1/x^2-7x+12 + 1/x^2-9x+20=2 – 1/x
ai giỏi toan 8 giúp vs ạ
giải phương trình
1/x.(x-1) + 1/x^2-3x+2 +1/x^2-5x+6 + 1/x^2-7x+12 + 1/x^2-9x+20=2 – 1/x
ai giỏi toan 8 giúp vs ạ
Đáp án:
\[x = \frac{{11}}{2}\]
Giải thích các bước giải:
Tổng quát:
\(\frac{1}{{n\left( {n + 1} \right)}} = \frac{{\left( {n + 1} \right) – n}}{{n\left( {n + 1} \right)}} = \frac{1}{n} – \frac{1}{{n + 1}}\)
Áp dụng ta có:
\(\begin{array}{l}
\frac{1}{{x\left( {x – 1} \right)}} + \frac{1}{{{x^2} – 3x + 2}} + \frac{1}{{{x^2} – 5x + 6}} + \frac{1}{{{x^2} – 7x + 12}} + \frac{1}{{{x^2} – 9x + 20}} = 2 – \frac{1}{x}\\
\Leftrightarrow \frac{1}{{\left( {x – 1} \right)x}} + \frac{1}{{\left( {x – 2} \right)\left( {x – 1} \right)}} + \frac{1}{{\left( {x – 3} \right)\left( {x – 2} \right)}} + \frac{1}{{\left( {x – 4} \right)\left( {x – 3} \right)}} + \frac{1}{{\left( {x – 5} \right)\left( {x – 4} \right)}} = 2 – \frac{1}{x}\\
\Leftrightarrow \frac{1}{{x – 1}} – \frac{1}{x} + \frac{1}{{x – 2}} – \frac{1}{{x – 1}} + \frac{1}{{x – 3}} – \frac{1}{{x – 2}} + \frac{1}{{x – 4}} – \frac{1}{{x – 3}} + \frac{1}{{x – 5}} – \frac{1}{{x – 4}} = 2 – \frac{1}{x}\\
\Leftrightarrow \frac{1}{{x – 5}} – \frac{1}{x} = 2 – \frac{1}{x}\\
\Leftrightarrow \frac{1}{{x – 5}} = 2\\
\Leftrightarrow x – 5 = \frac{1}{2}\\
\Leftrightarrow x = \frac{{11}}{2}
\end{array}\)