Giải phương trình sau
$\frac{1}{ x^2+2x}$ + $\frac{1}{ x^2+6x+8}$ + $\frac{1}{ x^2+10x+24}$ + $\frac{1}{ x^2+14x+48}$ = $\frac{4}{105}$
Giải phương trình sau $\frac{1}{ x^2+2x}$ + $\frac{1}{ x^2+6x+8}$ + $\frac{1}{ x^2+10x+24}$ + $\frac{1}{ x^2+14x+48}$ = $\frac{4}{105}$
By Margaret
Đáp án:
\(\left[ \begin{array}{l}x=7\\x=-15\end{array} \right.\)
Giải thích các bước giải:
PT đã cho ⇔ $\frac{1}{x(x+2)}$ +$\frac{1}{(x+2)(x+4)}$ +$\frac{1}{(x+4)(x+6)}$ +$\frac{1}{(x+6)(x+8)}$ =$\frac{4}{105}$
⇔ $\frac{2}{x(x+2)}$ +$\frac{2}{(x+2)(x+4)}$ +$\frac{2}{(x+4)(x+6)}$ +$\frac{2}{(x+6)(x+8)}$ =$\frac{8}{105}$
⇔ $\frac{1}{x}$ -$\frac{1}{x+2}$ +$\frac{1}{x+2}$-$\frac{1}{x+4}$ +…-$\frac{1}{x+8}$ =$\frac{8}{105}$
⇔ $\frac{1}{x}$ -$\frac{1}{x+8}$ =$\frac{8}{105}$
⇔ $\frac{8}{x(x+8)}$ =$\frac{8}{105}$
⇔ $x^{2}$ +$8x^{}$-$105^{}$ =$0^{}$
⇔ \(\left[ \begin{array}{l}x=7\\x=-15\end{array} \right.\) .
Đáp án:
\[\left[ \begin{array}{l}
x = 7\\
x = – 15
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{1}{{{x^2} + 2x}} + \frac{1}{{{x^2} + 6x + 8}} + \frac{1}{{{x^2} + 10x + 24}} + \frac{1}{{{x^2} + 14x + 48}} = \frac{4}{{105}}\\
\Leftrightarrow \frac{1}{{x\left( {x + 2} \right)}} + \frac{1}{{\left( {x + 2} \right)\left( {x + 4} \right)}} + \frac{1}{{\left( {x + 4} \right)\left( {x + 6} \right)}} + \frac{1}{{\left( {x + 6} \right)\left( {x + 8} \right)}} = \frac{4}{{105}}\\
\Leftrightarrow \frac{2}{{x\left( {x + 2} \right)}} + \frac{2}{{\left( {x + 2} \right)\left( {x + 4} \right)}} + \frac{2}{{\left( {x + 4} \right)\left( {x + 6} \right)}} + \frac{2}{{\left( {x + 6} \right)\left( {x + 8} \right)}} = \frac{8}{{105}}\\
\Leftrightarrow \frac{{\left( {x + 2} \right) – x}}{{x\left( {x + 2} \right)}} + \frac{{\left( {x + 4} \right) – \left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x + 4} \right)}} + \frac{{\left( {x + 6} \right) – \left( {x + 4} \right)}}{{\left( {x + 4} \right)\left( {x + 6} \right)}} + \frac{{\left( {x + 8} \right) – \left( {x + 6} \right)}}{{\left( {x + 6} \right)\left( {x + 8} \right)}} = \frac{8}{{105}}\\
\Leftrightarrow \frac{1}{x} – \frac{1}{{x + 2}} + \frac{1}{{x + 2}} – \frac{1}{{x + 4}} + \frac{1}{{x + 4}} – \frac{1}{{x + 6}} + \frac{1}{{x + 6}} – \frac{1}{{x + 8}} = \frac{8}{{105}}\\
\Leftrightarrow \frac{1}{x} – \frac{1}{{x + 8}} = \frac{8}{{105}}\\
\Leftrightarrow \frac{{x + 8 – x}}{{x\left( {x + 8} \right)}} = \frac{8}{{105}}\\
\Leftrightarrow \frac{8}{{x\left( {x + 8} \right)}} = \frac{8}{{105}}\\
\Leftrightarrow x\left( {x + 8} \right) = 105\\
\Leftrightarrow {x^2} + 8x – 105 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x = – 15
\end{array} \right.
\end{array}\)