cho x*2-5x+1=0. Tính x*2+1/x*2 Question cho x*2-5x+1=0. Tính x*2+1/x*2 in progress 0 Môn Toán Helga 6 years 2020-11-21T17:07:12+00:00 2020-11-21T17:07:12+00:00 2 Answers 110 views 0
Answers ( )
x*2-5x+1 =0
=>2x-5x +1+=0
=>2x-5x = 0 -1
=>2x-5x = -1
=> -3x = -1
=> x = 1/3
Từ đó ta thay x = 1/3 để tính x*2+1/x*2 => Ta được bài toán mới:
1/3*2+1/(1/3*2)
=> 2/3+ 1/(2/3)
=> 2/3+3/2
=> 4/6+9/6 (MC:6)
=> 13/6
Đáp án:
\({x_2} + \dfrac{1}{{{x_2}}} = 5\)
Giải thích các bước giải:
\(\begin{array}{l}
Xét:\Delta = 25 – 4.1 = 21 > 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{5 + \sqrt {21} }}{2}\\
x = \dfrac{{5 – \sqrt {21} }}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{5 + \sqrt {21} }}{2} + \dfrac{2}{{5 + \sqrt {21} }}\\
{x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{5 – \sqrt {21} }}{2} + \dfrac{2}{{5 – \sqrt {21} }}
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{25 + 10\sqrt {21} + 21 + 4}}{{2\left( {5 + \sqrt {21} } \right)}}\\
{x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{25 – 10\sqrt {21} + 21 + 4}}{{2\left( {5 – \sqrt {21} } \right)}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{50 + 10\sqrt {21} }}{{2\left( {5 + \sqrt {21} } \right)}}\\
{x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{50 – 10\sqrt {21} }}{{2\left( {5 – \sqrt {21} } \right)}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{10\left( {5 + \sqrt {21} } \right)}}{{2\left( {5 + \sqrt {21} } \right)}} = 5\\
{x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{10\left( {5 – \sqrt {21} } \right)}}{{2\left( {5 – \sqrt {21} } \right)}} = 5
\end{array} \right.\\
\to {x_2} + \dfrac{1}{{{x_2}}} = 5
\end{array}\)