Toán Lớp 8: giải các phương trình
a, x + 1/x=x^2 + 1/x^2
b. 3x-2/x+7=6x+1/2x-3
c, 2x- 2x^2/x+3=4x/x+3 + 2/7
mình cần gấp cần giải chi tiết nha
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Toán Lớp 8: giải các phương trình
a, x + 1/x=x^2 + 1/x^2
b. 3x-2/x+7=6x+1/2x-3
c, 2x- 2x^2/x+3=4x/x+3 + 2/7
mình cần gấp cần giải chi tiết nha
Comments ( 2 )
a)Dkxd:x\# 0\\
x + \dfrac{1}{x} = {x^2} + \dfrac{1}{{{x^2}}}\\
\Leftrightarrow x + \dfrac{1}{x} = {x^2} + 2.{x^2}.\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^2}}} – 2\\
\Leftrightarrow x + \dfrac{1}{x} = {\left( {x + \dfrac{1}{x}} \right)^2} – 2\\
Đặt:x + \dfrac{1}{x} = t\\
\Leftrightarrow t = {t^2} – 2\\
\Leftrightarrow {t^2} – t – 2 = 0\\
\Leftrightarrow \left( {t – 2} \right)\left( {t + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t – 2 = 0\\
t + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{x} – 2 = 0\\
x + \dfrac{1}{x} + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} – 2x + 1 = 0\\
{x^2} + x + 1 = 0\left( {ktm} \right)
\end{array} \right.\\
\Leftrightarrow {\left( {x – 1} \right)^2} = 0\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1\\
b)Dkxd:x\# \dfrac{2}{3};x\# \dfrac{3}{2}\\
\dfrac{{3x – 2}}{{x + 7}} = \dfrac{{6x + 1}}{{2x – 3}}\\
\Leftrightarrow \left( {3x – 2} \right).\left( {2x – 3} \right) = \left( {6x + 1} \right)\left( {x + 7} \right)\\
\Leftrightarrow 6{x^2} – 9x – 4x + 6 = 6{x^2} + 42x + 7x + 7\\
\Leftrightarrow 62x = – 1\\
\Leftrightarrow x = – \dfrac{1}{{62}}\left( {tmdk} \right)\\
Vậy\,x = – \dfrac{1}{{62}}\\
c)Dkxd:x\# – 3\\
\dfrac{{2x – 2{x^2}}}{{x + 3}} = \dfrac{{4x}}{{x + 3}} + \dfrac{2}{7}\\
\Leftrightarrow \dfrac{{7\left( {2x – 2{x^2}} \right)}}{{7\left( {x + 3} \right)}} = \dfrac{{7\left( {4x + 2} \right) + 2.\left( {x + 3} \right)}}{{7\left( {x + 3} \right)}}\\
\Leftrightarrow 14x – 14{x^2} = 28x + 14 + 2x + 6\\
\Leftrightarrow 14{x^2} + 16x + 20 = 0\\
\Leftrightarrow 7{x^2} + 8x + 10 = 0\left( {ktm} \right)
\end{array}$