Toán Lớp 8: 1, Thực hiện phép tính
a, (x+1/x-1-x-1/x+1):2x/5x-5
b, 3/x+3-x-6/x^2+3x
c, 2x^2-x/x-1 + x+1/1-x + 2-x^2/x-1
2, Tìm x biết
x^3-7x=0
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Toán Lớp 8: 1, Thực hiện phép tính
a, (x+1/x-1-x-1/x+1):2x/5x-5
b, 3/x+3-x-6/x^2+3x
c, 2x^2-x/x-1 + x+1/1-x + 2-x^2/x-1
2, Tìm x biết
x^3-7x=0
Comments ( 1 )
1)a)\left( {\dfrac{{x + 1}}{{x – 1}} – \dfrac{{x – 1}}{{x + 1}}} \right):\dfrac{{2x}}{{5x – 5}}\\
= \dfrac{{{{\left( {x + 1} \right)}^2} – {{\left( {x – 1} \right)}^2}}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{5\left( {x – 1} \right)}}{{2x}}\\
= \dfrac{{{x^2} + 2x + 1 – {x^2} + 2x – 1}}{{x + 1}}.\dfrac{5}{{2x}}\\
= \dfrac{{4x}}{{x + 1}}.\dfrac{5}{{2x}}\\
= \dfrac{{10}}{{x + 1}}\\
b)\dfrac{3}{{x + 3}} – \dfrac{{x – 6}}{{{x^2} + 3x}}\\
= \dfrac{{3.x – \left( {x – 6} \right)}}{{x\left( {x + 3} \right)}}\\
= \dfrac{{3x – x + 6}}{{x\left( {x + 3} \right)}}\\
= \dfrac{{2x + 6}}{{x\left( {x + 3} \right)}}\\
= \dfrac{{2\left( {x + 3} \right)}}{{x\left( {x + 3} \right)}}\\
= \dfrac{2}{x}\\
c)\dfrac{{2{x^2} – x}}{{x – 1}} + \dfrac{{x + 1}}{{1 – x}} + \dfrac{{2 – {x^2}}}{{x – 1}}\\
= \dfrac{{2{x^2} – x – x – 1 + 2 – {x^2}}}{{x – 1}}\\
= \dfrac{{{x^2} – 2x + 1}}{{x – 1}}\\
= \dfrac{{{{\left( {x – 1} \right)}^2}}}{{x – 1}}\\
= x – 1\\
2){x^3} – 7x = 0\\
\Leftrightarrow x\left( {{x^2} – 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 7
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \sqrt 7 \\
x = – \sqrt 7
\end{array} \right.\\
Vậy\,x = 0;x = \sqrt 7 ;x = – \sqrt 7
\end{array}$