Toán Lớp 8: bài 1 tìm x
a)(x+2)^2-(2x-5)^2=0
bài 2
a)x/x-1 + x+1/x^2
b)x^2/x^2-4 – x/x-2 + 2/x+2
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Toán Lớp 8: bài 1 tìm x
a)(x+2)^2-(2x-5)^2=0
bài 2
a)x/x-1 + x+1/x^2
b)x^2/x^2-4 – x/x-2 + 2/x+2
Comments ( 1 )
1)\\
{\left( {x + 2} \right)^2} – {\left( {2x – 5} \right)^2} = 0\\
\Leftrightarrow \left( {x + 2 – 2x + 5} \right)\left( {x + 2 + 2x – 5} \right) = 0\\
\Leftrightarrow \left( {7 – x} \right).\left( {3x – 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x = 1
\end{array} \right.\\
Vậy\,x = 7;x = 1\\
2)\\
a)\dfrac{x}{{x – 1}} + \dfrac{{x + 1}}{{{x^2}}}\\
= \dfrac{{{x^3} + \left( {x + 1} \right)\left( {x – 1} \right)}}{{{x^2}\left( {x – 1} \right)}}\\
= \dfrac{{{x^3} + {x^2} – 1}}{{{x^2}\left( {x – 1} \right)}}\\
b)\dfrac{{{x^2}}}{{{x^2} – 4}} – \dfrac{x}{{x – 2}} + \dfrac{2}{{x + 2}}\\
= \dfrac{{{x^2} – x\left( {x + 2} \right) + 2\left( {x – 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} – {x^2} – 2x + 2x – 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{ – 4}}{{{x^2} – 4}}\\
= \dfrac{4}{{4 – {x^2}}}
\end{array}$