Toán Lớp 8: Hãy tìm đa thức A trong mỗi hằng đẳng thức sau:
a)A/3x+1=9x^2-6x+1/3x-1
b)2x-3/A=6x^2-7x-3/12x+4
c)x^4+x^2/5x^2+5=x^2/A
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Toán Lớp 8: Hãy tìm đa thức A trong mỗi hằng đẳng thức sau:
a)A/3x+1=9x^2-6x+1/3x-1
b)2x-3/A=6x^2-7x-3/12x+4
c)x^4+x^2/5x^2+5=x^2/A
Comments ( 1 )
a)\dfrac{A}{{3x + 1}} = \dfrac{{9{x^2} – 6x + 1}}{{3x – 1}}\\
\Leftrightarrow \dfrac{A}{{3x + 1}} = \dfrac{{{{\left( {3x – 1} \right)}^2}}}{{3x – 1}}\\
\Leftrightarrow \dfrac{A}{{3x + 1}} = 3x – 1\\
\Leftrightarrow A = \left( {3x – 1} \right).\left( {3x + 1} \right)\\
\Leftrightarrow A = 9{x^2} – 1\\
b)\\
\dfrac{{2x – 3}}{A} = \dfrac{{6{x^2} – 7x – 3}}{{12x + 4}}\\
\Leftrightarrow \dfrac{{2x – 3}}{A} = \dfrac{{6{x^2} + 2x – 9x – 3}}{{4.\left( {3x + 1} \right)}}\\
\Leftrightarrow \dfrac{{2x – 3}}{A} = \dfrac{{\left( {3x + 1} \right)\left( {2x – 3} \right)}}{{4.\left( {3x + 1} \right)}}\\
\Leftrightarrow \dfrac{{2x – 3}}{A} = \dfrac{{2x – 3}}{4}\\
\Leftrightarrow A = 4\\
c)\dfrac{{{x^4} + {x^2}}}{{5{x^2} + 5}} = \dfrac{{{x^2}}}{A}\\
\Leftrightarrow \dfrac{{{x^2}\left( {{x^2} + 1} \right)}}{{5.\left( {{x^2} + 1} \right)}} = \dfrac{{{x^2}}}{A}\\
\Leftrightarrow \dfrac{{{x^2}}}{5} = \dfrac{{{x^2}}}{A}\\
\Leftrightarrow A = 5
\end{array}$