Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
a){\rm{ }}2x.\left( {3x{\rm{ }} + {\rm{ }}1} \right){\rm{ }} + {\rm{ }}\left( {x{\rm{ }} + {\rm{ }}3} \right)\left( {2x{\rm{ }} - {\rm{ }}5} \right)\\
= 6{{\rm{x}}^2} + 2{\rm{x}} + 2{{\rm{x}}^2} - 5{\rm{x}} + 6{\rm{x}} - 15\\
= 8{{\rm{x}}^2} + 3{\rm{x}} - 15\\
b){\rm{ }}{\left( {x{\rm{ }} + {\rm{ }}5} \right)^2} - {\rm{ }}\left( {4x{\rm{ }} - {\rm{ }}1} \right)\left( {4x{\rm{ }} + {\rm{ }}1} \right)\\
= {x^2} + 10{\rm{x}} + 25 - \left( {16{{\rm{x}}^2} - 1} \right)\\
= {x^2} + 10{\rm{x}} + 25 - 16{{\rm{x}}^2} + 1\\
= - 15{{\rm{x}}^2} + 10{\rm{x}} + 26\\
c)A = \frac{1}{{x + 1}} - \frac{1}{{x - 1}} + \frac{{2{\rm{x}}}}{{{x^2} - 1}}\\
dk:x \ne \pm 1\\
= \frac{1}{{x + 1}} - \frac{1}{{x - 1}} + \frac{{2{\rm{x}}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{x - 1 - \left( {x + 1} \right) + 2{\rm{x}}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{x - 1 - x - 1 + 2{\rm{x}}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{2{\rm{x}} - 2}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{2\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{2}{{x + 1}}\\
vay:A = \frac{2}{{x + 1}}\left( {x \ne \pm 1} \right)
\end{array}\]

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Answer 2

\[\begin{array}{l}
a){\rm{ }}2x.\left( {3x{\rm{ }} + {\rm{ }}1} \right){\rm{ }} + {\rm{ }}\left( {x{\rm{ }} + {\rm{ }}3} \right)\left( {2x{\rm{ }} - {\rm{ }}5} \right)\\
= 6{{\rm{x}}^2} + 2{\rm{x}} + 2{{\rm{x}}^2} - 5{\rm{x}} + 6{\rm{x}} - 15\\
= 8{{\rm{x}}^2} + 3{\rm{x}} - 15\\
b){\rm{ }}{\left( {x{\rm{ }} + {\rm{ }}5} \right)^2} - {\rm{ }}\left( {4x{\rm{ }} - {\rm{ }}1} \right)\left( {4x{\rm{ }} + {\rm{ }}1} \right)\\
= {x^2} + 10{\rm{x}} + 25 - \left( {16{{\rm{x}}^2} - 1} \right)\\
= {x^2} + 10{\rm{x}} + 25 - 16{{\rm{x}}^2} + 1\\
= - 15{{\rm{x}}^2} + 10{\rm{x}} + 26\\
c)A = \frac{1}{{x + 1}} - \frac{1}{{x - 1}} + \frac{{2{\rm{x}}}}{{{x^2} - 1}}\\
dk:x \ne \pm 1\\
= \frac{1}{{x + 1}} - \frac{1}{{x - 1}} + \frac{{2{\rm{x}}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{x - 1 - \left( {x + 1} \right) + 2{\rm{x}}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{x - 1 - x - 1 + 2{\rm{x}}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{2{\rm{x}} - 2}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{2\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{2}{{x + 1}}\\
vay:A = \frac{2}{{x + 1}}\left( {x \ne \pm 1} \right)
\end{array}\]

Answer 3

a)
= 6x^2 + 2x + 2x^2 - 5x + 6x - 15
= 8x^2 + 3x -15
b)
= (x + 5)^2 - ( 16x^2 - 1 )
= - 15x^2 + 10x + 26
c)
= 2x - 1

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