Toán Lớp 8: giúp mình với mn ơi huhu
2x(x^2-x+5)
(15x^3y^3+20x^4y^5) 5xy+6x^2y^2
5x+2/x^2-4 + x-5/x-2
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Toán Lớp 8: giúp mình với mn ơi huhu
2x(x^2-x+5)
(15x^3y^3+20x^4y^5) 5xy+6x^2y^2
5x+2/x^2-4 + x-5/x-2
Comments ( 1 )
a)2x\left( {{x^2} – x + 5} \right)\\
= 2x.{x^2} – 2x.x + 2x.5\\
= 2{x^3} – 2{x^2} + 10x\\
b)\\
\left( {15{x^3}{y^3} + 20{x^4}{y^5}} \right).5xy + 6{x^2}{y^2}\\
= 75{x^4}{y^4} + 100{x^5}{y^6} + 6{x^2}{y^2}\\
c)\\
\dfrac{{5x + 2}}{{{x^2} – 4}} + \dfrac{{x – 5}}{{x – 2}}\\
= \dfrac{{5x + 2 + \left( {x – 5} \right)\left( {x + 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{5x + 2 + {x^2} + 2x – 5x – 10}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} + 2x – 8}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{\left( {x + 4} \right)\left( {x – 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{x + 4}}{{x + 2}}
\end{array}$