Toán Lớp 8: THỰC HIỆN PHÉP TÍNH
a) (6x^2-xy):x+(2x^3 y+3xy^2):xy-(2x-1)x
b)(x^2-xy):x+(6x^2 y^5-9x^3 y^4+15x^4 y^2): 3/2 x^2 y^3
c)(3a^2 b)^3(ab^3)^2 / (a^2b^2)^4
d)(2xy^2)^3(3x^2 y)^2 / (2x^3 y^2)^2
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Toán Lớp 8: THỰC HIỆN PHÉP TÍNH
a) (6x^2-xy):x+(2x^3 y+3xy^2):xy-(2x-1)x
b)(x^2-xy):x+(6x^2 y^5-9x^3 y^4+15x^4 y^2): 3/2 x^2 y^3
c)(3a^2 b)^3(ab^3)^2 / (a^2b^2)^4
d)(2xy^2)^3(3x^2 y)^2 / (2x^3 y^2)^2
Comments ( 1 )
a)\left( {6{x^2} – xy} \right):x + \left( {2{x^3}y + 3x{y^2}} \right):xy – \left( {2x – 1} \right).x\\
= 6x – y + 2{x^2} + 3y – 2{x^2} + x\\
= 7x + 2y\\
b)\\
\left( {{x^2} – xy} \right):x + \left( {6{x^2}{y^5} – 9{x^3}{y^4} + 15{x^4}{y^2}} \right):\dfrac{3}{2}{x^2}{y^3}\\
= x – y + 4{y^2} – 6xy + \dfrac{{10{x^2}}}{y}\\
c)\\
\dfrac{{{{\left( {3{a^2}b} \right)}^3}{{\left( {a{b^3}} \right)}^2}}}{{{{\left( {{a^2}{b^2}} \right)}^4}}} = \dfrac{{27{a^6}{b^3}.{a^2}{b^6}}}{{{a^8}{b^8}}} = 27b\\
d)\\
\dfrac{{{{\left( {2x{y^2}} \right)}^3}.{{\left( {3{x^2}y} \right)}^2}}}{{{{\left( {2{x^3}{y^2}} \right)}^2}}} = \dfrac{{8{x^3}{y^6}.9{x^4}{y^2}}}{{4{x^6}{y^4}}} = 18x{y^4}
\end{array}$