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222-9+11+12:2*14+14 = ? ( )

## Toán Lớp 8: tính:1/(2x-y)^2 +2/4x^2-y^2 +1/(2x+y)^2 *4x^2 +4xy+y^2 /16x

Toán Lớp 8: tính:1/(2x-y)^2 +2/4x^2-y^2 +1/(2x+y)^2 *4x^2 +4xy+y^2 /16x

1. Giải đáp:
$\begin{array}{l} \dfrac{1}{{{{\left( {2x – y} \right)}^2}}} + \dfrac{2}{{4{x^2} – {y^2}}} + \dfrac{1}{{{{\left( {2x + y} \right)}^2}}}.\dfrac{{4{x^2} + 4xy + {y^2}}}{{16x}}\\ = \dfrac{1}{{{{\left( {2x – y} \right)}^2}}} + \dfrac{2}{{\left( {2x – y} \right)\left( {2x + y} \right)}} + \dfrac{{{{\left( {2x + y} \right)}^2}}}{{{{\left( {2x + y} \right)}^2}.16x}}\\ = \dfrac{1}{{{{\left( {2x – y} \right)}^2}}} + \dfrac{2}{{\left( {2x – y} \right)\left( {2x + y} \right)}} + \dfrac{1}{{16x}}\\ = \dfrac{{\left( {2x + y} \right).16x + 2.\left( {2x – y} \right).16x + {{\left( {2x – y} \right)}^2}.\left( {2x + y} \right)}}{{16x{{\left( {2x – y} \right)}^2}.\left( {2x + y} \right)}}\\ = \dfrac{{32{x^2} + 16xy + 64{x^2} – 32xy + \left( {2x – y} \right)\left( {4{x^2} – {y^2}} \right)}}{{16x{{\left( {2x – y} \right)}^2}.\left( {2x + y} \right)}}\\ = \dfrac{{96{x^2} – 16xy + 8{x^3} – 2x{y^2} – 4{x^2}y + {y^3}}}{{16x{{\left( {2x – y} \right)}^2}.\left( {2x + y} \right)}} \end{array}$