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

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1. Giải đáp:

   $=x\left(4x\left(\frac{1}{\left(2x-y\right)^2}+\frac{x^2}{2}-y^2+\frac{1}{\left(2x+y\right)^2}\right)+4y+\frac{y^2}{16}\right)$

   Lời giải và giải thích chi tiết:

   $\left(\frac{1}{\left(2x-y\right)^2}+\frac{2}{4}x^2-y^2+\frac{1}{\left(2x+y\right)^2}\right)\cdot \:4x^2+4xy+\frac{y^2}{16}x$

   $=x\left(\left(\frac{1}{\left(2x-y\right)^2}+\frac{2}{4}x^2-y^2+\frac{1}{\left(2x+y\right)^2}\right)\cdot \:4x+4y+\frac{y^2}{16}\right)$

   $=x\left(\left(\frac{1}{\left(2x-y\right)^2}+\frac{1}{2}x^2-y^2+\frac{1}{\left(2x+y\right)^2}\right)\cdot \:4x+4y+\frac{y^2}{16}\right)$

   $=x\left(4\left(\frac{1}{\left(2x-y\right)^2}+\frac{x^2}{2}-y^2+\frac{1}{\left(2x+y\right)^2}\right)x+4y+\frac{y^2}{16}\right)$

   $=x\left(4x\left(\frac{1}{\left(2x-y\right)^2}+\frac{x^2}{2}-y^2+\frac{1}{\left(2x+y\right)^2}\right)+4y+\frac{y^2}{16}\right)$

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