Thực hiện phép tính:
\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
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NN
3
19 tháng 4 2020
\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
\(=\frac{2x}{x\left(x+y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2\left(x-2y\right)+x+2y+4}{\left(x+2y\right)\left(x-2y\right)}\)
\(=\frac{3x-2y+4}{\left(x+2y\right)\left(x-2y\right)}\)
NN
19 tháng 4 2020
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\y\ne0\\x\ne\pm2y\end{cases}}\)
\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x+2y\right)\left(x-2y\right)}\)
\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x+2y\right)\left(x-2y\right)}\)\(=\frac{2\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}+\frac{x+2y}{\left(x+2y\right)\left(x-2y\right)}+\frac{4}{\left(x+2y\right)\left(x-2y\right)}\)
\(=\frac{2\left(x-2y\right)+x+2y+4}{\left(x+2y\right)\left(x-2y\right)}=\frac{2x-4y+x+2y+4}{\left(x+2y\right)\left(x-2y\right)}\)
\(=\frac{3x-2y+4}{\left(x+2y\right)\left(x-2y\right)}\)
OM
31 tháng 3 2020
Làmmmm
1/ \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)(ĐKXĐ:x\(\ne0\), x\(\ne\frac{1}{2}\))
= \(\frac{\left(1-2x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\frac{4x^2}{\left(2x-1\right)2x}-\frac{1}{2x\left(2x-1\right)}\)
\(=\frac{2x-1-4x^2+2x+4x^2-1}{2x\left(2x-1\right)}\)
\(=\frac{4x-2}{2x\left(2x-1\right)}=\frac{2\left(2x-1\right)}{2x\left(2x-1\right)}=\frac{1}{x}\)
KL:..............
OM
31 tháng 3 2020
2/\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)(ĐKXĐ : x\(\ne1\))
\(=\frac{x^2+2}{x^3-1}+\frac{2x-2}{x^3-1}-\frac{x^2+x+1}{x^3-1}\)
\(=\frac{x^2+2+2x-2-x^2-x-1}{x^3-1}=\frac{x-1}{x^3-1}=\frac{1}{x^2+x+1}\)
Kl:....................
3 tháng 9 2020
a, \(\frac{x+2y}{8x^2y^5}-\frac{3x^2+2}{12x^4y^4}\)
=\(\frac{\left(x+2y\right)3x^2}{24x^4y^5}-\frac{\left(3x^2+2\right)2y}{24x^4y^5}\)
=\(\frac{3x^3+6x^2y}{24x^4y^5}-\frac{6x^2y+4y}{24x^4y^5}\)
=\(\frac{3x^3+6x^2y-6x^2y-4y}{24x^4y^5}\)
=\(\frac{3x^3-4y}{24x^4y^5}\)
b,\(\frac{y}{xy-5x^2}-\frac{15y-25x}{y^2-25x^2}\)
=\(\frac{y}{x\left(y-5x\right)}-\frac{15y-25x}{\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\frac{\left(15y-25x\right)x}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2+5xy}{x\left(y-5x\right)\left(y+5x\right)}-\frac{15xy-25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2-10xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y-5x}{x\left(y+5x\right)}\)
c,\(\frac{4-x}{x^3+2x}-\frac{x+5}{x^3-x^2+2x-2}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x^3-x^2\right)+\left(2x-2\right)}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\frac{\left(x+5\right)x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{4x-4-x^2+x}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x^2+5x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{4x-4-x^2+x-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2x^2-4}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2\left(x^2+2\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2}{x\left(x-1\right)}\)
DV
20 tháng 12 2016
\(=\left[\frac{2xy}{\left(x-y\right).\left(x+y\right)}+\frac{x-y}{2.\left(x+y\right)}\right]:\frac{x+y}{2x}+\frac{x}{y-x}\)
\(=\frac{4xy+\left(x-y\right).\left(x-y\right)}{2.\left(x-y\right).\left(x+y\right)}.\frac{2x}{x+y}+\frac{x}{y-x}\)
\(=\frac{x^2+2xy+y^2}{\left(x-y\right).\left(x+y\right)^2}.x+\frac{x}{y-x}\)
\(=\frac{x.\left(x+y\right)^2}{\left(x-y\right).\left(x+y\right)^2}+\frac{x}{y-x}\)
\(=\frac{x}{x-y}-\frac{x}{x-y}=0\)
Bạn giùm mik nhé, tks bạn nhiều (:
2 tháng 12 2023
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
15 tháng 7 2017
a) ĐKXĐ: \(x;y\ne0,x\ne\frac{y}{2},y\ne\frac{x}{2}\)
\(\frac{y}{2x^2-xy}+\frac{4x}{y^2-2xy}=\frac{y}{x\left(2x-y\right)}-\frac{4x}{y\left(2x-y\right)}\)\(=\frac{y^2-4x^2}{xy\left(2x-y\right)}=\frac{\left(y-2x\right)\left(y+2x\right)}{xy\left(2x-y\right)}\)
\(=\frac{-\left(y+2x\right)}{xy}\)
b) ĐKXĐ: \(x\ne2;x\ne-2\)
\(\frac{1}{x+2}+\frac{3}{x^2-4}+\frac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}\)\(=\frac{1}{x+2}+\frac{3}{\left(x-2\right)\left(x+2\right)}+\frac{x-14}{\left(x+2\right)^2\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x^2+4x+4\right)-16}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x+2\right)^2-16}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x+2-4\right)\left(x+2+4\right)}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\frac{x+6}{\left(x+2\right)^2}\)
\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\) (điều kiện: \(x;y\ne0\); \(x\ne\pm2y\))
\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2\left(x-2y\right)+\left(x+2y\right)+4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)