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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: \(x\ne2;x\ne-2\)
\(\frac{1}{x+2}+\frac{3}{x^2-4}+\frac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}\)\(=\frac{1}{x+2}+\frac{3}{\left(x-2\right)\left(x+2\right)}+\frac{x-14}{\left(x+2\right)^2\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x^2+4x+4\right)-16}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x+2\right)^2-16}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x+2-4\right)\left(x+2+4\right)}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\frac{x+6}{\left(x+2\right)^2}\)
HH
4
9 tháng 9 2020
\(\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(4x-1\right)\)
Áp dụng hằng đẳng thức thứ 3 => (A + B)(A - B) = A2 - B2
=> \(\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=x^2-\left(\frac{1}{2}\right)^2=x^2-\frac{1}{4}\)
=> \(\left(x^2-\frac{1}{4}\right)\left(4x-1\right)=x^2\left(4x-1\right)-\frac{1}{4}\left(4x-1\right)\)
\(=4x^3-x^2-x+\frac{1}{4}\)
Vậy : ....
15 tháng 12 2017
a) \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)
b) \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}=\frac{3+6x}{2x+8}\)
NH
0
U
15 tháng 3 2020
1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)
TM
3
6 tháng 12 2018
\(\frac{4}{x-1}-\frac{2}{1-x}-\frac{x}{x-1}\)
\(=\frac{4}{x-1}+\frac{-2}{x-1}-\frac{x}{x-1}\)
\(=\frac{2-x}{x-1}\)
6 tháng 12 2018
ĐKXĐ: \(x\ne1\)
\(\frac{4}{x-1}-\frac{2}{1-x}-\frac{x}{x-1}\)
\(=\frac{4}{x-1}+\frac{2}{x-1}-\frac{x}{x-1}\)
\(=\frac{4+2-x}{x-1}\)
\(=\frac{6-x}{x-1}\)
26 tháng 7 2017
Ta có \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(=\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{2x}{\left(x+2\right)\left(x-2\right)}\)
\(\frac{-4.2x}{\left(x+2\right)^2\left(x-2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{2x}=\frac{-4}{\left(x+2\right)\left(x-2\right)}\)
NQ
2
14 tháng 4 2019
\(=\frac{\left(2x-4\right)\left(2x+4\right)}{\left(1-x\right)^2}.\frac{1-x}{3x-6}=\frac{\left(2x-4\right)\left(2x+4\right)}{\left(1-x\right)\left(3x-6\right)}\)
28 tháng 3 2020
\(\frac{4x^2-16}{1-2x+x^2}:\frac{3x-6}{1-x}\)
\(=\frac{4x^2-16}{1-2x+x^2}\cdot\frac{1-x}{3x-6}\)
\(=\frac{4\left(x^2-4\right)}{\left(1-x\right)^2}\cdot\frac{1-x}{3\left(x-2\right)}\)
\(=\frac{4\left(x-2\right)\left(x+2\right)}{\left(1-x\right)^2}\cdot\frac{1-x}{3\left(x-2\right)}\)
\(=\frac{4\left(x-2\right)\left(x+2\right)\left(1-x\right)}{\left(1-x\right)^2\cdot3\left(x-2\right)}\)
\(=\frac{4\left(x+2\right)}{3\left(1-x\right)}\)
BM
0
Bài làm:
\(\frac{1}{x+2}+\frac{1}{4x^2+15x+14}=\frac{1}{x+2}+\frac{1}{\left(x+2\right)\left(4x+7\right)}\)
\(=\frac{4x+7}{\left(x+2\right)\left(4x+7\right)}+\frac{1}{\left(x+2\right)\left(4x+7\right)}\)
\(=\frac{4x+8}{\left(x+2\right)\left(4x+7\right)}\)
\(=\frac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\frac{4}{4x+7}\left(x\ne-2;x\ne-\frac{7}{4}\right)\)