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\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
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21 tháng 7 2019
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)
= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x}{10\left(x+y\right)}\)
16 tháng 3 2020
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)
\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)
\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)
\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)
16 tháng 3 2020
chỗ cuối tớ sai
\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)
đây nha , e xin lỗi
DT
1
KL
7 tháng 12 2020
a,\(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)
\(=\frac{3}{2x\left(x+1\right)}+\frac{2x-1}{\left(x-1\right)\left(x+1\right)}-\frac{2}{x}\)
\(=\frac{3\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}+\frac{\left(2x-1\right).2x}{2x\left(x-1\right)\left(x+1\right)}-\frac{2.2\left(x+1\right)\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{3x-3}{2x\left(x+1\right)\left(x-1\right)}+\frac{4x^2-2x}{2x\left(x-1\right)\left(x+1\right)}-\frac{4x^2-4}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{3x-3+4x^2-2x-4x^2+4}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x+1}{2x\left(x+1\right)\left(x-1\right)}=\frac{1}{2x\left(x-1\right)}\)
\(b,\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)
\(=\frac{3x.2\left(x-y\right)}{10\left(x+y\right).\left(x-y\right)}-\frac{x.\left(x+y\right)}{10\left(x-y\right).\left(x+y\right)}\)
\(=\frac{6x^2-6xy}{10\left(x+y\right)\left(x-y\right)}-\frac{x^2+xy}{10\left(x-y\right)\left(x+y\right)}\)
\(=\frac{6x^2-6xy-x^2+xy}{10\left(x+y\right)\left(x-y\right)}\)
\(=\frac{5x^2-5xy}{10\left(x+y\right)\left(x+y\right)}\)
\(=\frac{5x\left(x-y\right)}{10\left(x-y\right)\left(x+y\right)}=\frac{x}{2\left(x+y\right)}\)
NL
1
27 tháng 11 2018
\(\frac{5x^2+y^2}{xy}-\frac{3x-2y}{xy}\)
\(=\frac{5x^2+y^2-3x-2y}{xy}\)
Tham khảo nhé~
AT
16 tháng 3 2020
a, \(\frac{4x+1}{2}-\frac{3x+2}{3}=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{12x+3-6x-4}{6}=\frac{6x-1}{6}\)
b, \(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}=\frac{x+3}{\left(x-1\right)\left(x+2\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)}{x\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{x\left(x-1\right)}\)
16 tháng 3 2020
\(\frac{4x+1}{2}-\frac{3x+2}{3}\)
\(=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}\)
tương tự đến hết nha a hay cj gì đps !
NT
0
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5.\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)
\(=\frac{2.3x\left(x-y\right)}{2.5.\left(x+y\right)\left(x-y\right)}-\frac{\left(x+y\right).x}{10.\left(x+y\right)\left(x-y\right)}\)
\(=\frac{6x^2-6xy-x^2-xy}{10\left(x+y\right)\left(x-y\right)}\)
\(=\frac{5x^2-7xy}{10\left(x+y\right)\left(x-y\right)}\)
Tham khảo nhé~