\(\frac{x}{5x+5}-\frac{x}{10x-10}\)
Tính
K
Khách
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12 tháng 8 2016
\(\frac{x}{5x+5}-\frac{x}{10x-10}\)
\(=\frac{x}{5\left(x+1\right)}-\frac{x}{10\left(x-1\right)}\)
\(=\frac{2x\left(x-1\right)-x\left(x+1\right)}{10\left(x+1\right)\left(x-1\right)}\)
\(=\frac{2x^2-2x-x^2-x}{10\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2-3x}{10\left(x+1\right)\left(x-1\right)}\)
Z
1
LD
19 tháng 7 2017
\(\frac{x^3+2x^2}{2x^2+10x}\)+\(\frac{2x^2-10x+10x-50}{2x^2-10x}\)+\(\frac{50-5x}{2x^2+10x}\)=\(\frac{x^3+4x^2-5x}{2x^2-10x}\)=\(\frac{x\left(x^2+4x-5\right)}{2x\left(x-5\right)}\)=\(\frac{x\left(x-1\right)\left(x-5\right)}{2x\left(x-5\right)}\)=\(\frac{x-1}{2}\)
TD
0
BT
27 tháng 11 2017
\(=\frac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}.\frac{5\left(x-1\right)}{3\left(x+1\right)}=\frac{x\left(x+1\right).5\left(x-1\right)}{5\left(x-1\right)^2.3\left(x+1\right)}=\frac{x}{3x-3}\)
27 tháng 11 2017
\(\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\)
=\(\frac{x\left(x+1\right)}{5\left(x^2-2+1\right)}:\frac{3\left(x+1\right)}{5\left(x-1\right)}\)
=\(\frac{x\left(x+1\right)}{5\left(x-1\right)^2}:\frac{3\left(x+1\right)}{5\left(x-1\right)}\)
=\(\frac{x\left(x+1\right)}{5\left(x-1\right)^2}\cdot\frac{5\left(x-1\right)}{3\left(x+1\right)}\)
=\(\frac{x}{3\left(x-1\right)}\)
5 tháng 3 2017
1.Tính giá trị của biểu thức: A=\(\frac{5x^2+3y^2}{10x^2-3y^2}\left(1\right)biết\frac{x}{3}=\frac{y}{5}suyra:5x=3y;suyra:x=\frac{3y}{5};thayvào\left(1\right)taco:\frac{5\left(\frac{3y}{5}\right)^2+3y^2}{10\left(\frac{3y}{5}\right)^2-3y^2}=\frac{\frac{9y^2}{5}+3y^2}{\frac{18y^2}{5}-3y^2}=\frac{24y^2}{5}\cdot\frac{5}{3y^2}=8\)
5 tháng 3 2017
2.\(\frac{x}{y}=\frac{7}{10}suyra;\frac{x}{7}=\frac{y}{10}\left(1\right)và\frac{y}{z}=\frac{5}{8}suyra;\frac{y}{5}=\frac{z}{8}suyra;\frac{y}{5}\cdot\frac{1}{2}=\frac{z}{8}\cdot\frac{1}{2}suyra;\frac{y}{10}=\frac{z}{16}\left(2\right)Tù\left(1\right)và\left(2\right)suyra\frac{x}{7}=\frac{y}{10}=\frac{z}{16}và2x+5y-2z=9;suyra:\frac{2x}{14}=\frac{5y}{50}=\frac{2z}{32}ápdụngtínhchấtcủadãytỉsốbằngnhautacó\frac{2x}{14}=\frac{5y}{50}=\frac{2z}{32}=\frac{2x+5y-2z}{14+50-32}=\frac{9}{32}suyra;x=\frac{63}{32};y=\frac{45}{16};z=\frac{9}{2}\)
bằng
\(\frac{x}{5\left(x+1\right)}\)-\(\frac{x}{10\left(x-1\right)}\)=\(x\left(\frac{1}{5\left(x+1\right)}-\frac{1}{10\left(x-1\right)}\right)\)=\(x\left(\frac{2\left(x-1\right)}{10\left(x+1\right)\left(x-1\right)}-\frac{x+1}{10\left(x-1\right)\left(x+1\right)}\right)\)
=\(x\left(\frac{2\left(x-1\right)-\left(x+1\right)}{10\left(x+1\right)\left(x-1\right)}\right)\)=\(x\left(\frac{2x-2-x+1}{10\left(x-1\right)\left(x+1\right)}\right)\)=\(x\left(\frac{x-1}{10\left(x-1\right)\left(x+1\right)}\right)\)=\(x\left(\frac{1}{10\left(x+1\right)}\right)\)=\(\frac{x}{10x+10}\)
\(\Rightarrow\frac{x}{5\left(x+1\right)}-\frac{x}{10\left(x-1\right)}\)
\(\Leftrightarrow\frac{2x\left(x-1\right)}{10\left(x+1\right)\left(x-1\right)}-\frac{x\left(x+1\right)}{10\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{2x\left(x-1\right)-x\left(x+1\right)}{10\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\frac{2x^2-2x-x^2-x}{10\left(x+1\right)\left(x-1\right)}\)\(\Leftrightarrow\frac{x^2-3x}{10\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow\frac{x}{5x+5}-\frac{x}{10x-10}=\frac{x^2-3x}{10\left(x+1\right)\left(x-1\right)}\)