\(\frac{x}{2}\)=\(\frac{y}{3}\);\(\frac{y}{5}\)=\(\frac{z}{4}\)va x-y+z=-49
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KN
20 tháng 11 2019
a) \(\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)
\(=\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}.\frac{x^3+y^3}{10x-10y}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5\left(x^2-xy+y^2\right)}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{10\left(x-y\right)}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)^2}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)}{5}.\frac{x+y}{10}\)
\(=\frac{3x^2-3y^2}{50}\)
KN
20 tháng 11 2019
c) \(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)-\frac{x^2-y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\frac{y-x}{xy}-\frac{\left(x+y\right)\left(x-y\right)}{\left(x-y\right)^2}\)
\(=\frac{2}{y-x}-\frac{x+y}{x-y}\)
\(=\frac{2}{y-x}+\frac{x+y}{y-x}\)
\(=\frac{x+y+2}{y-x}\)
TA
22 tháng 8 2017
\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)
\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)
\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(=\left(\frac{x+y}{2.\left(x-y\right)}-\frac{x-y}{2.\left(x+y\right)}+\frac{2y^2}{x-y}\right).\frac{x-y}{2y}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{\left(x+y+x-y\right)\left(x+y-x+y\right)+4y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{4xy+4xy^2+4y^3}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}=\frac{4y.\left(x+xy+y^2\right).\left(x-y\right)}{4y.\left(x-y\right)\left(x+y\right)}=\frac{x+xy+y^2}{x+y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)
\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)
\(=3x:\left(\frac{x-y}{x^2-xy+y^2}.\frac{xy.\left(x^2-xy+y^2\right)}{y.\left(x-y\right)}\right)\)
\(=3x:\frac{xy.\left(x-y\right)\left(x^2-xy+y^2\right)}{y.\left(x-y\right)\left(x^2-xy+y^2\right)}=3x:x=3\)
\(E=\frac{2}{x.\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
\(=2.\left(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)\)
\(=2.\frac{\left(x+2\right)\left(x+3\right)+x.\left(x+3\right)+x.\left(x+1\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)
Ta có: \(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\\x-y+z=49\end{cases}\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{49}{7}=7}\)
Vậy \(x=7.10=70\)
\(y=7.15=105\)
\(z=7.12=84\)
Ta có : \(\frac{x}{2}=\frac{y}{3}\text{ }\Leftrightarrow\text{ }\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{4}\text{ }\Leftrightarrow\text{ }\frac{y}{15}=\frac{z}{12}\)
Nên \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\) và x - y + z = -49
Ta có: \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=-\frac{49}{7}=-7\)
Nên \(\frac{x}{10}=-7\Rightarrow x=-7.10=-70\)
\(\frac{y}{15}=-7\Rightarrow y=-7.15=-105\)
\(\frac{z}{12}=-7\Rightarrow z=-84\)
Vậy ..............................................