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Đặt B\(=\frac{y^2}{\left(x-y\right)^2}-\frac{2x^2y}{\left(x^2-y^2\right)^2}+\frac{x^2}{\left(y^2-x^2\right)}\)
\(B=\frac{y^2}{\left(x-y\right)^2}-\frac{2x^2y}{\left[\left(x-y\right)\left(x+y\right)\right]^2}-\frac{x^2}{\left(x-y\right)\left(x+y\right)}\) (làm tắt đấy x^2/(y^2 - x^2) = - x^2 /(x^2 - y^2)
Thay x + y = 1 vào B ta có
\(B=\frac{y^2}{\left(x-y\right)^2}-\frac{2x^2y}{\left(x-y\right)^2}-\frac{x^2}{x-y}\)
\(B=\frac{y^2-2x^2y-x^2\left(x-y\right)}{\left(x-y\right)^2}=\frac{y^2-x^2y-x^3}{\left(x-y\right)^2}\)
A = \(\frac{y-x}{xy}:B=\frac{y-x}{xy}\cdot\frac{\left(x-y\right)^2}{\left(y^2-x^2y-x^3\right)}=\frac{\left(x-y\right)^3}{-xy\left(y^2-x^2y-x^3\right)}\)
Sorry mình không giúp đc bạn
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
a: \(=\left(\dfrac{x}{y\left(x-y\right)}-\dfrac{2x-y}{x\left(x-y\right)}\right):\dfrac{x+y}{xy}\)
\(=\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)}\cdot\dfrac{xy}{x+y}\)
\(=\dfrac{\left(x-y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x-y}{x+y}\)
b: \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x-y}{2y}\)
\(=\dfrac{4xy+4y^2}{2\left(x+y\right)}\cdot\dfrac{1}{2y}=\dfrac{4y\left(x+y\right)}{4y\left(x+y\right)}=1\)
1) Đặt \(B=x^2+y^2+z^2\)
\(C=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)
Ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)
\(\Leftrightarrow-2\left(xy+yz+xz\right)=x^2+y^2+z^2\)
Suy ra: \(C=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow A=\dfrac{B}{C}=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
2) \(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\)
\(\Leftrightarrow x^2+xy-2xy-2y^2=0\)
\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)
Do \(x+y\ne0\) nên \(x-2y=0\Leftrightarrow x=2y\)
Do đó: \(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
1)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)
d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)
e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)
f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)
\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)
2)
a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)
b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )
d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)
e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)