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\(B=\left(\frac{x}{y^2+xy}-\frac{x-y}{x^2+xy}\right):\left(\frac{y^2}{x^3-xy^2}+\frac{1}{x+y}\right):\frac{3y}{y}\)
\(=\frac{x^2-xy+y^2}{xy\left(x+y\right)}\cdot\frac{x\left(x^2-y^2\right)}{x^2-xy+y^2}\cdot\frac{y}{3x}\)\(=\frac{x-y}{y}\cdot\frac{y}{3x}=\frac{x-y}{3x}\)
\(a,ĐKXĐ:x\ne-;y\ne0\)
\(P=\frac{2}{x}-\left(\frac{x^2}{x^2+xy}+\frac{y^2-x^2}{xy}-\frac{y^2}{xy+y^2}\right)\cdot\frac{x+y}{x^2+xy+y^2}\)
\(P=\frac{2}{x}-\left(\frac{x^2}{x\left(x+y\right)}+\frac{y^2-x^2}{xy}-\frac{y^2}{y\left(x+y\right)}\right)\cdot\frac{x+y}{x^2+xy+y^2}\)
\(P=\frac{2}{x}-\left(\frac{x^2y}{xy\left(x+y\right)}+\frac{\left(x+y\right)\left(y^2-x^2\right)}{xy\left(x+y\right)}-\frac{xy^2}{xy\left(x+y\right)}\right)\cdot\frac{x+y}{x^2+xy+y^2}\)
\(P=\frac{2}{x}-\left(\frac{x^2y+xy^2-x^3+y^3-x^2y-xy^2}{xy\left(x+y\right)}\right)\cdot\frac{x+y}{x^2+xy+y^2}\)
\(P=\frac{2}{x}+\frac{x^3-y^3}{xy\left(x+y\right)}\cdot\frac{x+y}{x^2+xy+y^2}\)
\(P=\frac{2}{x}-\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy}\cdot\frac{1}{x^2+xy+y^2}\)
\(P=\frac{2}{x}-\frac{x-y}{xy}=\frac{2y-x+y}{xy}=\frac{3y-x}{xy}\)
\(b,x^2+y^2+10=2\left(x-3y\right)\)
\(\Leftrightarrow x^2+y^2+10=2x-6y\)
\(\Leftrightarrow x^2-2x+1+y^2+6y+9=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
thay vào P được : \(P=\frac{3\left(-3\right)-1}{-3\cdot1}=\frac{-10}{-3}=\frac{10}{3}\)
a, Rút gọn A
b,Tìm giá trị P, biết x,y thỏa mãn đẳng thức
x^2+y^2+10=2(x-3y)
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\frac{4y^2-\left(x-y\right)^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{x\left(x-2y\right)-2\left(x^2-xy\right)}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{3y^2+2xy-x^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{-x^2}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{\left(x+y\right)\left(3y-x\right)}{y^2\left(x-y\right)}.\frac{y\left(y-x\right)}{x-3y}-\frac{x^2}{2\left(x-2y\right)}.\frac{2\left(x-2y\right)}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)}{y}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{2xy+y^2}{y\left(x+y\right)}=\frac{2x+y}{x+y}\)
Giờ chỉ cần thế x, y vô nữa là xong nhé.
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y\left(y-x\right)}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x\left(x-y\right)}{x-2y}\right):\frac{y\left(x+y\right)}{2\left(x-2y\right)}\)
\(=\frac{4y\left(y-x\right)}{\left(x-y\right)\left(x-3y\right)}-\frac{\left(x-y\right)y\left(y-x\right)}{y^2\left(x-3y\right)}\)\(+\frac{x.2\left(x-2y\right)}{2.y\left(x+y\right)}-\frac{x\left(x-y\right).2\left(x-2y\right)}{\left(x-2y\right).y\left(x+y\right)}\)
\(=\frac{-4y}{x-3y}+\frac{\left(x-y\right)^2}{y\left(x-3y\right)}+\frac{x\left(x-2y\right)}{y\left(x+y\right)}-\frac{2x\left(x-y\right)}{y\left(x+y\right)}\)
\(=\frac{-4y^2+x^2-2xy+y^2}{y\left(x-3y\right)}+\frac{x^2-2xy-2x^2+2xy}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy-3y^2}{y\left(x-3y\right)}+\frac{-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2+xy-3xy-3y^2}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x\left(x+y\right)-3y\left(x+y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(\frac{\left(x+y\right)\left(x-3y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x+y}{y}-\frac{x^2}{y\left(x+y\right)}=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy+y^2-x^2}{y\left(x+y\right)}=\frac{-2xy+y^2}{y\left(x+y\right)}\)
\(=\frac{y\left(y-2x\right)}{y\left(x+y\right)}=\frac{y-2x}{x+y}\)
Thay \(x=\frac{1}{2};y=\frac{1}{3}\)vào A ta có :
\(A=\frac{\frac{1}{3}-2.\frac{1}{2}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}-1}{\frac{3}{6}+\frac{2}{6}}=\frac{2}{3}:\frac{5}{6}=\frac{2.6}{3.5}=\frac{4}{5}\)
Vậy \(A=\frac{4}{5}\)tại \(x=\frac{1}{2};y=\frac{1}{3}\)
Xét tích : \(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)
=\(x^3\left(z-y\right)+x^2\left(z-y\right)\left(z+y\right)+y^3\left(x-z\right)+y^2\left(x-z\right)\left(x+z\right)\)
\(+z^3\left(y-x\right)+z^2\left(y-x\right)\left(y+x\right)\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2\left(z^2-y^2\right)+y^2\left(x^2-z^2\right)+z^2\left(y^2-x^2\right)\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2z^2-x^2y^2+y^2x^2-y^2z^2+z^2y^2-z^2x^2\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)
Như vậy:
\(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)
<=> \(\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)
Ta có: \(\frac{\frac{x^2\left(z-y\right)}{yz}+\frac{y^2\left(x-z\right)}{xz}+\frac{z^2\left(y-x\right)}{xy}}{\frac{x\left(z-y\right)}{yz}+\frac{y\left(x-z\right)}{xz}+\frac{z\left(y-x\right)}{xy}}\)
\(=\frac{\frac{x^3\left(z-y\right)}{xyz}+\frac{y^3\left(x-z\right)}{xyz}+\frac{z^3\left(y-x\right)}{xyz}}{\frac{x^2\left(z-y\right)}{xyz}+\frac{y^2\left(x-z\right)}{xyz}+\frac{z^2\left(y-x\right)}{xyz}}\)
\(=\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)
\(\frac{2}{x}-\left(\frac{x^2}{x^2+xy}-\frac{x^2-y^2}{xy}-\frac{y^2}{xy+y^2}\right)\)\(\left(\frac{x+y}{x^2+xy+y^2}\right)\)
ĐK: \(\hept{\begin{cases}x,y\ne0\\x\ne-y\end{cases}}\)
\(A=\frac{2}{x}-\frac{x^2y-\left(x-y\right)\left(x+y\right)^2-xy^2}{xy\left(x+y\right)}.\frac{x+y}{x^2+xy+y^2}\)
\(A=\frac{2}{x}+\frac{x^3-y^3}{xy\left(x+y\right)}.\frac{x+y}{x^2+xy+y^2}\)
\(A=\frac{2}{x}+\frac{x-y}{xy}\)
\(A=\frac{2y+x-y}{xy}\)
\(A=\frac{x+y}{xy}\)