Cho P = \(\dfrac{y-x}{xy}:\left(\dfrac{y^2}{\left(x-y\right)^2}-\dfrac{2x^2y}{\left(x^2-y^2\right)^2}+\dfrac{x^2}{y^2-x^2}\right)\)
Voi x>0>y va x+y = 1
a/ Rut gon P
b/ Chung minh P < -4
Cho x,y là hai số thay đổi thoat mãn x>0,y<0,x+y=1
A=\(\dfrac{\text{y}-x}{\text{xy}}:\left(\dfrac{y^2}{\left(x-y\right)^2}-\dfrac{2x^2y}{\left(x^2-y^2\right)^2}+\dfrac{x^2}{y^2-x^2}\right)\)
a) Rút gọn A
b) cmr A<-4
Tìm tập xác định, rồi rút gọn biểu thức:
B = \(\dfrac{y-x}{xy}\) : [\(\dfrac{y^2}{\left(x-y\right)^2\left(x+y\right)}\) - \(\dfrac{2x^2y}{x^4-2x^2y^2+y^4}\) + \(\dfrac{x^2}{\left(y^2-x^2\right)\left(x+y\right)}\)]
Tính giá trị của B với x = -\(\dfrac{1}{2}\), y = 2
Rút gọn biểu thức:
\(a,\left(\dfrac{x}{xy-y^2}+\dfrac{2x-y}{xy-x^2}\right):\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(b,\left(\dfrac{x+y}{2x-2y}-\dfrac{x-y}{2x+2y}-\dfrac{2y^2}{y-x}\right):\dfrac{2y}{x-y}\)
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: \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x-y}{2y}\)
\(=\dfrac{4xy+4y^2}{2\left(x+y\right)}\cdot\dfrac{1}{2y}=\dfrac{4y\left(x+y\right)}{4y\left(x+y\right)}=1\)
Rút gọn biểu thức:
\(a,\left(\dfrac{x}{xy-y^2}+\dfrac{2x-y}{xy-x^2}\right):\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(b,\left(\dfrac{x+y}{2x-2y}-\dfrac{x-y}{2x+2y}-\dfrac{2y^2}{y-x}\right):\dfrac{2y}{x-y}\)
Rút gọn biểu thức:
\(a,\left(\dfrac{x}{xy-y^2}+\dfrac{2x-y}{xy-x^2}\right):\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(b,\left(\dfrac{x+y}{2x-2y}-\dfrac{x-y}{2x+2y}-\dfrac{2y^2}{y-x}\right):\dfrac{2y}{x-y}\)
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
cho 2 số thực `x,y` thỏa mãn `x>0,y>2,x`\(\ne\)`2y`. CMR: \(\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-2}{x^2-xy-2y^2}\right)\left(2x^2+y+2\right):\dfrac{x^4+4x^2y^2+y^4-4}{x^2+y+xy+x}=\dfrac{x+1}{2y-x}\)
Đề bài sai, đề đúng thì phân thức đằng sau dấu chia phải là:
\(\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{3x+2}{x-1}-\dfrac{3y-1}{y+2}=0\\\dfrac{2}{x-1}+\dfrac{3}{y+2}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{4x-5}{x+1}+\dfrac{2y-3}{y-5}=8\\\dfrac{3}{x+1}-\dfrac{2}{y-5}=-1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{x+y-2}{x+1}+\dfrac{3-x}{y+1}=\dfrac{5}{4}\\\dfrac{3\left(x+y-2\right)}{x+1}-\dfrac{5-x+2y}{y+1}=\dfrac{3}{4}\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x-y+1}{x-3}+\dfrac{x+1}{y-3}=\dfrac{-7}{2}\\\dfrac{2\left(x-y+1\right)}{x-3}-\dfrac{x+y-2}{y-3}=-\dfrac{9}{2}\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}x^2-y^2+2y=1\\\left(x+y\right)^2-2x-2y=0\end{matrix}\right.\)
f)\(\left\{{}\begin{matrix}4x^2+y^2-4xy=4\\x^2+y^2-2\left(xy+8\right)=0\end{matrix}\right.\)
Cho A = \(\dfrac{\left(x-y\right)^2+xy}{\left(x+y\right)^2-xy}.\left[1:\dfrac{x^5+y^5+x^3y^2+x^2y^3}{\left(x^3-y^3\right)\left(x^3+y^3+x^2y+xy^2\right)}\right]\)
B = x - y
Chứng minh đẳng thức A = B
Tính giá trị của A, B tại x = 0; y = 0 và giải thích vì sao A ≠ B
\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)
\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)
\(x=0;y=0\Leftrightarrow B=0\)
Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)
Vậy \(A\ne B\)
Giải hệ phương trình
a. \(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}\dfrac{3x+5}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5y+9}{y+4}=9\end{matrix}\right.\)
c. \(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x-3y-3=0\end{matrix}\right.\)
d.\(\left\{{}\begin{matrix}\left(x-y\right)^2-3x-3y=4\\2x+y=3\end{matrix}\right.\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)-xy=100\\xy-\left(x-2\right)\left(y-2\right)=64\end{matrix}\right.\)
=>xy+3x+2y+6-xy=100 và xy-xy+2x+2y-4=64
=>3x+2y=94 và 2x+2y=68
=>x=26 và x+y=34
=>x=26 và y=8
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x+3+2}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5y+20-11}{y+4}=9\end{matrix}\right.\)
=>\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+1}-\dfrac{2}{y+4}=4-3=1\\\dfrac{-2}{x+1}+\dfrac{11}{y+4}=9+5-2=12\end{matrix}\right.\)
=>x+1=18/35; y+4=9/13
=>x=-17/35; y=-43/18