Rút gọn : \(\left(x+y-\frac{4xy}{x+y}\right):\left(\frac{x}{x+y}-\frac{y}{y-x}-\frac{2xy}{x^2-y^2}\right)\)
Bài 1 rút gọn biểu thức
A=\(\left(x-\frac{4xy}{x+y}+y\right)\):\(\left(\frac{x}{x+y}-\frac{y}{x-y}-\frac{2xy}{x^2-y^2}\right)\)
B=\(\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right)\):\(\left(\frac{x^2+4x^2y^2+y^4}{x^2+y+xy+x}\right):\left(\frac{1}{2x^2+y+2}\right)\)
Rút gọn \(\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right):\frac{4xy}{y^2-x^2}\)
ĐKXĐ: x2-y2\(\ne\)0 4xy\(\ne\)0
\(\Leftrightarrow\)\(\left(x-y\right)\left(x+y\right)\ne0\) <=>x\(\ne\)0 và y \(\ne\)0
\(\Leftrightarrow x\ne y\) và \(x\ne-y\)
Đặt P= \(\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right):\frac{4xy}{y^2-x^2}\)
<=>\(\left(\frac{1}{\left(x+y\right)^2}-\frac{1}{\left(x+y\right)\left(x-y\right)}\right).\frac{y^2-x^2}{4xy}\)
<=>\(\left(\frac{x-y}{\left(x+y\right)^2\left(x-y\right)}-\frac{x+y}{\left(x+y\right)^2\left(x-y\right)}\right).\frac{-\left(x^2-y^2\right)}{4xy}\)
<=>\(\frac{x-y-x-y}{\left(x+y\right)^2\left(x-y\right)}.\frac{-\left(x-y\right)\left(x+y\right)}{4xy}=\frac{-2y}{\left(x+y\right)^2\left(x-y\right)}.\frac{-\left(x-y\right)\left(x+y\right)}{4xy}\)
<=>\(\frac{1}{2x\left(x+y\right)}=\frac{1}{2x^2+2xy}\)
Rút gọn biểu thức:
\(A=\left[\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right]:\frac{4xy}{y^2-x^2}\)
Giúp nha m.n! thanks!!!
Ta có: \(A=\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right):\frac{4xy}{y^2-x^2}\)
\(=\left[\frac{1}{\left(x+y\right)^2}-\frac{1}{\left(x+y\right)\left(x-y\right)}\right].\frac{\left(y+x\right)\left(y-x\right)}{4xy}\)
\(=\frac{1}{x+y}\left(\frac{1}{x+y}-\frac{1}{x-y}\right).\frac{\left(x+y\right)\left(y-x\right)}{4xy}\)
\(=\frac{-2y}{\left(x+y\right)\left(x-y\right)}.\frac{x-y}{-4xy}\)
\(=\frac{1}{\left(x+y\right).2x}\)
Kb với mình nha mn!
Rút gọn :
A=\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{z}{x^2-1}\right)\)
B= \([\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right):\frac{2x+y}{x^2+2xy+y^2}].\frac{x-9}{3}\)
a) ĐKXĐ: \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\left(\frac{1-x}{\left(1+x\right)\left(1-x\right)}-\frac{x\left(1+x\right)}{\left(1-x\right)\left(1+x\right)}+\frac{x}{x^2-1}\right)\)
\(=\frac{4x-1}{x^2-1}:\left(\frac{-x^2-2x+1}{1-x^2}-\frac{x}{1-x^2}\right)=\frac{4x-1}{x^2-1}:\frac{-x^2-3x+1}{1-x^2}\)
\(=\frac{1-4x}{1-x^2}:\frac{-x^2-3x+1}{1-x^2}=\frac{\left(1-4x\right)\left(1-x^2\right)}{\left(1-x^2\right)\left(-x^2-3x+1\right)}\)
\(=\frac{1-4x}{-x^2-3x+1}=\frac{4x-1}{x^2+3x-1}\) (chắc hết rút gọn được rồi)
Ơ sao câu trả lời của mình có khung màu vàng nhỉ?
Rút gọn \(\frac{1}{\left(x+y\right)^3}.\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^5}.\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}.\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(\frac{4xy}{y^2-x^2}:\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right)\)
\(\left(\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right).\frac{4x^2+4xy+y^2}{16x}\)
B1: Rút gọn A=\(\left(\frac{x}{x-1}-\frac{1}{x^2}\right):\left(\frac{1}{x+1}+\frac{2}{x^2-1}\right)\)
B2: Rút gọn A=\(\left(\frac{x-y}{x+y}-\frac{x+y}{x-y}\right):\frac{-4y^2}{x-y}\)
Cho biểu thức :
\(A=\frac{4xy}{y^2-x^2}:\left(\frac{1}{y^2+2xy+x^2}-\frac{x^3+y^3}{x^4-y^4}\right)\left(x\ne\pm y;y\ne0\right)\)
a) Rút gọn A và tìm giá trị x,y để A = 0
b ) tìm giá trị x,y nguyên thỏa mãn \(A=x^3+xy+x+y+1\)
\(A=\frac{4xy}{y^2-x^2}:\left(\frac{1}{y^2+2xy+x^2}-\frac{x^3+y^3}{x^4-y^4}\right)\left(x\ne\pm y;y\ne0\right)\)
\(\Leftrightarrow A=\frac{4xy}{\left(y^2-x^2\right)\left(y^2+x^2\right)}:\left(\frac{1}{\left(y+x\right)^2}-\frac{x^3+y^3}{\left(x^2-y^2\right)\left(x^2+y^2\right)}\right)\)
Rút gọn biểu thức rồi tính giá trị:
a) \(\frac{x^2y\left(y-x\right)+xy^2\left(x-y\right)}{3y^2-3x^2}\) ,với x = -3 ; y =\(\frac{1}{2}\)
b) \(\frac{\left(8x^3-y^3\right)\left(4x^2-y^2\right)}{\left(2x+y\right)\left(4x^2-4xy+y^2\right)}\)với x = 2; y =\(\frac{-1}{2}\)