HL
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TT
2 tháng 9 2020
\(ĐKXĐ:x\ne y,x\ne0,y\ne0\)
Ta có : \(\frac{3xy^2+x^2y}{xy\left(x-y\right)}-\frac{3x^2y+xy^2}{xy.\left(x-y\right)}\)
\(=\frac{3xy^2+x^2y-3x^2y-xy^2}{xy.\left(x-y\right)}\)
\(=\frac{-3xy.\left(x-y\right)+xy.\left(x-y\right)}{xy.\left(x-y\right)}=\frac{-2xy.\left(x-y\right)}{xy.\left(x-y\right)}=-2\)
VM
2 tháng 9 2020
\(\frac{3xy^2+x^2y}{xy\left(x-y\right)}-\frac{3x^2y+xy^2}{xy.\left(x-y\right)}\)
\(=\frac{3xy^2+x^2y}{xy\left(x-y\right)}+\frac{-\left(3x^2y+xy^2\right)}{xy.\left(x-y\right)}\)
\(=\frac{3xy^2+x^2y-3x^2y-xy^2}{xy.\left(x-y\right)}\)
\(=\frac{\left(3xy^2-3x^2y\right)+\left(x^2y-xy^2\right)}{xy.\left(x-y\right)}\)
\(=\frac{3xy.\left(y-x\right)+xy.\left(x-y\right)}{xy.\left(x-y\right)}\)
\(=\frac{-3xy.\left(x-y\right)+xy.\left(x-y\right)}{xy.\left(x-y\right)}\)
\(=\frac{\left(x-y\right).\left(-3xy+xy\right)}{xy.\left(x-y\right)}\)
\(=\frac{-3xy+xy}{xy}\)
\(=\frac{-2xy}{xy}\)
\(=-2.\)
22 tháng 8
a: \(\frac{x+1}{x-y}+\frac{x-1}{x-y}+\frac{x+3}{x-y}\)
\(=\frac{x+1+x-1+x+3}{x-y}=\frac{3x+3}{x-y}\)
b: \(\frac{5xy^2-3z}{3xy}+\frac{4x^2y+3z}{3xy}=\frac{5xy^2-3z+4x^2y+3z}{3xy}\)
\(=\frac{5xy^2+4x^2y}{3xy}=\frac{xy\left(5y+4x\right)}{3xy}=\frac{4x+5y}{3}\)
c: \(\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}\)
\(=\frac{x+9}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x\left(x+3\right)}\)
\(=\frac{x\left(x+9\right)-3\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2+9x-3x+9}{x\left(x+3\right)\left(x-3\right)}\)
\(=\frac{x^2+6x+9}{x\left(x+3\right)\left(x-3\right)}=\frac{\left(x+3\right)^2}{x\left(x+3\right)\left(x-3\right)}=\frac{x+3}{x\left(x-3\right)}\)
d: \(\frac{x}{5x+5}-\frac{x-10}{10x-10}\)
\(=\frac{x}{5\left(x+1\right)}-\frac{x-10}{10\left(x-1\right)}\)
\(=\frac{2x\left(x-1\right)-\left(x-10\right)\left(x+1\right)}{10\left(x-1\right)\left(x+1\right)}=\frac{2x^2-2x-\left(x^2+x-10x-10\right)}{10\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x^2-2x-x^2+9x+10}{10\left(x-1\right)\left(x+1\right)}=\frac{x^2+7x+10}{10\left(x-1\right)\left(x+1\right)}\)
16 tháng 3 2020
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)
\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)
\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)
\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)
16 tháng 3 2020
chỗ cuối tớ sai
\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)
đây nha , e xin lỗi
11 tháng 2 2020
\(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}\)
\(=\frac{2\left(x^2-y^2\right)+2y^2}{x}\)
\(=\frac{2x^2-2y^2+2y^2}{x}\)
\(=\frac{2x^2}{x}\)
\(=2x\)
11 tháng 12 2019
a)\(\frac{4x+1}{3x}\)+\(\frac{2x-3}{6x}\)=\(\frac{2\left(4x+1\right)}{2.3x}\)+\(\frac{2x-3}{6x}\)=\(\frac{8x+2}{6x}\)+\(\frac{2x-3}{6x}\)=\(\frac{8x+2+2x-3}{6x}\)=\(\frac{10x-1}{6x}\)
b)\(\frac{x^2-y^2}{6x^2y^2}\):\(\frac{x+y}{3xy}\)=\(\frac{\left(x+y\right)\left(x-y\right)}{6x^2y^2}\) . \(\frac{3xy}{x+y}\)=\(\frac{\left(x+y\right)\left(x-y\right)3xy}{6x^2y^2\left(x+y\right)}\)=\(\frac{x-y}{2xy}\)
DK: \(x\ne0;y\ne0\)
\(\left(\frac{x^2-y^2}{6x^2y^2}\right):\left(\frac{x+y}{3xy}\right)=\left(\frac{\left(x-y\right)\left(x+y\right)}{6x^2y^2}\right).\left(\frac{3xy}{\left(x+y\right)}\right)=\frac{x-y}{2xy}\)