Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. Đặt \(A=\left[\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-1}{x^2-xy-2y^2}\right):\dfrac{4x^4+4x^2y+y^2-4}{x^2+x+xy+y}\right]:\dfrac{x+1}{2x^2+y+2}\) \(A=\left[\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-1}{\left(x+y\right).\left(x-2y\right)}\right):\dfrac{\left(2x^2+y+2\right).\left(2x^2+y-2\right)}{\left(x+y\right).\left(x+1\right)}\right]:\dfrac{x+1}{2x^2+y+2}\) \(A=\left(\dfrac{\left(x-y\right).\left(x+y\right)+x^2+y^2+y-2}{\left(x+y\right).\left(2y-x\right)}.\dfrac{\left(x+y\right).\left(x+1\right)}{\left(2x^2+y+2\right).\left(2x^2+y-2\right)}\right):\dfrac{2x^2+y+2}{x+1}\) \(A=\left(\dfrac{2x^2+y-2}{2y-x}.\dfrac{x+1}{2x^2+y-2}\right).\dfrac{1}{x+1}\) \(A=\dfrac{1}{2y-x}\) Thay \(x=-1,76\) và \(y=\dfrac{3}{25}\) vào biểu thức ta được: \(A=\dfrac{1}{2.\dfrac{3}{25}-\left(-1,76\right)}\) \(A=\dfrac{1}{2}\) \(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\right]:\frac{x+1}{2x^2+y+2}\) \(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right):\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+y\right)\left(x+1\right)}\right]:\frac{x+1}{2x^2+y+2}\) \(P=\left(\frac{\left(x-y\right)\left(x+y\right)+x^2+y^2+y-2}{\left(x+y\right)\left(2y-x\right)}.\frac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\right):\frac{2x^2+y+2}{x+1}\) \(P=\left(\frac{2x^2+y-2}{2y-x}.\frac{x+1}{2x^2+y-2}\right).\frac{1}{x+1}\) \(P=\frac{1}{2y-x}\) Tại \(x=-1,76\) và \(y=\frac{3}{25}\) thì giá trị của \(Q=\frac{1}{2}\) \(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\) \(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\) \(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\) \(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\) Đặt \(A=\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\) \(B=\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\) \(C=\frac{x+1}{2x^2+y+2}\) Ta có: A = \(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-y^2-xy-y^2}=\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{\left(x-2y\right)\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)+x^2+y^2+y-2}{\left(2y-x\right)\left(x+y\right)}\) =>A=\(\frac{x^2-y^2+x^2+y^2+y-2}{\left(2y-x\right)\left(x+y\right)}=\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}\) B=\(\frac{\left(2x^2\right)^2+2.2x^2.y+y^2-4}{x^2+xy+x+y}=\frac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}=\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\) =>\(P=\left(A:B\right):C\) \(=\left[\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}:\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+y\right)\left(x+1\right)}\right]:\frac{x+1}{2x^2+y+2}\) \(=\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}.\frac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}.\frac{2x^2+y+2}{x+1}\) \(=\frac{1}{2y-x}\) =>\(P=\frac{1}{2y-x}\) Thế x=-1,76 và y=3/25 vào P =>\(P=\frac{1}{2.\frac{3}{25}-1,76}=\frac{1}{2}\) 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: \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x-y}{2y}\) \(=\dfrac{4xy+4y^2}{2\left(x+y\right)}\cdot\dfrac{1}{2y}=\dfrac{4y\left(x+y\right)}{4y\left(x+y\right)}=1\) \(a,\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\) \(=\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\) \(=\frac{2\left(x-2\right)}{x+2}\) Với \(x=\frac{1}{2}\) \(\Rightarrow\frac{2\left(x-2\right)}{x+2}=\frac{2\left(\frac{1}{2}-2\right)}{\frac{1}{2}+2}=\frac{2.-\frac{3}{2}}{\frac{5}{2}}=-3.\frac{2}{5}=\frac{-6}{5}\) b,Do x = -5; y = 10=> y = -2x Thay y = -2x vào biểu thức ta được \(\frac{x^3-x^2\left(-2x\right)+x\left(-2x\right)^2}{x^3+\left(-2x\right)^3}\) \(=\frac{x^3+2x^3+2x^2}{x^3-8x^3}\) \(=\frac{3x^3+2x^2}{-7x^3}=\frac{3}{-7}+\frac{2}{-7x}\) Thay x = -5 là đc 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) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\) \(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)
K
Khách
NT
Nguyen Thuy Hoa
24 tháng 6 2017
Đúng(0)
Các câu hỏi dưới đây có thể giống với câu hỏi trên
PT
12 tháng 9 2017
8 tháng 7 2016
H
14 tháng 12 2018
8 tháng 7 2016
7 tháng 12 2022
29 tháng 11 2018
HL
7 tháng 11 2017
