Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)
\(=\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)\left(x+y\right)^2}\)
\(=\frac{10y}{15\left(x+y\right)^2}\)
\(\frac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\frac{\left(x^2-x\right)-\left(xy-y\right)}{\left(x^2-x\right)+\left(xy-y\right)}\)
\(=\frac{x\left(x-1\right)-y\left(x-1\right)}{x\left(x-1\right)+y\left(x-1\right)}\)
\(=\frac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)
\(=\frac{x-y}{x+y}\)
a)\(\frac{2xy}{3\left(x+y\right)^2}\)
b)=\(\frac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)=\(\frac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)
=\(\frac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)=\(\frac{\left(x-y\right)}{\left(x+y\right)}\)
a, =-3y/5
b,A=x2-2x.1/2+1/4+3/4=(x-1/2)2+3/4 > hoặc=3/4 suy ra >0 với mọi x thuộc R
1)\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}=\frac{2y}{5\left(x+y\right)^2}\)
2) \(\frac{15x\left(x+y\right)^2}{20x^2\left(x+5\right)}=\frac{3\left(x^2+2xy+y^2\right)}{4x\left(x+5\right)}=\frac{3\left(x+y\right)^2}{4x^2+20x}\)
3) \(\frac{15x\left(x-y\right)}{3\left(y-x\right)}=\frac{5x\left(x-y\right)}{-3\left(x-y\right)}=-\frac{5x}{3}\)
4)\(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\frac{\left(y-x\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x+y\right)}{\left(x-y\right)^2}\)
Ta có
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)= \(\frac{2y}{3\left(x+y\right)^2}\)
\(\frac{x^2+2x+1}{5x^3+5x^2}=\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}=\frac{x+1}{5x^2}\)
1/
x2 - 3x - 4
= \(x^2-3x+\frac{9}{4}-\frac{9}{4}-4\)
\(=\left(x^2-3x+\frac{9}{4}\right)-\frac{25}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\left(\frac{5}{2}\right)^2\)
\(=\left(x-\frac{3}{2}-\frac{5}{2}\right)\left(x-\frac{3}{2}+\frac{5}{2}\right)\)
\(=\left(x-4\right)\left(x+1\right)\)
Bài 1 :
\(x^2-3x-4\)
\(=x^2+x-4x-4\)
\(=x\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x+1\right)\left(x-4\right)\)