Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,=\dfrac{2y^4}{3x\left(2x-3y\right)}\\ b,=-\dfrac{2y\left(3x-1\right)^2}{3x^2}\\ c,=\dfrac{5\left(4x^2-9\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)\left(2x+3\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)}{2x+3}\\ d,=\dfrac{5x\left(x-2y\right)}{-2\left(x-2y\right)^3}=-\dfrac{5x}{2\left(x-2y\right)^2}\)
\(a,\dfrac{21x^2y^3}{24x^3y^2}=\dfrac{7y}{8x}\)
\(b,\dfrac{15xy^3\left(x^2-y^2\right)}{20x^2y\left(x+y\right)^2}=\dfrac{15xy^3\left(x-y\right)\left(x+y\right)}{20x^2y\left(x+y\right)^2}=\dfrac{3y^2\left(x-y\right)}{4x\left(x+y\right)}=\dfrac{3xy^2-3y^3}{4x^2+4xy}\)
a) Ta có: \(\dfrac{21x^2y^3}{24x^3y^2}\)
\(=\dfrac{21x^2y^3:3x^2y^2}{24x^3y^2:3x^2y^2}\)
\(=\dfrac{7y}{8x}\)
Ta có: \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)
\(=\dfrac{\left(x^3+y^3\right)\left(x^3+y^3\right)}{x\left(x^3+y^3\right)\left(x^3-y^3\right)}\)
\(=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)
\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{\left(x^2-y^2\right)\left(x^2+y^2\right)}=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}=\dfrac{x^2-xy+y^2}{x^3+xy^2-x^2y-y^3}\)
\(\dfrac{2a\cdot x^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)
\(=\dfrac{-2a\left(x-1\right)^2}{5b\left(x-1\right)\left(x+1\right)}=\dfrac{-2a\left(x-1\right)}{5b\left(x+1\right)}\)
\(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)
\(=\dfrac{4x\cdot x-4x\cdot y}{5x^2\cdot x-5x^2\cdot y}\)
\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)
\(=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}\)
=x+y-z
\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3+y^3\right)\left(x^3-y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)
\(\frac{14xy^5\left(2x-3y\right)}{21x^2y\left(2x-3y\right)^2}=\frac{2y^4}{3x\left(2x-3y\right)}\)
\(=\frac{2.7.xy.y^5\left(2x-3y\right)}{3.7.xy.x\left(2x-3y\right)^2}=\frac{2y^5}{3x\left(2x-3y\right)}\)
\(\frac{14xy^5\left(2x-3y\right)}{21x^2y\left(2x-3y\right)^2}\)
\(=\frac{2y^4}{3x\left(2x-3y\right)}\)
Câu 2:
\(=\dfrac{x^2\left(2x-5\right)+3\left(2x-5\right)}{2x-5}=x^2+3\)
Câu 3:
\(\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{3;1;5;-1\right\}\)
Ta có:
\(\dfrac{21x^5y^6}{-56x^3y^7}=\dfrac{3.7.x^3.x^2.y^6}{-8.7.x^3.y^6.y^1}=\dfrac{3x^2}{-8y}\)