Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
) \(\dfrac{x^3+8y^3}{2y+x}\)
\(=\dfrac{x^3+\left(2y\right)^3}{x+2y}\)
\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)
\(=x^2+2xy+4y^2\)
b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)
\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)
\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)
\(=\dfrac{3a-1}{2\left(a-4\right)}\)
c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)
\(=\dfrac{\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2}\)
d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)
\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)
\(=x^2-10x+25+7x+14-x^2-2x\)
\(=39-5x\)
e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)
\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)
\(=\dfrac{3x+2x+1}{x-2}\)
\(=\dfrac{5x+1}{x-2}\)
h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)
\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
a.=\(\frac{7x+2}{3xy^2}.\frac{x^2y}{14x+4}\)
=\(\frac{7x+2}{3y}.\frac{x^2y}{2\left(7x+2\right)}\)
=\(\frac{1}{3y}.\frac{x}{2}\)
=\(\frac{x}{6y}\)
b.=\(\frac{8xy}{3x-1}.\frac{5-15x}{12xy^3}\)
=\(\frac{2}{3x-1}.\frac{-15x+5}{3y^2}\)
=\(\frac{2}{3x-1}.\frac{-5\left(3x-1\right)}{3y^2}\)
=\(\frac{-10}{3y^2}\)
c.=\(\frac{3\left(x^3+1\right)}{x-1}.\frac{1}{x^2-x+1}\)
=\(\frac{3\left(x+1\right).\left(x^2-x+1\right)}{x-1}.\frac{1}{x^2-x+1}\)
=\(\frac{3x+3}{x-1}\)
d.=\(\frac{4\left(x+3\right)}{.\left(3x-1\right)}.\frac{1-3x}{x^2+3x}\)
=\(\frac{4\left(x+3\right)}{x.\left(3x-1\right)}.\frac{-\left(3x-1\right)}{x\left(x+3\right)}\)
=\(\frac{-4}{x^2}\)
e.=\(\frac{2\left(2x+3y\right)}{x-1}.\frac{1-x^3}{4x^2+12xy+9y^2}\)
=\(2.\frac{-\left(1+x+x^2\right)}{2x+3y}\)
=\(-\frac{2x^2+2x+2}{2x+3y}\)
a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
1. a) Ta có: \(x^2-2y^2=xy\) \(\Leftrightarrow\) \(x^2-xy-2y^2=0\)
\(\Leftrightarrow\) \(x^2+xy-2xy-2y^2=0\)
\(\Leftrightarrow\) \(x\left(x+y\right)-2y\left(x+y\right)=0\)
\(\Leftrightarrow\) \(\left(x+y\right)\left(x-2y\right)=0\)
Vì \(\left(x+y\right)\ne0\) nên \(x-2y=0\) hay \(x=2y\). Thay \(x=2y\) vào A, ta được:
\(A=\dfrac{\left(2y\right)^2-y^2}{\left(2y\right)^2+y^2}=\dfrac{4y^2-y^2}{4y^2+y^2}=\dfrac{3y^2}{5y^2}=\dfrac{3}{5}\)
\(a,3x^3y^3-15x^2y^2=3x^2y^2\left(xy-5\right)\)
\(b,5x^3y^2-25x^2y^3+40xy^4\)
\(=5xy^2\left(x^2-5xy+8y^2\right)\)
\(c,-4x^3y^2+6x^2y^2-8x^4y^3\)
\(=-2x^2y^2\left(2x-3+4x^2y\right)\)
\(d,a^3x^2y-\frac{5}{2}a^3x^4+\frac{2}{3}a^4x^2y\)
\(=a^3x^2\left(y-\frac{5}{2}x^2+\frac{2}{3}ay\right)\)
\(e,a\left(x+1\right)-b\left(x+1\right)=\left(x+1\right)\left(a-b\right)\)
\(f,2x\left(x-5y\right)+8y\left(5y-x\right)\)
\(=2x\left(x-5y\right)-8y\left(x-5y\right)=\left(x-5y\right)\left(2x-8y\right)\)
\(g,a\left(x^2+1\right)+b\left(-1-x^2\right)-c\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(a-b-c\right)\)
\(h,9\left(x-y\right)^2-27\left(y-x\right)^3\)
\(=9\left(x-y\right)^2+27\left(x-y\right)^3\)
\(=9\left(x-y\right)^2\left(1+3x-3y\right)\)
a,3x3y3−15x2y2=3x2y2(xy−5)a,3x3y3−15x2y2=3x2y2(xy−5)
b,5x3y2−25x2y3+40xy4b,5x3y2−25x2y3+40xy4
=5xy2(x2−5xy+8y2)=5xy2(x2−5xy+8y2)
c,−4x3y2+6x2y2−8x4y3c,−4x3y2+6x2y2−8x4y3
=−2x2y2(2x−3+4x2y)=−2x2y2(2x−3+4x2y)
d,a3x2y−52a3x4+23a4x2yd,a3x2y−52a3x4+23a4x2y
=a3x2(y−52x2+23ay)=a3x2(y−52x2+23ay)
e,a(x+1)−b(x+1)=(x+1)(a−b)e,a(x+1)−b(x+1)=(x+1)(a−b)
f,2x(x−5y)+8y(5y−x)f,2x(x−5y)+8y(5y−x)
=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)
g,a(x2+1)+b(−1−x2)−c(x2+1)g,a(x2+1)+b(−1−x2)−c(x2+1)
=(x2+1)(a−b−c)=(x2+1)(a−b−c)
h,9(x−y)2−27(y−x)3h,9(x−y)2−27(y−x)3
=9(x−y)2+27(x−y)3
a/ \(\frac{x}{2}=\frac{y}{4}\)
\(\Rightarrow\frac{x^2}{4}=\frac{y^2}{16}=\frac{x^2+y^2}{20}=\frac{2000}{20}=100\)
\(\Rightarrow\orbr{\begin{cases}x=-20\\x=20\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}y=-40\\y=40\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}z=-50\\z=50\end{cases}}\)
b/ \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{x-2y+3z-1+4-9}{2-6+12}=1\)
\(\Rightarrow\hept{\begin{cases}x=3\\y=5\\z=7\end{cases}}\)