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a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)
b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)\left(2x-y\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)}{\left(2x-y\right)^2}\)
c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)
=1/3
d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{2x\left(2x-3\right)}{3\left(x+1\right)}=\dfrac{2x\left(2x-3\right)}{3\left(x+2\right)}\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
a/ \(\dfrac{4x+2}{3x^2-x}:\dfrac{x^2+3x}{1-3x}=-\dfrac{4x+2}{x\left(1-3x\right)}\cdot\dfrac{1-3x}{x^2+3x}=-\dfrac{4x^2+2}{x\left(x^2+3x\right)}\)
b/ \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2-12xy+9y^2}{1-x^2}=-\dfrac{2\left(2x+3y\right)}{1-x}\cdot\dfrac{\left(1-x\right)\left(1+x\right)}{\left(2x+3y\right)^2}=\dfrac{-2\left(x+1\right)}{2x+3y}=\dfrac{-2x-2}{2x+3y}\)
c/ \(\dfrac{x^4-xy^3}{2xy+y^2}:\dfrac{x^3+x^2y+xy^2}{2x+y}=\dfrac{x\left(x^3-y^3\right)}{y\left(2x+y\right)}\cdot\dfrac{2x+y}{x\left(x^2+xy+y^2\right)}=\dfrac{x\left(x-y\right)\left(x^2+xy+y^2\right)}{y}\cdot\dfrac{1}{x\left(x^2+xy+y^2\right)}=\dfrac{x-y}{y}\)
\(a,\left(2x-5\right)\left(5-x\right)=5\left(2x-5\right)-x\left(2x-5\right)=10x-25-2x^2+5x=15x-2x^2-25\\ b,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}=\dfrac{3x+2-3x+2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{4}{\left(3x-2\right)\left(3x+2\right)}\)
\(c,\dfrac{3}{x-3}-\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9-6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
a)\(\dfrac{x^2}{x-1}+\dfrac{1-2x}{x-1}\)
=\(\dfrac{x^2+1-2x}{x-1}\)
=\(\dfrac{x^2-2x+1}{x-1}\)
=\(\dfrac{\left(x-1\right)^2}{x-1}\)
= x - 1
b) \(\dfrac{x}{x-3}\) + \(\dfrac{-9}{x^2-3x}\)
=\(\dfrac{x}{x-3}\)+ \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x.x}{x\left(x-3\right)}\) + \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x^2+3^2}{x\left(x-3\right)}\)
=\(\dfrac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}\)
=\(\dfrac{x+3}{x}\)
#Fiona