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a: \(=\dfrac{2x^4+x^3-5x^2-3x-3}{x^2-3}\)
\(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)
\(=2x^2+x+1\)
b: \(=\dfrac{x^5+x^2+x^3+1}{x^3+1}=x^2+1\)
c: \(=\dfrac{2x^3-x^2-x+6x^2-3x-3+2x+6}{2x^2-x-1}\)
\(=x+3+\dfrac{2x+6}{2x^2-x-1}\)
d: \(=\dfrac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)
\(=x^2-2x-5\)
\(\frac{1}{6}\)(2x-3) = \(\frac{-1}{2}\)(x-\(\frac{1}{4}\))-\(\frac{2}{3}\)
\(\frac{1}{3}\)x - \(\frac{1}{2}\)=\(\frac{-1x}{2}\)-\(\frac{-1}{8}\)-\(\frac{2}{3}\)
\(\frac{1x}{3}\)-\(\frac{1}{2}\)=\(\frac{-1x}{2}\) - \(\frac{19}{24}\)
\(\frac{1x}{3}\) - \(\frac{1}{2}\) - \(\frac{-1x}{2}\) - \(\frac{19}{24}\) =0
\(\frac{5x}{6}\) - \(\frac{7}{24}\)=0
\(\frac{5x}{6}\) = \(\frac{7}{24}\)
x = \(\frac{7}{20}\)
\(5x\left(x-2\right)-3\left(x-1\right)=20x^2-15x\left(2x+1\right)-24\)
\(\Rightarrow5x^2-10x-3x+3=20x^2-30x^2-15x-24\)
\(\Rightarrow5x^2-13x+3=-10x^2-15x-24\)
\(\Rightarrow5x^2+10x^2-13x+15x+3+24=0\)
\(\Rightarrow15x^2+2x+27=0\)
Ta có:
\(\Delta=2^2-4\cdot15\cdot27==-1616< 0\)
Nên pt vô nghiệm
\(5x\left(x-2\right)-3\left(x-1\right)=20x^2-15x\left(2x+1\right)-24\\ \Leftrightarrow5x^2-10x-3x+3=20x^2-30x^2-15x-24\\ \Leftrightarrow5x^2-20x^2+30x^2-10x-3x+15x+3+24=0\\ \Leftrightarrow15x^2+2x+27=0\\ \Leftrightarrow15x^2-2.x.\sqrt{15}+\dfrac{2}{15}+\dfrac{403}{15}=0\\ \Leftrightarrow\left(\sqrt{15}x+\dfrac{\sqrt{30}}{15}\right)^2+\dfrac{403}{15}=0\left(Vô.lí\right)\\ Vậy:Không.có.x.thoả\)
a: \(=\left(4xy^2+2xy^2\right)+\left(3x^2y-3x^2y\right)=6xy^2\)
b: \(=xy\left(\dfrac{1}{5}+\dfrac{1}{3}\right)+xy^2\left(\dfrac{4}{3}-\dfrac{2}{5}\right)=\dfrac{8}{15}xy+\dfrac{14}{15}xy^2\)
d: \(=\dfrac{-4}{9}\cdot\dfrac{3}{2}\cdot xy^2\cdot xy^3=-\dfrac{2}{3}x^2y^5\)
-2/5x-1/3=3/4
-2/5x=3/4+1/3
-2/5x = 13/12
x=13/12:2/5
x=-65/24
-13/20x+1/2=-2/5
-13/20x=-2/5-1/2
-13/20x=-9/10
x=-9/10:13/20
x=-18/13
4/3-|2x-1/2|=1
|2x-1/2|=1-4/3
|2x-1/2|=-1/3
<=> 2x-1/2=1/3 hoặc 2x-1/2=-1/3
TH1:
2x-1/2=1/3
2x=1/3+1/2
2x=5/6
x=5/6:2
x=5/12
TH2:
2x-1/2=-1/3
2x=-1/3+1/2
2x=1/6
x=1/6:2
x=1/12