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a: =3x^3+12x-9-2x^4-8x^2+6x
=-2x^4+3x^3-8x^2+18x-9
b: \(=\dfrac{-2x^4+x^3+6x^3-3x^2}{2x+1}=-x^3+3x^2\)
\(=-\dfrac{7}{2}x+1+\dfrac{5}{4}x-3-\dfrac{1}{2}x\left(2x^2+x-2x-1\right)\)
\(=\dfrac{-9}{4}x-2-x^3-\dfrac{1}{2}x^2+x^2+\dfrac{1}{2}x\)
\(=-x^3+\dfrac{1}{2}x^2-\dfrac{7}{4}x-2\)
a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)
=-4/3x^2+8/3-10/3
=-4/3x^2-2/3
d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)
\(=3x^2+9x+22+\dfrac{68}{x-3}\)
\(1,\frac{x+1}{x-2}=\frac{3}{4}\)
\(\Rightarrow3x-6=4x+4\)
\(\Rightarrow3x-4x=4+6\)
\(\Rightarrow-x=10\Leftrightarrow x=-10\)
\(2,\frac{x-1}{3}=\frac{x+3}{5}\)
\(\Rightarrow5x-5=3x+9\)
\(\Rightarrow5x-3x=9+5\)
\(\Rightarrow2x=14\Leftrightarrow x=7\)
\(3,\frac{2x+3}{24}=\frac{3x-1}{32}\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow72x-64x=24+96\)
\(\Rightarrow8x=120\)
\(\Rightarrow x=15\)
Ta có: \(\frac{2x-2}{3}=\frac{7x+3}{2-1}\)
\(\Leftrightarrow\frac{2x-2}{3}=7x+3\)
\(\Leftrightarrow2x-2=21x+9\)
\(\Leftrightarrow19x=-11\)
\(\Rightarrow x=-\frac{11}{19}\)
\(\frac{2x-2}{3}=\frac{7x+3}{2-1}\Leftrightarrow\frac{2x-2}{3}=7x+3\)
\(\Leftrightarrow2x-2=21x+9\Leftrightarrow-19x=11\Leftrightarrow x=-\frac{11}{19}\)