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\(\dfrac{2x+4}{2015}-\dfrac{2x+4}{2016}=\dfrac{2x+4}{2017}-\dfrac{2x+4}{2018}\)
\(\Rightarrow\left(2x+4\right)\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)=\left(2x+4\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
Vì \(\dfrac{1}{2015}-\dfrac{1}{2016}\ne\dfrac{1}{2016}-\dfrac{1}{2017}\) nên 2x + 4 = 0
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
Vậy, x = -2
\(\dfrac{2x+4}{2015}-\dfrac{2x+4}{2016}=\dfrac{2x+4}{2017}-\dfrac{2x+4}{2018}\)
\(\Rightarrow\left(2x+4\right)\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)=\left(2x+4\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
Vì \(\dfrac{1}{2015}-\dfrac{1}{2016}\ne\dfrac{1}{2016}-\dfrac{1}{2017}\) nên \(2x+4=0\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
Vậy, x = -2
18, \(\frac{x}{2}+\frac{x^2}{8}=0\Leftrightarrow4x+x^2=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow x=-4;x=0\)
19, \(4-x=2\left(x-4\right)^2\Leftrightarrow\left(4-x\right)-2\left(4-x\right)^2=0\)
\(\Leftrightarrow\left(4-x\right)\left[1-2\left(4-x\right)\right]=0\Leftrightarrow\left(4-x\right)\left(-7+2x\right)=0\Leftrightarrow x=4;x=\frac{7}{2}\)
20, \(\left(x^2+1\right)\left(x-2\right)+2x-4=0\Leftrightarrow\left(x^2+1\right)\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3>0\right)=0\Leftrightarrow x=2\)
21, \(x^4-16x^2=0\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\Leftrightarrow x=0;x=\pm4\)
22, \(\left(x-5\right)^3-x+5=0\Leftrightarrow\left(x-5\right)^3-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\Leftrightarrow\left(x-5\right)\left(x-6\right)\left(x-4\right)=0\Leftrightarrow x=4;x=5;x=6\)
23, \(5\left(x-2\right)-x^2+4=0\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\Leftrightarrow x=2;x=3\)
h) \(=8-12y+6y^2-y^3\)
i) \(=8y^3-125\)
j) \(=27y^3+64\)
k) \(=x^3-9x^2+27x-27+8-12x+6x-x^3=-9x^2+21x-19\)