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\(x^4-x^3+1+x^3=0\)

\(x^4+1=0\)

mà \(x^4\ge0\) với mọi x

      1 > 0

=> \(x^4+1>0\) với mọi x

=> Đa thức Q(x) vô nghiệm

\(4x^4+9=0\)

\(4x^4=-9\)

mà \(4x^4\ge0\)  với mọi x

=> Vô nghiệm

\(x^2-3x-4=x^2+x-4x-4=x\left(x+1\right)-4\left(x+1\right)=\left(x-4\right)\left(x+1\right)\)

\(M=\dfrac{3.20182018.20182020-5.20182017-2.20182018^2-5}{20182018}\)

\(M=\dfrac{3.20182018.20182020-2.20182018^2-5.20182017-5}{20182018}\)

\(M=\dfrac{20182018\left(60546060-40364036\right)-5\left(20182017+1\right)}{20182018}\)

\(M=\dfrac{20182018.20182024-5.20182018}{20182018}\)

\(M=\dfrac{20182018\left(20182024-5\right)}{20182018}\)

\(M=20182019\)

9158-(300+158)

=9158-300-158

=(9158-158)-300

=9000-300

=8700

m>n

=> 3m>3n

\(\Leftrightarrow\) 3m+2>3n

\(\dfrac{2x+1}{5}-\dfrac{3+x}{10}=6\)

\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{10}-\dfrac{3+x}{10}=\dfrac{60}{10}\)

\(\Leftrightarrow4x+2-3-x=60\)

\(\Leftrightarrow3x-1=60\)

\(\Leftrightarrow3x=61\)

\(\Leftrightarrow x=\dfrac{61}{3}\)

Vậy....

\(\left|x-2\right|=3x+5\)

Nếu \(x-2\ge0\) thì \(x\ge2\)

\(\Leftrightarrow x-2=3x+5\)

\(\Leftrightarrow x-3x=5+2\)

\(\Leftrightarrow-2x=7\)

\(\Leftrightarrow x=-\dfrac{7}{2}\) (ktm)

Nếu \(x-2< 0\) thì \(x< 2\) 

\(\Leftrightarrow x-2=-3x-5\)

\(\Leftrightarrow x+3x=-5+2\)

\(\Leftrightarrow4x=-3\)

\(\Leftrightarrow x=-\dfrac{3}{4}\left(tm\right)\)

Vậy.....