a: \(\Leftrightarrow\left(x+12-3x\right)\left(x+12+3x\right)=0\)

=>(-2x+12)(4x+12)=0

=>x=-3 hoặc x=6

b: \(\Leftrightarrow20x^3-15x^2+45x-45=0\)

=>\(x\simeq0.93\)

d: =>-4x+28+11x=-x+3x+15

=>7x+28=2x+15

=>5x=-13

=>x=-13/5

e: \(\Leftrightarrow4x^3-12x+x=4x^3-3x+5\)

=>-9x=-3x+5

=>-6x=5

=>x=-5/6

\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=5x\left(1\right)\)

Ta có :

\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\ge\left|x+1+x+2+x+3+x+4\right|=\left|4x+10\right|\)

\(pt\left(1\right)\Leftrightarrow\left|4x+10\right|=5x\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+10=5x\\4x+10=-5x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\9x=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-\dfrac{10}{9}\end{matrix}\right.\) \(\left(thỏa.mãnx\inℚ\right)\)

`Answer:`

`1/5+2/7-1<x<\frac{13}{3}+6/5+\frac{4}{15}`

`VT =1/5+2/7-1=\frac{17}{35}-1=\frac{-18}{35}`

`VP=\frac{13}{3}+6/5+\frac{4}{15}=\frac{83}{15}+\frac{4}{15}=\frac{203}{35}`

`=>\frac{-18}{35}<x<\frac{203}{35}`

`=>-18<x<203`

Vậy `-18<x<203` với `x\inZZ`

\(\dfrac{1}{5}+\dfrac{2}{7}-1< x< \dfrac{13}{3}+\dfrac{6}{5}+\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{7}{35}+\dfrac{10}{35}-\dfrac{35}{35}< x< \dfrac{65}{15}+\dfrac{18}{15}+\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{-18}{35}< x< \dfrac{29}{5}\)

\(\Leftrightarrow\dfrac{-18}{35}< \dfrac{35x}{35}< \dfrac{203}{35}\)

\(\Leftrightarrow-18< 35x< 203\)

\(\Leftrightarrow x\in\left\{0;1;2;3;4;5\right\}\)

tick cho mk lên 180 đi