\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4.5\right)=3.5\)

\(=>2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4.5-3.5=0\)

\(=>-6x-8=0\)

\(=>-2\left(3x+4\right)=0\)

\(=>3x+4=0\)(vì \(-2\ne0\))

\(=>x=\frac{-4}{3}\)

\(x=\frac{-4}{3}\)

#)Giải :

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Leftrightarrow\left(2x^2-x\right)\left(x+5\right)-2x^3-9x^2-x-4,5=3,5\)

\(\Leftrightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)

\(\Leftrightarrow-5x-4,5=3,5\)

\(\Leftrightarrow-5x=8\)

\(\Leftrightarrow x=-\frac{8}{5}\)

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Leftrightarrow\left(2x^2-x\right)\left(x+5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\)

\(\Leftrightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)

\(\Leftrightarrow-6x=8\)

\(\Leftrightarrow x=\frac{-8}{6}=\frac{-4}{3}\)

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Rightarrow\left(2x^2-x\right)\left(x+5\right)-2x^3-9x^2-x-4,5=3,5\)

\(\Rightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)

\(\Rightarrow-5x-4,5=3,5\)

\(\Rightarrow-5x=8\)

\(\Rightarrow x=-\dfrac{8}{5}\)

\(3x^2-3x\left(x-2\right)=36\\ \Rightarrow3x\left(x-x+2\right)=36\\ \Rightarrow6x=36\\ \Rightarrow x=6\)

\(\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\\ \Rightarrow3x^3-4x^2+2x-1+\left(4x^2-3x^3\right)=\dfrac{5}{2}\\ \Rightarrow2x-1=\dfrac{5}{2}\\ \Rightarrow x=\dfrac{7}{4}\)

 3/4-2x| 1|=7/8

7.5-3|5-2x|=-4.5

|x +4/15|+ 1/2=3.5       

Ai giúp mình với, Mình đang cần gấp

(Ai làm nhanh mình k nha)

\(2x-3+3|x-1|=4x+1.\)

\(\Leftrightarrow3|x-1|=2x+4\)

*Với x < 1 ta có phương trình: 

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

\(\Leftrightarrow-3x+3=2x+4\)

\(\Leftrightarrow5x+1=0\)

\(\Leftrightarrow x=-\frac{1}{5}\)(TM)

*Với \(x\ge1\)ta có phương trình: 

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

\(\Leftrightarrow2x-3+3x-3=4x+1\)

\(\Leftrightarrow x-7=0\)

\(\Leftrightarrow x=7\)(TM)

Vậy ............