gio con noc ha ?!

<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x

      2x^2-8x-17-2x^2-2=0

     -8x-19=0

x=-19/8

(x-2).(2x+1)-5(x+3)=2x(x-3)+4(1+2x)-2(1+x)

3x-4x+x-2-5x-15=3x-6x+4+8x-2-2x

-5x-17=3x+2

-19=8x

-19/8=x

Vậy x=-19/8

=> \(\left|2x+1\right|=3x-2\)

TH1 : với   \(x\ge\frac{-1}{2}\) ta có :

\(2x+1=3x-2\)

<=> \(x=3\)( thoả mãn ) 

TH2 : với   \(x< \frac{-1}{2}\) ta có :

\(-2x-1=3x-2\)

<=> \(5x=1\) <=> \(x=\frac{1}{5}\) ( ko thoả mãn ) 

Vậy \(x=3\)

\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)

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

\(2x^2+x-4x-2-5x-15=2x^2-6x+4+8x-2-2x\)

\(x-4x-2-5x-15=-6x+4+8x-2-2x\)

\(\Rightarrow-8x-17=2\)

\(-8x=19\Rightarrow x=-\dfrac{19}{8}\)

Vậy \(x=-\dfrac{19}{8}\)

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

\(\Rightarrow2x^2+x-4x-2-5x-15-2x^2+6x-4-8x+2+2x=0\\ \Leftrightarrow-8x-19=0\\ \Leftrightarrow-8x=19\\ \Leftrightarrow x=-\dfrac{19}{8}\)

=> 4x^2  - 12x + 4 = 2x^2 - 2x - 2 - 2x^2 - 2x - 13

=> 4x^2 - 12x + 4 = - 4x - 15

=> 4x^2 - 12x + 4x + 4 + 15 = 0 

=> 4x^2 - 8x + 19 = 0

Đề sai 

|x-1,5| + 2|x-1,5| =7,5

3|x-1,5| =7,5

|x-1,5| = 2,5

+ x -1,5 = 2,5 => x =4

+ x-1,5 = -2,5 => x =-1

                                                          \(GIẢI\)

\((\frac{9}{2}-2x)\times\frac{-11}{7}=\frac{11}{14}\)

\(\frac{9}{2}-2x=\frac{11}{14}:\frac{-11}{7}\)

\(\frac{9}{2}-2x=\frac{-1}{2}\)

            \(2x=\frac{9}{2}-(\frac{-1}{2})\)

            \(2x=5\)

            \(x=5:2\)       

            \(x=2,5\)

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