\(\Leftrightarrow\frac{1}{2}x+\frac{2}{3}x=x+4\)

\(\Leftrightarrow\frac{7}{6}x=x+4\)

\(\Leftrightarrow\frac{7}{6}x-x=4\)

\(\Leftrightarrow\frac{1}{6}x=4\)

\(\Leftrightarrow x=24\)

50%x + \(\frac{2}{3}\)x = x + 4 

=> x = 24

\(\text{Ta có: }\) \(50\%x+\frac{2}{3}x=x+4\)

\(\Leftrightarrow\frac{1}{2}x+\frac{2}{3}x=x+4\)

\(\Leftrightarrow x\left(\frac{1}{2}+\frac{2}{3}\right)=x+4\)

\(\Rightarrow\frac{7}{6}x=x+4\)

50%x + 2/3x = x+4

1/2x + 2/3x = x+4

(1/2 + 2/3)x = x+4

7/6x = x+4

7/6x-x =4

(7/6-1)x =4

1/6x=4

x=24

\(\frac{1}{2}\)x   +   \(\frac{2}{3}\)x =x+4

7/6x=x+4

1/6x +x=x+4

1/6x =4

x=4:1/6

x=24

=>1/2x+2/3x=x+4

x(1/2+2/3)=x+4

x7/6=x+4

x7/6-x-4=0

x(7/6-1)-4=0

=>x1/6=4

=>x=24

e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)

\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)

\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)

\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)

2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1

a: =>10|x-2|=50

=>|x-2|=5

=>x-2=5 hoặc x-2=-5

=>x=7 hoặc x=-3

b: =>|3x-12|+|5x-20|=56

=>8|x-4|=56

=>|x-4|=7

=>x-4=7 hoặc x-4=-7

=>x=11 hoặc x=-3