=>5x-19+3x+24=125

=>8x+5=125

hay x=15

( 5 x − 19 ) + ( 3 x + 24 ) = 125

. 5 x − 19 + 3 x + 24 = 125

8 x + 5 = 125

8 x = 120

x = 15

\(\frac{24+x}{50+x}=\frac{3}{5}\Rightarrow\left(24+x\right)\cdot5=\left(50+x\right)\cdot3\)3 

hay \(120+24x=150+50x\)

suy ra 50x-24x=150-120 hay 26x = 30 \(\Rightarrow x=\frac{26}{30}\)

\(a)x=\dfrac{1}{4}+\dfrac{5}{13}=\dfrac{33}{52}.\\ b)\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}.\\ \Leftrightarrow\dfrac{x}{3}=\dfrac{11}{21}.\\ \Leftrightarrow\dfrac{7x}{21}=\dfrac{11}{21}.\\ \Rightarrow7x=11.\\ \Leftrightarrow x=\dfrac{11}{7}.\\ c)\dfrac{x}{3}=\dfrac{16}{24}+\dfrac{24}{36}=\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}.\\ \Rightarrow x=4.\\ d)\dfrac{x}{15}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{13}{15}.\\ \Rightarrow x=13.\)

x=18 nha

x+x/3=24

3x/3+x/3=24

3x+x/3=24

x(3+1)=72

4x=72

x=72:4

x=18

x + \(\frac{x}{3}=24\)

x( 1 + \(\frac{1}{3}\)) = 24

X x \(\frac{4}{3}=24\)

          X  = 24 : \(\frac{4}{3}\)

          X = \(24x\frac{3}{4}\)

          X = 18

\(\Rightarrow x\left(1+\frac{1}{3}\right)=24\)

\(\Rightarrow x\times\frac{4}{3}=24\)

\(\Rightarrow x=18\)

mọi số tự nhiên đều có thể viết dưới dạng phân số có mâu số là 1 

vậy x/1 = x/3 = 24 

=> 3x/3 + x/3 = 24

                4x/3 = 24 

                  4x = 72 

                    x =18

\(x+\frac{x}{3}=24\)

\(\frac{4x}{3}=24\)

=> \(x=24:\frac{4}{3}=18\)

\(\left(5x-19\right)+\left(3x+24\right)=125\)

.\(5x-19+3x+24=125\)

\(8x+5=125\)

\(8x=120\)

\(x=15\)

ta có x *8 =125+19-24

        x*8=120

        x=120\8

       x=15

\(\Rightarrow x\left(x+3\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Rightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=24\)

Đặt \(x^2+3x+1=t\)

\(\Rightarrow\left(t-1\right)\left(t+1\right)=24\)

\(\Rightarrow t^2-1=24\Rightarrow t^2=25\Rightarrow t=5;-5\)

Xét t=5 thì \(x^2+3x+1=5\Rightarrow x^2+3x-4=0\)

\(\Rightarrow x^2-x+4x-4=0\)

\(\Rightarrow x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x-1\right)=0\Rightarrow x=-4;1\)

Xét t=-5 ta có

\(x^2+3x+1=-5\Rightarrow x^2+3x+6=0\)

\(\Rightarrow x_1=\frac{-3+\sqrt{15}i}{2};x_2=\frac{-3-\sqrt{15}i}{2}\)

mà \(x\in Z\)nên x=-4;1