\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}=\dfrac{x-1990}{5}+\dfrac{x-1980}{15}\\ =>\dfrac{x-5}{1990}-1+\dfrac{x-15}{1980}-1=\dfrac{x-1990}{5}-1+\dfrac{x-1980}{15}-1\\ =>\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}-\dfrac{x-1995}{5}-\dfrac{x-1995}{15}=0\\ =>\left(x-1995\right).\left(\dfrac{1}{1990}+\dfrac{1}{1980}-\dfrac{1}{5}-\dfrac{1}{15}\right)=0\\ =>x-1995=0\\ =>x=1995\)

`x/3=-5/21`

`=>x=(-5/21).3`

`=>x=-15/21=-5/7`
Vậy : `x=-5/7`

\(đk:x\ne-5\\ \dfrac{x-3}{x+5}=\dfrac{5}{7}\\ =>7\left(x-3\right)=5.\left(x+5\right)\\ =>7x-21=5x+25\\ =>7x-5x=25+21\\ =>2x=46\\ =>x=23\left(t/m\right)\)

\(\dfrac{x-3}{x+5}=\dfrac{5}{7}\left(x\ne-5\right)\)

\(\Rightarrow7\left(x-3\right)=5\left(x+5\right)\)

\(\Rightarrow7x-21=5x+25\)

\(\Rightarrow7x-21-5x-25=0\)

\(7x-5x-21-25=0\)

\(2x-46=0\)

\(2x=0+46=46\)

\(x=\dfrac{46}{2}=23\left(tmđk\right)\)

Áp dụng dãy tỉ số bằng nhau 

\(\dfrac{x}{12}=\dfrac{y}{13}=\dfrac{z}{15}=\dfrac{3x}{36}=\dfrac{2y}{26}=\dfrac{3x+2y}{36+26}=\dfrac{62}{62}=1\)

Khi đó x = 12 ; y = 13 ; y = 15

`x : y : z= 3:4:5`

`=> x/3 = y/4 = z/5 <=> x^2/9 = y^2/16 = z^2/25`

Áp dụng dãy tỉ số bằng nhau:

`x^2/9 = y^2/16 = z^2/25 = (2x^2 + 2y^2 - 3z^2)/(18 + 32 - 75) = -100/-25 = 4`.

`=> {(x^2/9 = 4 => x = +-6), (y^2/16 =4 <=> x = +-8), (z^2/25 = 4 => z = +-10):}`

Vậy ...