Question

Tìm x, y, z, biết:

\(x+y=\dfrac{1}{2};y+z=\dfrac{1}{3};z+x=\dfrac{1}{4}\)

Answer 1

Ta có: x+y=\(\dfrac{1}{2}\); y+z=\(\dfrac{1}{3}\); z+x=\(\dfrac{1}{4}\)

=> 2.(x+y+z)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)=\(\dfrac{13}{12}\)

=>x+y+z=\(\dfrac{13}{24}\)

=> x=\(\dfrac{13}{24}\) - \(\dfrac{1}{3}\)=\(\dfrac{5}{24}\)

y=\(\dfrac{13}{24}\) - \(\dfrac{1}{4}\)=\(\dfrac{7}{24}\)

z=\(\dfrac{13}{24}\) - \(\dfrac{1}{2}\)=\(\dfrac{1}{24}\)

Vậy x, y, z lần lượt là: \(\dfrac{5}{24}\), \(\dfrac{7}{24}\), \(\dfrac{1}{24}\).

Answer 2

Cộng theo vế:

\(2\left(x+y+z\right)=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{13}{12}\Leftrightarrow x+y+z=\dfrac{13}{24}\)

Trừ ra là được