Question

Tìm x,y,z biết:

\(x+y=\dfrac{1}{2}\)

\(y+z=\dfrac{1}{3}\)

\(z+x=\dfrac{1}{4}\)

mong các bn giúp mik!

Answer 1

Ta có: \(\left\{{}\begin{matrix}x+y=\dfrac{1}{2}\left(1\right)\\y+z=\dfrac{1}{3}\left(2\right)\\z+x=\dfrac{1}{4}\left(3\right)\end{matrix}\right.\)

Cộng (1); (2); (3) vế theo vế ta được:

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

=> \(2\left(x+y+z\right)=\dfrac{13}{12}\)

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

+) Mà \(x+y=\dfrac{1}{2}\) => \(z=\dfrac{13}{24}-\dfrac{1}{2}\) = \(\dfrac{1}{24}\)

+) Mà y + z = \(\dfrac{1}{3}\) => \(\left\{{}\begin{matrix}y=\dfrac{1}{3}-\dfrac{1}{24}\\x=\dfrac{13}{24}-\dfrac{1}{3}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=\dfrac{7}{24}\\x=\dfrac{5}{24}\end{matrix}\right.\) (TM)

Vậy \(x=\dfrac{5}{24};y=\dfrac{7}{24};z=\dfrac{1}{24}\)

P/s: Bài này có nhiều cách giải lắm!

Answer 2

x + y=1/2

y + z=1/3

z + x=1/4

=> x + y + y + z + z + x = 1/2 + 1/3 + 1/4 = 13/12

hay: 2(x + y + z ) = 13/12

x + y + z = 13/12 :2

x + y + z = 13/24

x = 13/24 - 1/3 = 5/24

y = 13/24 - 1/4 = 7/24

z = 13/24 - 1/2 = 1/24

Vậy ...

Answer 3

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

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

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

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

=>\(z=\dfrac{1}{24}\) (1)

Thay (1) vào z+x=\(\dfrac{1}{4}\)

=>x=\(\dfrac{5}{24}\)

Thay (1) vào y+z=\(\dfrac{1}{3}\)

=>y=\(\dfrac{7}{24}\)

Answer 4

Thôi ! Để khác biệt tui sẽ lm 1 cách nx!

+) \(x+y=\dfrac{1}{2}\) => \(x=\dfrac{1}{2}-y\) (1)

+) \(y+z=\dfrac{1}{3}\) => \(z=\dfrac{1}{3}-y\) (2)

Mà \(x+z=\dfrac{1}{4}\) (3)

Từ (1); (2); (3) => \(\dfrac{1}{2}-y+\dfrac{1}{3}-y=\dfrac{1}{4}\)

=> \(-2y=\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\)

=> \(-2y=\dfrac{-7}{12}\) => \(y=\dfrac{-7}{12}:\left(-2\right)=\dfrac{7}{24}\)

=> \(\left\{{}\begin{matrix}x=\dfrac{1}{2}-\dfrac{7}{24}=\dfrac{5}{24}\\z=\dfrac{1}{3}-\dfrac{7}{24}=\dfrac{1}{24}\end{matrix}\right.\)

Vậy .......................