Question

[1/2x-4/3y-1/4z]²
Can anybody tell me the ans step by step please ​

Answer 1

Step-by-step explanation:

1/2x+2/3y-1/4z=4 (P1)

3/4x-1/2y-1/3z=23 (P2)

1/3x+1/4y+2/3z=-15 (P3)

Kita pakai metode eliminasi.

Kita eliminasi x pada P1 dan P2 :

1/2x+2/3y-1/4z=4 ( x 3)

3/4x-1/2y-1/3z=23 (x2)

-------------------------

3/2x+6/3y-3/4z = 12

6/4x-2/2y-2/3z = 46

----------------------------_

3y -1/12z = -34 (P4)

Kita eliminasi x dari P1 dan P3

1/2x+2/3y-1/4z=4 (x1/3)

1/3x+1/4y+2/3z=-15 (x1/2)

--------------------------

1/6x+2/9y-1/12z=4/3

1/6x+1/8y+2/6z=-15/2

------------------------------_

7/72y-5/12z = 53/6 (P5)

Kita eliminasikan z dari P4 dan P5

3y - 1/12z = -34 (x5)

7/72y-5/12z = 53/6

--------------------------

15y - 5/12z = -170

7/72y-5/12z=53/6

--------------------------_

1073/72y = -1073/6

y = -1073/6 x 72/1073

y = -12

Masukkan nilai y diatas ke dalam P4

3y -1/12z = -34

3(-12) - 1/12z = -34

-36 - 1/12z = -34

-1/12z = 2

z = -24

Masukkan nilai y dan z diatas ke dalam P1

1/2x+2/3y-1/4z=4

1/2x + 2/3(-12)-1/4(-24) = 4

1/2x -8 + 6 = 4

1/2x = 6

x = 12

Jadi nilai x = 12

y = -12

z = -24

Maka hp : {x,y,z} = {12,-12,-24}

Answer 2

Answer:

using the identity( a+b+c)^2 = a^2 +b^2+c^2+2×x×y+2yz+2zx