Question

Help me please I need this for my test tomorrow pleaseeeee!!!!!!

Answer 1

st

Just substitute in the given values

Answer 2

Explanation:

Hey, there!!

You can solve it simply,

1.

Given,

x= 1 , y=2 , z= 3

To solve:

`= \frac{1}{ {x}^(2) } - (x)/(y + z)`

`= \frac{1}{ {1}^(2) } - (1)/(2 + 3)`

`= 1 - (1)/(5)`

Taking LCM,

`= (5 - 1)/(5)`

Therefore, the final answer is 4/5.

2.

Given,

x= 1 , y= -2, z= -1.

To solve:

`= \frac{1}{ {x}^(2) } - (x)/(y + z)`

`= \frac{1}{ {1}^(2) } - (1)/( - 2 - 1)`

Taking LCM,

`= ( - 3 - 1)/( - 3)`

`= ( - 4)/( - 3)`

Cancelling the minus sign,

`(4)/( 3)`

is the final answer.

3.

Given,

x= -3, y= -1, z= 4

To solve:

`= \frac{1}{ {x}^(2) } - (x)/(y + z)`

`= \frac{1}{ {( - 3)}^(2) } - ( - 3)/( - 1 + 4)`

`= (1)/(9) - ( - 1)`

`= (1 + 9)/(9)`

Therefore, the answer is 10/9.

4.

Given,

x= 1/2, y=1, z=1.

To solve:

`= \frac{1}{ {x}^(2) } - (x)/(y + z)`

`= \frac{1}{ ( { (1)/(2)) }^(2) } - ( (1)/(2) )/(1 + 1)`

Simplifying them ,

`1 * (4)/(1) - (1)/(2) * 2`

`= 4 - 1`

Therefore, the answer is 3.

Hope it helps...