10 points best answer, help?

You need to use parentheses.

7/(5-√3) = 7(5+√3)/((5-√3)(5+√3))
= (35+7√3)/(5²-√3²)
= (35+7√3)/22
= 35/22 + (7/22)√3
:::::
√(3x-2) + 7 = 20
√(3x-2) = 13

square both sides of equation
3x-2 = 169
3x = 171
x = 57
:::::
3x² + 7 = -29
3x² = -36
x² = -12
x = ±√(-12)
 = ±√12√(-1)
 = ±(2√3)i

1.

7 [ 5 + √3 ]
-------------------------
[ 5 - √3 ] [ 5 + √3 ]

7 [ 5 + √3 ]
----------------
22

2.
√ [3x - 2 ] = 13
3x - 2 = 169
3x = 171
x = 57

3.
3x² = - 36
x² = - 12
x = 12 i²
x = ± 2√3 i

1) Write the expression in simplified radical form.
7/5-\sqrt, 3

7/(5 - √3)

= [ 7(5 + √3)]/[5 - √3)(5 + √3)]

= 7(5 + √3)/(25 - 3)

= 7/22(5 + √3)


2) Solve the equation for x.
\sqrt(3x-2)+7=20

√(3x - 2) = 20 - 7

√(3x - 2) = 13

3x - 2 = 169

3x = 171

x = 57


3) Solve the equation for x.
3x^(2)+7=-29

3x^2 = - 36

x^2 = - 12

x^2 = 12i^2

x = √12i^2

= ±2√3 i

1)
7/(5 - sqrt(3))

Apply the conjugate top & bottom .
The conjugate is '5 + sqrt(3))'

Hence
1(5 + sqrt(3)) / ( 5 - sqrt(3))(5 + sqrt(3))
Apply FOIL to the bottom.
Hence
(5 + sqrt(3)) / ( 25 +5sqrt(3) - 5 sqrt(30) - (sqrt(3))^2 )
Collect terms
(5 + sqrt(3)) / ( 25 - 3) NB +5sqrt(3) & -5sqrt(3) add out!!!
(5 + sqrt(3))/22
or
5/22 + sqrt(3)/22

2) sqrt(3 - 2) + 7 = 20
Subtract '7' from both sides.

sqrt(3x - 2) = 13
Square up both sides
(3x - 2) = 13^3 = 169
3x - 2 = 169
Add '2' to both sides
3x = 171
Divide both sides by '3'
x = 171/3 = 57
x = 57

3)
3x^2 + 7 = -29
Add 29 to both sides
3x^2 + 36 = 0
Factor '3'
3(x^2 + 12) = 0
x^2 + 12 = 0
x^2 = -12
'Unresolved' as you cannot take the 'square root of a negative number.