The square of a + root sign 7a = - 1, find the value of a + 1 / A

The square of a + root sign 7a = - 1, find the value of a + 1 / A

The square of a + the root sign 7a = - 1,
The square of a + 1 = - radical 7a
Divide both sides by a at the same time
A + 1 / a = - radical 7

The square of a is multiplied by the square of root 8a, and then 3a is multiplied by the square of root 50A? The square of a times the root number 8a, and then 3A times the root number 50a ＾ 2

Is your topic (a ^ 2 * (8a) + 3A * (50a)) ^ 2? (a ^ 2 * (8a) + 3A * (50a)) ^ 2 = (2a ^ 2 * (2a) + 15A * (2a)) ^ 2 = (2a ^ 2 * (2A) + 15A * (2a)) ^ 2 = (a * (2a) (2a)) ^ 2 * (2a + 15) ^ 2 = (2a ^ 3) * (4a ^ 2 + 60A + 225) = 8A ^ 5 + 120a ^ 4 + 450A ^ 3! Add: is "a ^ 2 * (8a) + 3A * * (50a ^ 2)" ^ 2 = 2 = (a ^ 2 * (8a) + 3A * * (50a ^ 2) "add: is" a ^ 2 * (8a) + 3A * * (50a ^ 2) "or" a ^ 2 * (8a) +

1. Under the root sign 18 - 9 / 2 = 2,2 / 3 times the root sign 9x + 6 times the root sign x / 4 = 3, a square times the root sign 8a, and + 3A times the root sign 50A cube= After answering the question, I will give you some points

1、√18-√9/2=3√2-3√2/2=3/2√2
2、3√9x﹢6√x/4=9√x+3√x=12√x
3、a²√8a＋3a√50a³=2a²√2a+15a²√2a=17a²√a

What is the value of root sign (25 + a) - root sign (16-3a) + root sign (8a + 4) - root sign (- a square)? By the way, please teach me how to do this kind of topic. Thank you

Because there is a root sign (- a square), so that the root is meaningful, to - A ^ 2 > = 0, so a = 0
therefore
Radical (25 + a) - radical (16-3a) + radical (8a + 4) - radical (- a squared)
=Root 25 - root 16 + root 4 - 0
= 5 - 4 + 2 - 0
= 3.

Detailed process is required for (a 2 × root number 8a) + (3a root number 50A 3)

The original formula = a 2 × 2 √ (2a) + 3a × 5A √ (2a)
=2a²√(2a)+15a²√(2a)
=17a²√(2a)

If the root sign 2x + y + absolute value x? - 9 = 0, find the cube root of 3x + 6y

When x = 3 y = - 6 or x = - 3 y = 3, y = - 6, when 3x + 6y = 9-36 = - 27 3 √ (3x + 6y) = - 3x = - 3 y = 6, 3x + 6y = - 3, y = 6, y = - 6, respectively

If the absolute value of 2x - radical 2 + the square of (y + radical 5) is 0, then XY? - radical 2 =?

Absolute values and complete squares are nonnegative
that
2x-√2=0
x=√2/2
y+√5=0
y=-√5
xy^2-√2=√2/2×(-√5)^2-√2=5√2/2-√2=3√2/2

If (2x-1) 2 and Y-2 are opposite numbers to each other, then the value of root XY is——————

y=|√7-3|=3-√7
(2x-1) 2 and radical Y-2 are opposite numbers to each other,
Then 2x-1 = 0, x = 1 / 2
y-2=0 y=2
Radical xy = 1

If the radical x + Y-3 + absolute value XY + X + Y-2 = 0, find X / y + Y / X Root x + Y-3 + absolute value XY + X + Y-2 = 0, find X / y + Y / X

There is a question
x+y-3>=0
xy+x+y-2>=0
Because the radical x + Y-3 + absolute value XY + X + Y-2 = 0
Then x + Y-3 = 0
xy+x+y-2=0
xy=-1
x+y=3
x^2+y^2=11
x/y+y/x =(x^2+y^2)/xy=-11

Given the absolute value of the radical x-2y-4 + 2X-4 = 0, find the cube root of x-6y

√(x-2y-4)+|2x-4|=0
If two nonnegative numbers are added to get 0, then the two nonnegative numbers are both 0
(x-2y-4) = 0 and | 2X-4 | = 0
X-2y-4 = 0 and 2X-4 = 0
x=2,y= -1
therefore
Cube root of (x-6y) = (2 + 6) = cube root of 8 = 2