## It is known that a > 0b > 0. Verify that a + B + 2 > = 2 (root a + root b) is urgent!

A + B + 2-2 (root a + root b)

=(√a) ²- 2√a+1+(√b) ²- 2√b+1

=(√a-1) ²+ (√b-1) ²>= 0

‡ a + B + 2 ＞ = 2 (root sign a + root sign b)

Compare sizes by subtraction

### Senior two mathematics problem: the inequality with the same solution as inequality LG (x + 1) < 0 is () a (x + 1) 2 < 1 B x + 1 < 1 C x + 1 > 1 D root sign (x + 1) < 1 Is there a solution? I can't seem to find the answer. Please be more detailed

lg（x+1）<0

It's 0. Well, there's no answer

### If the real numbers a and B satisfy the square of (A-2) + root sign (b-2a) = 0, then a + B =?

Because the square of (A-2) > = 0

Root sign (b-2a) > = 0

And the square of (A-2) + root sign (b-2a) = 0

So, only

It is satisfied when A-2 = 0 and b-2a = 0

So a = 2, B = 2A = 4

So a + B = 2 + 4 = 6

### Given that a and B are real numbers, and the root sign 2A + 6 and | b-root sign 2 | are opposite to each other, solve the equation about X (a + 2) x + b square = A-1

Root 2A + 6 and | b-root 2 | are opposite numbers to each other

However, neither root 2A + 6 nor | B - root 2 | can be less than zero

So radical 2A + 6 = |b - radical 2| = 0

Then a = - 3, B = root 2

On the equation of X (a + 2) the square of X + B = A-1

That is - x + 2 = - 4

x=6

### It is known that the real number A.B satisfies (square of A-2) + radical b-2a + 10 = 0

a-2=0 a=2

b-2a+10=0 b=-6

### Square of X - 4 square of radical 3ax + 12a = 0

(x-2√3a) ²

x-2√3a=0

x=2√3a