Given a > 0, b > O, then the minimum value of 1 / A + 1 / B + 2 √ AB is? two 2√2 C.4 D.5

Given a > 0, b > O, then the minimum value of 1 / A + 1 / B + 2 √ AB is? two 2√2 C.4 D.5

Sorry, I made a mistake
1/a+1/b+2√ab
>=2√(1/ab)+2√ab
>=2√(2√(1/ab)*2√ab)
=4
The minimum value is 4

The mathematician Gauss once studied such a problem in school, 1 + 2 + 3 + +10=？ After research, the general conclusion of this problem is 1 + 2 + 3 + +N = 12n (n + 1), where n is a positive integer. Now let's study a similar problem: 1 × 2 + 2 × 3 + +n（n+1）=？ Observe the following three special equations: 1 × 2 = n (1 × 2 × 3-0 × 1 × 2) 2 × 3 = x (2 × 3 × 4-1 × 2 × 3) 3 × 4 = n (3 × 4 × 5-2 × 3 × 4). Add the two sides of the three equations to get 1 × 2 + 2 × 3 + 3 × 4 = m × 3 × 4 × 5 = 20 +100×101=______ (2) 1 × 2 + 2 × 3 + +N (n + 1); (write the calculation process) (3) 1 × 2 × 3 + 2 × 3 × 4 + +n（n+1）（n+2）=______ ．

（1）∵1×2+2×3+3×4=m×3×4×5=13×4×5=20，∴1×2+2×3+… +100×101=13×100×101×102=343400；（2）∵1×2=n（1×2×3-0×1×2）=13（1×2×3-0×1×2），2×3=x（2×3×4-1×2×3）=13（2×3×4-1×2×3）...

N is a positive integer, and the lengths of three sides of a triangle are 2n & # 178; + 2n + 1,2n & # 178; + 2n, 2n + 1
And explain the reason

Note
2n²+2n+1>2n²+2n>2n+1
If it is a right triangle, there must be
(2n²+2n+1)²-(2n²+2n)²
=4n²+4n+1
=(2n+1)²
So (2n & # 178; + 2n + 1) & # 178; = (2n & # 178; + 2n) & # 178; + (2n + 1) & # 178;
From Pythagorean theorem
The triangle is a right triangle

N is a positive integer, and the lengths of three sides of a triangle are 2n ^ 2 + 2n + 1, 2n ^ 2 + 2n, 2n + 1 respectively. Judge whether the triangle is a right triangle and explain the reason

Because n is a positive integer, so 2n ^ 2 + 2n + 1 > 2n ^ 2 + 2n > 2n + 1
If the triangle is a right triangle, it can only have:
（2n^2+2n)^2+( 2n+1)^2=(2n^2+2n+1)^2
Right = (2n ^ 2 + 2n + 1) ^ 2
=(2n^2+2n)^2+2(2n^2+2n)+1
=(2n^2+2n)^2+4n^2+4n+1
=(2n^2+2n)^2+(2n+1)^2
=Right
So the triangle is a right triangle

If the length m, N and P of three line segments satisfy that the square of P = the square of M - the square of N, is the triangle formed by these three line segments a right triangle? Please explain the reason

P^2=M^2-N^2
P^2+N^2=M^2
Satisfy the Pythagorean theorem, is a right triangle

Can a, B, C form a right triangle if a = m square - n square, B = 2Mn, C = m square + n square?
There is a reward for the right answer

c²-a²=(c+a)(c-a)
=(m²+n²+m²-n²)(m²+n²-m²+n²)
=2m²×2n²
=(2mn)²=b²
So a & sup2; + B & sup2; = C & sup2;
So you can make right triangles

Prove: M & sup2; - N & sup2;, 2Mn, M & sup2; + n & sup2; (m, n are natural numbers, and M > n > 0) is the major side length of right triangle

Because (m ^ 2 + n ^ 2) ^ 2 = (m ^ 2-N ^ 2) ^ 2 + (2Mn) ^ 2
And then it's spread out on the left and right
Step by step, turn it upside down

The three sides of a triangle are 2n (n + 1), 2n + 1, 2n ^ 2 + 2n + 1 (n > 0). Try to judge whether the triangle is a right triangle

Square these three sides directly, and then see if a ^ 2 + B ^ 2 = C ^ 2 is satisfied
If satisfied, it is a right triangle

Under what condition is the point Q [2,3] on the curve where the equation is the square of AX + the square of ay = B [A is not equal to o, B is not equal to 0]?

I understand that your expression is (AX) ^ 2 + (Ay) ^ 2 = B, which is reduced to (x) ^ 2 + (y) ^ 2 = B / (a ^ 2)
This curve is a circle whose center is at (0,0) radius r = B ^ (0.5) / A
When radius r = (2 ^ 2 + 3 ^ 2) ^ (0.5) = (13) ^ (0.5), the given condition is satisfied

One part of X-1 is equal to the square of x minus one part
Please answer the following question again: the square of X + 5 parts of X + 2 = 3 parts of X + 1

1/(x-1)=1/(x^2-1)
1/(x-1)-1/(x^2-1)=0
(x+1)/(x-1)(x+1)-1/(x-1)(x+1)=0
(x+1-1)/(x-1)(x+1)=0
x/(x-1)(x+1)=0
x=0
It is proved that x = 0 is the solution of the equation