Let A.B be a rational number and A.B satisfy the equation a + 2B + a root sign 5 = 8 + 2 radical sign 5. Find the square of 3A + B

Let A.B be a rational number and A.B satisfy the equation a + 2B + a root sign 5 = 8 + 2 radical sign 5. Find the square of 3A + B

A + 2B + a root sign 5 = 8 + 2 root sign 5 A + 2b-8 = (2-A) root sign 5, because rational number does not = irrational number
So a + B-8 = 0 2-A = 0 solution a = 2 b = 3

The teaching of addition, subtraction, multiplication and division of square roots in Volume 1 of grade 2 is as follows: root 50 - root 8; root 50 times root 18

The addition and subtraction operation of quadratic radical is to reduce each quadratic radical to the simplest quadratic radical, and then merge the same kind of quadratic radical,
The multiplication and division operation is carried out according to the following rules:
(√a)×(√b)=√(a×b),(a≥0,b≥0)
(√a)÷(√b)=√(a÷b) (a≥0,b＞0)
The result should be reduced to the simplest quadratic root
√50-√8
＝5√2-2√2
＝3√2
√50×√18
＝√（50×18）
＝√900
＝30

How to use the root sign, that is to say, how to calculate the root sign, what is its function? For example, addition, subtraction, multiplication and division all have their own skills, so what is the root used? A little more detailed, building master, primary school students

Corresponding to square, for example: 3 square = 3 x 3 = 9; root 9 = 3; example: 5 square = 5 x 5 = 25; root 25 = 5;
Example: given the square garden area of 25 square meters, solve the length of each side;
Length of each side = root number (area) = root number (25) = 5 meters
Example: given the circular garden area of 25 square meters, solve the length of radius. Area = π x length of radius x length of radius;
Length of radius x length of radius = area / π
The length of radius = root number (area / π) = root number (25 / 3.1416) = root number (7.957747) = 2.8209479 meters (because 2.820947 x 2.820947 = 7.957747)
Hand held calculators are usually used to calculate the root value

Let P = 1 / 2 (a + B + C), then according to the formula s = radical (P-A) (P-B) (P-C) It is known that a = 11, B = 4, C = 9 Find: the length of the line on side B

a=11,b=4,c=9.
P=1/2(a+b+c)=12
p-a=1
p-b=8
p-c=3
therefore
Area = root number 1 × 3 × 8 = 2 root number 6
therefore
The height of edge B = 2 × 2 root sign 6 △ 4 = root 6

In the formula C = the square of root a + the square of B, given a = 3, B = 4, find the length of C

c=√3²+4²=√9+16=√25=5
If there is anything you don't understand, you can ask,

In the formula a = the square of root c - the square of B, given C = 13, B = 12, find the length of A

a=√(c²-b²)=√(13²-12²)=√25=5

Does a, B satisfy that root a plus root B equals root 275?

√a+√b=5√11
therefore
√a+√b=√11+4√11
√a+√b=2√11+3√11
Namely
√a+√b=√11+√176
√a+√b=√44+√99
So there is
a=11,b=176
a=44,b=99
a=99,b=44
a=176,b=11

What is the root sign ((a + b) ^ 2 - (a ^ 2-B ^ 2) ^ 2) equal to

√((a+b)²-(a²-b²)² )
=√﹛﹙a＋b﹚﹙1﹢a-b﹚﹙a+b﹚﹙1-a+b﹚﹜
=|a+b|√﹛1－﹙a﹣b﹚²﹜

Given that 1-B plus B times root 1-A is equal to 1, find a + B

The title is incomplete, isn't it
a. If b > 0, a-fold root 1 minus b-fold root 1 minus A-Square equals 1, proving that A-plus b-square equals 1
Let x = (a, b), y = (√ (1-B ^ 2), √ (1-A ^ 2))
X*Y=a√(1-b^2)+b√(1-a^2)=√(a^2+b^2)*√(1-b^2+1-a^2)*cosθ
Because θ∈ [0, π / 2], 0 < = cos, θ < = 1
So the above equation, right side (remove cos θ) > = left side, that is (both sides are squared at the same time, and then the known equation is replaced with the following equation)
(a^2+b^2)(2-a^2-b^2)>=1
Let t = a ^ 2 + B ^ 2
t(2-t)>=1
t^2-2t+1<=0
(t-1)^2<=0
So t = 1, that is a ^ 2 + B ^ 2 = 1
(the solution before t can also be done by Cauchy inequality)

9 / 4 of the root sign B (a is greater than or equal to 0, B is greater than or equal to 0)

Root 9 / 4 (square a) B
=3/2a √b
If there is anything you don't understand, you can ask,