If ABC is the triangle ABC three sides, simplify the square of (a + B-C) under the root sign - the square of (a-b-c) under the root sign

If ABC is the triangle ABC three sides, simplify the square of (a + B-C) under the root sign - the square of (a-b-c) under the root sign

The square of (a + B-C) under the root sign is its absolute value, that is, a + b-c. because the sum of the two sides of the triangle is greater than the third side, the latter half is reduced to - A + B + C
So the answer is 2a-2c

Let a, B, C be the three sides of the triangle ABC, and simplify the square of root sign (a + B + C) Root number (a + B + C) 2 - root sign (a-b-c) 2 + root sign (a-b + C) 2 + root sign (a + B-C) 2

Because the relationship between the three sides of a triangle is: the sum of any two sides is greater than the third side
Therefore, a + B + C is greater than 0; a-b-c is less than 0; A-B + C is greater than 0; a + B-C is greater than 0; a + B-C is greater than 0; a-b-c is greater than 0; a-b-c is greater than 0; a-b-c is greater than 0; a-b-;
Then the original formula can be simplified according to the arithmetic square root of the square of a number = the absolute value of the number, and then remove the sign of the absolute value
The original formula = (a + B + C) - (- (a-b-c)) + (a-b + C) + (a + B-C) (the double bracket in the formula is bracket, I can't type it)
=a+b+c+a-b-c+a-b+c+a+b-c
=4a

3 − x is known =a. Y2 = B (y < 0), and (4a−b)2=8（b＞4a），3 (a+b)3 =18, find the value of XY

(4a−b)2=8（b＞4a），3 (a+b)3
=18，
b−4a＝8
a+b＝18 ，
The solution
a＝2
b＝16 ，
3 −x
=a，y2=b（y＜0），
-x=23=8，y=-
16=-4，
x=-8，y=-4，
xy=-8×（-4）=32．

Given a = 4 / Radix 5-1, find the square - 4A of cube - 2A of a It is known that: a = 4 / 5-1 of the root sign, find the cube of a - 2 × a squared - 4A

a=4/(√5-1)
=4(√5+1)/(√5-1)(√5+1)
=√5+1
a^3-2a^2-4a
=a(a^2-2a-4)
= a〔(a-1)^2-5 〕
=(√5+1)〔(√5+1-1)^2-5 〕
=(√5+1)(5-5 )
=0

Radical a + 2-radix 8-4a + radix-a squared

√(a+2)-√(8-4a)+√(-a^2)
Therefore, a ^ 2 = 0, a = 0
a+2=2 8-4a=8
√(a+2)-√(8-4a)+√(-a^2)
=√2-√8
=√2-2√2
=-√2

Simplify the cube of root - A (- the cube of a is in the root)

A ^ 3 ≥ 0
∴a≤0
∴-a≥0
∴√（-a^3）=√(（-a）²·(-a)）=-a√（-a）

2B root sign b fraction a + 3 / a root sign a cubic B - (4a root sign a fraction B + root number 9ab)

2B root sign b fraction a + 3 / a root sign a cubic B - (4a root sign a fraction B + root number 9ab)
=2b*1/b√ab+3/a*a*√ab-4a*1/a*√ab-3√ab
=2√ab+3√ab-4√ab-3√ab
=-2√ab

Simplification: a + 1 / (root A-1) square - 4A + 4 root sign a / (A-1) square Supplement the title again Root a + 1 / (root A-1) square - 4A + 4 root sign a / (A-1) square

T = radical a
t+1/(t-1)^2 - 4t^2+4t /（t^2-1）^2
t+1/(t-1)^2 - 4t^2+4t/(t+1) ^2(t-1)^2
(t+1)^3/(t+1) ^2(t-1)^2 - 4t^2+4t/(t+1) ^2(t-1)^2
(t+1)^3-4t^2-4t/(t+1) ^2(t-1)^2
t^3-t^2-t+1/(t+1) ^2(t-1)^2
(t-1)(t^2-1)/(t+1) ^2(t-1)^2
1/t+1
1 / radical a + 1
Answer: the same question as root A-1 / A-1~

First simplify and then evaluate the 2 / 1 of a + the square of a - 4A + 4 △ a + 1 / 2 of a, where a = 1 + radical 2

2/(a-1) +(a-2)²/(a+1)(a-1)×(a+1)/(a-2)
=2/(a-1) +(a-2)/(a-1)
=a/(a-1)
When a = 1 + √ 2
The original formula = (1 + √ 2) / √ 2
=√2/2 +1

Simplify the following questions: (1) the square of the root sign a + the square of the root sign (4-4a + a); (2) the square of the root sign a + the square of the root sign (4 + 4A + a)

√a²+√（4-4a+a²）=√a²+√（2-a）²
When a < 0, the original formula = (- a) + (2-A) = 2-2a;
When 0 ≤ a ≤ 2, the original formula = a + (2-A) = 2;
When 2 < A, the original formula = a + (a - 2) = 2a-2
（2.√a²+√（4+4a+a²）=√a²+√（2+a）²
When 0 < A, the original formula = a + (2 + a) = 2A + 2;
When - 2 ≤ a ≤ 0, the original formula = - A + (2 + a) = 2;
When a ＜ - 2, the original formula = - A - (2 + a) = - 2a-2