If x, y satisfy y

If x, y satisfy y

∵ the number of the square root must be greater than or equal to 0
ν and (x) ≥ 0
∴x+1=0 ,x=-1
∴y﹤0+0+5 ,y﹤5
The original formula = √ (Y-5) 2 + | x + 1|
=|y-5|+|x+1|
=-(y-5)+0
=5-y

If x is less than - 5, what is the absolute value of 6-A + the square of 2A + 1 under the root sign and (a + 5) under the root sign

If a is less than - 5, the absolute value of 6-A + the square of 2A + 1 under the root sign + (a + 5) under the root sign is equal to
Because (a + 5) under the radical is meaningless, there is no result in the range of real numbers

If AB < 0, then the square B of the root of the algebraic expression a can be reduced to ()

By the condition: if √ (a 2 b) holds, then there must be: b > 0, a < 0;
So: √ (a? B) = √ (a?) √ B = | a | B = - a √ B

When x = radical 3-2, find the value of the algebraic formula x cube + 4x squared + X + 3

Three
calculation:
x^3+4x^2+x+3
=x(x^2+4x+1)+3
=x[(x+2)^2-3]+3
=0+3
=3

In this paper, the square of the cube - 2x of X / x under the root sign x - 2 / X - 2 △ X / X is simplified

X-2 / X-2 ÷ the square of X / X cube-2x under radical
=1/√(x-2)÷1/√(x-2)
=1;
It's my pleasure to answer your questions and skyhunter 002 to answer your questions
If there is anything you don't understand, you can ask,

Simplify the cube of negative a under the root sign and add the square of a under the root sign

The absolute value of a multiplied by (1 + root - a)
Because the cube of - A is > = 0
-a>=0
a<=0
So, - A (1 + radical - a)

Simplification: under radical (cube of negative a) / under radical (4 / 2 of a)

Under radical (cube of negative a) / under radical (4 / 2 of a)
=Under the radical (the cube of negative a) / the square of (A / 2)]
=Under the root sign (4 times the cube of negative a) / under the root sign (a square)
=Under double root sign (negative a)
Note: because the root sign (the cube of negative a) holds, so a is not a positive number; there is a in the denominator, so a is not 0. Therefore, a is a negative number

In △ ABC, if a, B and C are the opposite sides of ∠ a, ∠ B and ∠ C respectively, then it is simplified (a-b-c)2+ (b-a-c)2+ (C-A-B) 2___ .

According to the meaning of the title: a-b-c < 0, b-a-c < 0, C-A-B < 0,
Then the original formula = | a-b-c | + | b-a-c | + | C-A-B | = - A + B + C-B + A + C-C + A + B = a + B + C
So the answer is: a + B + C

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

Three sides of ABC feeding triangle
a+b>c
a+b-c>0
a+c>b
a+c-b>0
b+c>a
b+c-a>0
Simplify the square of (a + B-C) under the root sign + the square of (b-c-a) under the root sign + the square of (B + C-A) under the root sign
=a+b-c-(b-c-a）+b+c-a
=a+b-c+a+c-b+b+c-a
=a+b+c

In △ ABC, a, B, C are the three sides of a triangle (a-b+c)2-2|c-a-b|．

∵ in ᙽ ABC, a, B, C are the three sides of a triangle,
∴a-b+c＞0，c-a-b＜0，
The original formula = A-B + C-2 [- (C-A-B)]
=a-b+c+2c-2a-2b
=-a-3b+3c．