# The condition that (x + 1) under the root sign multiplied by (x-1) = under the root sign (x square - 1) holds

## The condition that (x + 1) under the root sign multiplied by (x-1) = under the root sign (x square - 1) holds

∴x+1≥0；
x≥-1;
x-1≥0;
x≥1;
x ²- 1≥0;
X ≥ 1 or X ≤ - 1;
The condition is x ≥ 1
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### If root sign (1-2a) square = 1-2a Then the value range of a is

Root sign (1-2a) square = |1-2a| = 1-2a
So 1-2a ≥ 0
a≤1/2

### Given a = 1 / (root sign 5 + 2), find the value of square / (A-1) of 1-2a + a - square-2a + 1 / (square-a of a)

a=1/(√5+2)=(√5-2)/(√5+2)(√5-2)=√5-2;
∴a-1=√5-3＜0
Original formula = (A-1) ²/ (a-1)-√(a-1) ²/ a(a-1)
=a-1-1/a
=√5-2-1-1/(√5-2)
=√5-3-√5-2
=-5;
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### If the root sign a square - 2A + 1 = 1-A and the root sign a square + 4A + 4 = a + 2, find the absolute value of a + the square of the root sign (1-A)

Root sign a square - 2A + 1 = 1-A, it can be seen that a ≤ 1;
Root sign a square + 4A + 4 = a + 2, it can be seen that a + 2 ≥ 0, that is, a ≥ - 2
According to the above two conclusions, we have - 2 ≤ a ≤ 1
Then, the absolute value of a + the square of the root sign (1-A) = |a| + (1-A). The answer is as follows:
When - 2 ≤ a ≤ 0, the original formula of the question is equal to - A + 1-A = 1-2a;
When 0 ≤ a ≤ 1, the original formula of the question is equal to a + (1-A) = 1

### Is the image of positive scale function y = KX, (k is not equal to 0) fixed through the origin (0,0) and point (1, K)~

yes
Because if you bring in x = 0 and x = 1 respectively, you can get the value of Y, no matter why K is 0 and K

three