The 2011 power of (2 + Radix 3) is multiplied by the 2012 power of (2-radix 3) Thank you very much····

The 2011 power of (2 + Radix 3) is multiplied by the 2012 power of (2-radix 3) Thank you very much····

=2011 power of (2 + √ 3) × (2 - √ 3) 2011 power × (2 - √ 3)
=The 2011 power of [(2 + √ 3) × (2 - √ 3)] is × (2 - √ 3)
=The 2011 power of (4-3) × (2 - √ 3)
=2011 power of 1 × (2 - √ 3)
=2-√3

The 2011 power of (root 2-1) and the 2012 power of (root 2 + 1)

The 2011 power of (root 2-1) * the 2012 power of (Radix 2 + 1)
=2011 power of [(Radix 2-1) * (Radix 2 + 1)] (Radix 2 + 1)
=2011 power of 1 * (root 2 + 1)
=Radical 2 + 1

Radix 18 minus (π - 1) to the power of 0 minus 4 times Radix 1 / 2 plus 1 / 2 (Radix 2-1)

√18-(π-1)^0-4√(1/2)+(√2-1)/2
=3√2 - 1 - 2√2 + √2/2 - 1/2
=3√2 - 3/2

The third power of minus a under the root sign minus a times that of a under the root sign is minus 1

solution
The third power of minus a under the root indicates that a is a negative number
√（-a ³）-a √（-1/a）
=-a√（-a）+√a² （-1/a）
=-a√（-a）+√（-a）
=（-a+1）√（-a）

Radical 18 minus (π - 1) zero minus 2cos45 degree plus quarter minus minus negative first power

Radical 18 - (π - 1) ^ 0-2cos45 ° + (1 / 4) ^ (- 1)
=3 root sign 2-1 root sign 2 + 4
=2 root sign 2 + 3

Let a = (B + 2 of b-2) + (B + 2 of 2-B) + 1B3 of 2, and the root (X-Y + 2) = - absolute value x + y-6, find the cube root of ABXY

In fact, consider the conditions of the former two equations. First, B + 2 of B-2 and B + 2 of 2-B must be meaningful, then B + 2 / (b-2) and B + 2 / - (b-2) must be greater than or equal to 0, then only B + 2 / (b-2) = 0 can be satisfied. So B = 2, then B = 2 is brought into equation 1 and a = 4

The real number ^ + 2C + 1 ^ - 2C + 1 B is satisfied Find the value of a + B + C

C ^ 2 + 2 | A-1 | + radical {(2B + C + 1) / (4-C) = 0
C ^ 2 ≥ 0, 2 | A-1 | 0, radical {(2B + C + 1) / (4-C) ≥ 0
c^2=0,a-1=0,(2b+c+1)/(4-c) =0
a=1,b=-1/2,c=0
a+b+c=1-1/2+0=1/2

Given that the real number a, B, C satisfies 1 / A-B / + (radical 2B + C) + C? - C + 1 / 4 = 0, find the value of a (B + C)

A is 1 "4. B is - 1" 4. C is 1 "2

Given that real numbers a and B satisfy a−1 4 + | 2B + 1| = 0, find B The value of A

According to the meaning of the title:
a−1
4＝0
2b+1＝0 ，
The solution is as follows:
a＝1
Four
b＝−1
2 ，
Then B
a=（-1
2）×
One
4=-1
4．

It is known that a and B satisfy B= a2−4+ 4−a2+4 A − 2. Find | a − 2B|+ The value of ab

According to the meaning of the title:
a2−4≥0
4−a2≥0
a−2≠0 ，
2
Then B = - 1
Then the original formula = | - 2 + 2|+
2=
2．