The absolute value of (x-radical 7) is 3. Find the value of X, The absolute value of (x-radical 7) = 3, what is the value of X |What is X - √ 7 | = 3 x?

The absolute value of (x-radical 7) is 3. Find the value of X, The absolute value of (x-radical 7) = 3, what is the value of X |What is X - √ 7 | = 3 x?

x=√7±3

The real number a satisfies | 2005 − a|+ If a − 2006 = a, then the value of the algebraic formula a-20052 is______ .

A kind of
A − 2006 makes sense,
∴a-2006≥0，
The original formula = a-2005+
A − 2006 = a, i.e
a−2006=2005，
∴a-2006=20052，
∴a-20052=2006．
So the answer is: 2006

Given that a satisfies the root sign (2006-a) ^ 2 + root sign (A-2007) = a, calculate the value of a-2006 ^ 2 1) Given that a satisfies the root sign (2006-a ^ 2) + the root sign (A-2007) = a, calculate the value of a-2006 ^ 2 2) If the integer part of root 2a is a and the fractional part is B, find the value of a + A / b 3) Know (radical 5-x) + 2 (radical X-5) + y = 2 to find the value of x ^ y

Your supplement is wrong
Should be root (2006-a) ^ 2 + root (A-2007) = a
So | 2006-a | + √ (A-2007) = a
Greater than or equal to 0 under root sign
So A-2007 > = 0
a>=2007
So 2006-a

Absolute value 2006-a absolute value + root sign A-2007 = a, find a - [2006 square] =? most urgent,

A > 2007 is obtained from A-2007 under the root sign, so the original formula is a-2006 + A-2007 = a under the root sign, that is: A-2007 = 2006 under the root sign, the square of both sides, A-2007 = 2006 square, so a - [2006 square] = 2007

Given the absolute value 2004-A + Radix a-2006 = a, find a-2004 ^ 2

|2004-a|+√(a-2006)=a
A-2006 ≥ 0 is obtained from the significance of quadratic radical
Therefore, a ≥ 2006
The original formula = = = > (a-2004) + √ (a-2006) = a
===> √(a-2006)=2004
===> a-2006=2004^2
===> a-2004^2=2006

If a square - 3A + 1, then + b square + 2B + 1 = 0, then a square + a square 1 - B =?

The original formula is changed into
√(a²-3a+1)+(b+1)²=0
The arithmetic square root and square number are both nonnegative
The sum of two nonnegative numbers is 0, both of which are 0
a²-3a+1=0
At the same time, divide by a to get:
a-3+1/a=0
a+1/a=3
Square, get:
a²+2+1/a²=9
a²+1/a²=7
b+1=0
b=-1
|a²+1/a²-b|
=|7+1|
=8

if If A2 − 3A + 1 + B2 + 2B + 1 = 0, then A2 + 1 a2−|b|=______ .

A kind of
a2−3a+1+b2+2b+1＝0，
Qi
a2−3a+1+（b+1）2=0，
∴a2-3a+1=0，b+1=0，
∴a+1
a=3，
∴（a+1
a）2=32，
∴a2+1
a2=7；
b=-1．
∴a2+1
a2−|b|=7-1=6．
So the answer is: 6

If the absolute value of a + B + [(C-1) - 1] is = (A-2 under 4-radix) + (B + 1-4 under 2-radical), find the value of a + 2b-3c

a+b+|√(c-1)-1|=4√(a-2)+2√(b+1)-4
(a-2)-4√(a-2)+4+(b+1)-2√(b+1)+1+|√(c-1)-1|=0
(√(a-2)-2)^2+(√(b+1)-1)^2+|√(c-1)-1|=0
∴√(a-2)=2 -> a=6
√(b+1)=1 -> b=0
√(c-1)=1 -> c=2
∴a+2b-3c=0

A + B + absolute value [[radical (C-1) - 1] = 4 radical (A-2) + 2 radical (B + 1) - 4, calculate the value of a + 2b-3c

A + B + | root sign (C-1) - 1 | = 4 * root sign (A-2) + 2 * root sign (B + 1) - 4
A-2 + 2-4 * radical (A-2) + 4 + B + 1-1-2 * radical (B + 1) + | radical (C-1) - 1| = 0
[root (A-2)] ^ 2-4 * root (A-2) + 4 + [root (B + 1)] ^ 2-2 * root (B + 1) + 1 + | root (C-1) - 1 | = 0
[radical (A-2) - 2] ^ 2 + [radical (B + 1) - 1] ^ 2 + | radical (C-1) - 1| = 0
So: radical (A-2) - 2 = 0
Radical (B + 1) - 1 = 0
Radical (C-1) - 1 = 0
a=6 ,b=0 ,c=2
a+2b-3c=6+0-3*2=0

(under the root sign a 2 - 3a) + B 2 + 2B + 1 = 0, then the absolute value of a 2 + 1 / 2 - B =?

∵ (under root sign a ∵ - 3a) + B ∵ + 2B + 1 = 0
∴a²-2a=0 b+1=0
A = 0 or a = 2 b = - 1
Absolute value of a 2 + 1 / 4-1 = 4 + 1 / 4-1 = 3 + 1 / 4 = 13 / 4