What is the absolute value of 3-radical 7

What is the absolute value of 3-radical 7

l3-√7l=l√9-√7l=√9-√7=3-√7

Given that f (x) = a ^ X / (a ^ x + root a), a is a constant greater than 0, find f (1 / 2007) + F (2 / 2007) +... " +The value of F (2005 / 2007) + F (2006 / 2007)

f(x)=a^x/(a^x+√a)
f(1-x)=a^(1-x)/(a^(1-x)+√a)=a/(a+√a*a^x)=√a/(a^x+√a)
So f (x) + F (1-x) = a ^ X / (a ^ x + √ a) + √ A / (a ^ x + √ a) = (a ^ x + √ a) = 1
So f (1 / 2007) + F (2 / 2007) +... + F (2006 / 2007) = [f (1 / 2007) + F (2006 / 2007)] + [f (2 / 2007) + F (2005 / 2007)] +... + [f (1003 / 2007) + F (1004 / 2007)]
=1+1+...+1
=1003
To do this kind of problem, it must be to find the law, and it is impossible to make a strong calculation

Comparison of Radix 2006-gen-2005 and gen-2004-gen-2003

∵(√2006-√2005)/(√2004-√2003)=[(√2006-√2005)(√2006+√2005)(√2004+√2003)]/([√2004-√2003)(√2004+√2003)(√2006+√2005)]=(√2004+√2003)/(√2006+√2005)＜1
∴(√2006-√2005)＜(√2004-√2003)

If (a square - 3A - b) + b square + 2B + 1 = 0, then a square - 3A - | B|=

√ (a? - 3a-b) + B? + 2B + 1 = 0 ∵ ∵ a? - 3a-b ≥ 0b ﹤ 2B + 1 = (B + 1) mm2 ≥ 0

If the absolute value of a + B + is (C-1) - 1) = 4 (A-2) + 2 (B + 1) - 4, then a + 2b-3c =?

Organized
[(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
The sum of bungalows and absolute values equals 0
So it's all equal to zero
So √ (A-2) - 2 = 0, A-2 = 4, a = 6
√(b+1)-1=0,b+1=1,b=0
√(c-1)-1=0,c-1=1,c=2
a+2b-3c=0

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

Root number (C-1) - 1| = 4 * root number (A-2) + 2 * root number (B + 1) - 4 A-2 + 2-4 * root number (A-2) + 4 + B + 1-1-2 * root number (B + 1) + 124root number (C-1) - 1-1 = 0 [root number (A-2)] ^ 2-4 * root number (A-2) + 4 + [root number (B + 1)] ^ 2-2 * root number (B + 1) + 1 + 1 + (C-1) - 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + root number (C-1) - 1-1-1-1-1-1-2 * root number (A-2), 2-2 * root number (B + 2 + 1 + 1 + 1 + 1 + 1| = 0 [root (A-2) - 2] ^ 2 + [root (B + 1) - 1] ^ 2 + | root (C-1) - 1 | = 0, so: root (A-2) - 2 = 0, root (B + 1) - 1 = 0, root (C-1) - 1 = 0, a = 6, B = 0, c=2 a+2b-3c=6+0-3*2=0

The absolute root of + 3c-1b + 2b-1 is the absolute root value of 3c-1b

There is something wrong with the title

If under the root sign a 2 - 3A + 1 + B + 2B + 1 = 0, then a 2 + 1 / a 2 - | B | =?

Indeed, there is a problem in the expression of this topic. You need to use brackets to clarify the range of the root sign, and there is B + 2B, which is b square + 2bk. The title is not clear, so you can't do it

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

Given that a is equal to the root 2 plus 1, find the cube of a minus the square of a minus 3A plus 2008

a=√2+1
a-1=√2
Square on both sides
a²-2a+1=2
a²-2a=1
So the original formula = a 3 - 2A 2 + a 2 - 2A - A + 2008
=a(a²-2a)+(a²-2a)-a+2008
=a+1-a+2008
=2009