3 (2x + 1) trisquare = 192, find the value of X

3 (2x + 1) trisquare = 192, find the value of X

Is three square a cubic?
(2x+1)³=192÷3=64
(2x+1)³=4³
2x+1=4
x=3/2

Known: 22x + 3-22x + 1 = 192, find X

22x+3-22x+1=22x+1（4-1）=3×22x+1=192，
∴22x+1=64，
∴2x+1=6，
X = 5
2．

Given x ^ (2x + 3) - x ^ (2x + 1) = 192, find the value of X It's 2x + 3 power of X! I was also wondering if there was a problem with the problem. Power function should not have been learned in Grade 7, but there was this problem on the exercise paper of the little girl, wondering whether there was a clever solution,

X ^ (2x + 3) - x ^ (2x + 1) = 192, where x ^ (2x + 3) is x × (2x + 3), not the power handle of 2x + 3 of X. it seems that grade 7 should not learn this power function. It is easy to misunderstand
X ^ (2x + 3) - x ^ (2x + 1) = 192 is converted into 2x square + 3x-2x square - 1 x = 192, that is, 2x = 192
x＝96

X ^ 2 + 4x + 1 = 0, find the value of x ^ 2 + x ^ - 2, x ^ 2-x-1 = 0, and the value of - x ^ 3 + 2x ^ 2 + 2008

X ^ 2 + 4x + 1 = 0 both sides of the equation are divided by XX + 4 + 1 / x = 0x + 1 / x = - 4x ^ 2 + x ^ (- 2) = (x + 1 / x) ^ 2-2 = (- 4) ^ 2-2 = 16-2 = 14x ^ 2-x = 1-x ^ 3 + 2x ^ 2 + 2008 = - x ^ 3 + x ^ 2 + 2008 = - x (x ^ 2-x) + x ^ 2 + 2008 = - x + x ^ 2 + 2008 = 1 + 2008 = 2009

If the value of 2x 2 + X-2 is 3, what is 4x 2 + 2x + 1?

2x²+x-2=3
2x²+x=54x²+2x=2(2x²+x)=2X5=104x²+2x+1=10+1=11

If the absolute values of 2x + 3 and - 4x + 21 are equal, then the value of X is

If the absolute values of 2x + 3 and - 4x + 21 are equal
Then 2x + 3 = - 4x + 21 x = 3
Or - (2x + 3) = - 4x + 21 x = 12
X = 3 or 12

Given that x + 3 / x + 2 = 1 / radical 3 + radical 2 + 1, find x-3 / 2X-4 / [(5 / X-2) - X-2]

Because (x + 3) / (x + 2) = 1 / (√ 3 + √ 2 + 1)
So x + 2 = (x + 3) (√ 3 + √ 2 + 1) = (root 3 + root 2 + 1) x + 3 (root 3 + root 2 + 1)
So x + 3 = - (√ 3 - √ 2)
Therefore, the original formula = (x-3) / (2X-4) * (X-2) / (9-x ^ 2) = - 1 / 2 (x + 3) = (√ 3 + √ 2) / 2

Given x = 2008-5 radical sign 3, find the value of the algebraic formula (x ^ 2-2x + 1) / (x ^ 2-1) / (1 + x-3) / (x + 3) After simplification, it is X-1 / X-2, and then

After simplification, (x ^ 2 + 2x-3) / (x ^ 2-x-2) should be like this

Solve equation 1) root x + 2 times root 2x-1-radical 3 = 0 2) root x + 5 + root x-3 = 4

1. Root x + 2 times root 2x-1 = root 3
Square both sides to get (x + 2) (2x-1) = 3
(2x + 5) (x-1) = 0
X = - 5 / 2 or x = 1
Because x + 2 > = 0,2x-1 > = 0, i.e. > = 1 / 2
So x = 1
2. (root x + 5 + root x-3) / 2 = 2
1 / (root x + 5-radical x-3) = 2
(root x + 5-radical x-3) = 1 / 2, the following method is the same as (1)

Given that x2 + X-1 = 0, then the value of x3-2x + 4 is______ .

∵x2+x-1=0，
∴x2=1-x，x2+x=1，
∵x3-2x+4，
=x（x2-2）+4
=x（1-x-2）+4
=x（-1-x）+4
=-x2-x+4，
=-（x2+x）+4
=3．
So the answer is: 3