Let x = 2 / 2 root sign 5 minus 1, find the value of the fourth power of X + the second power of X + 2x-1

Let x = 2 / 2 root sign 5 minus 1, find the value of the fourth power of X + the second power of X + 2x-1

x=(√5-1)/2
2x+1=√5
Square on both sides
4x²+4x+1=5
x²=1-x
So x & # 8308; = (1-x) &# 178;
=x²-2x+1
So the original formula = x & # 178; - 2x + 1 + X & # 178; + 2x-1
=2x²
=2(1-x)
=2-2x
=2-(√5-1)
=3-√5

The fourth power of (X-2) under 4 times root minus x2-2x + 1 under root=

4 √ (X-2) ^ 4 - √ (X & # 178; - 2x + 1) = 4 (X-2) & # 178; - √ (x-1) & # 178; = 4 (X-2) & # 178; - (x-1) = 4x & # 178; - 17x + 17, X ≥ 2; 1 ≤ x < 2, primitive = 4 (2-x) & # 178; - (1-x) = 4x & # 178; - 15x + 15; x < 1, primitive = 4x & # 178; - 1

Use the formula method to solve the following equation: x ^ 2 = 2 root sign 2x-1

X ^ 2-2 radical 2x = - 1
X ^ 2-2 radical 2x + 2 = 1
(x-radical 2) square = 1
X1 = radical 2-1
X2 = root 2 + 1

The vertex of the image with the function y = 1-5 / 2x + X squared is the intersection point of the Y axis, and the coordinate is
If the parabola y = - 1 / 10x square - 9 / 5x + C passes through the point (2, - 12), then C is K

C = - 8 the third quadrant B = - 12 y = - 3 × x × x y = - x × x-2x + M = - (x × x + 2x-m) = - [(x + 1) × (x + 1) - M-1], then the vertex of the parabola is (- 1, M + 1). Substitute this point into the linear equation to get m = - 5 / 2

As shown in the figure, BF is parallel to the diagonal AC of square ABCD, point E is on BF, and AE = AC, CF ‖ AE, then the degree of ∠ BCF is______ ．

Through point a, make the extension line of Ao ⊥ FB at point O, connect BD, intersect AC at point Q, ∵ quadrilateral ABCD is a square, ∵ BQ ⊥ AC ∵ BF ∥ AC, ∵ Ao ∥ BQ Moreover, qAB = QBA = 45 °, Ao = BQ = AQ = 12ac, ∵ AE = AC, ∵ Ao = 12ae, ∵ AEO = 30 °, ∵ BF ∥ AC, ∵ CAE = AEO = 30 °, ∵ BF ∥ AC, CF ∥ AE, ∫ CFE = 30 °,