A simple algorithm of 56 + 78-46 + 12

78+12+（56-46）
=90+10
=100

5/4-9/20+11/30-13/42+15/56-17/72+19/90=

5/4-9/20+11/30-13/42+15/56-17/72+19/90=5/4-(4+5)/4*5+(5+6)/5*6-(6+7)/6*7+(7+8)/7*8-(8+9)/8*9+(9+10)/9*10=5/4-1/4-1/5+1/5+1/6-1/6-1/7+1/7+1/8-1/8-1/9+1/9+1/10=1+1/10=11/10.

42 / 1 + 56 / 1 + 90 / 1 + 110 / 1 + 132 / 1
(/ is a semicolon)

1/42+1/56+1/90+1/110+1/132
=1/6-1/7+1/7-1/8+1/9-1/10+1/10-1/11+1/11-1/12
=1/6-1/8+1/9-1/12
=12/72-9/72+8/72-6/72
=5/72

Quadrilateral (20 17:31:56)
1. In RT △ ABC, & ang; ACB = 90 & deg;, & ang; BAC = 60 & deg;, de bisects BC vertically, the perpendicular foot is D, intersects AB at point D, and the point F is on the extension line of De, and AF = CE

(1) Certification:
∵ ed vertical bisection BC
∴∠EDC=∠EDB=90°,CD=DB
It can be proved that: △ CDE ≌ △ EDB (SAS) (which three elements need not be mentioned)
∴∠ECD=∠B=30°,CE=DB
∴∠ACE=∠CAE=60
The ace is an equilateral triangle
∴AE=AC=CE=EB
The AED is an isosceles triangle
The angle AED is 60 degrees
So △ AED is an equilateral triangle
Just prove that the quadrilateral ACEF is a diamond
∵△ CDE is an equilateral triangle
The quadrilateral ABCD is a square
∴AB=BC=CD=DE=CE
The ∧ ade is an isosceles triangle with 150 ° vertex angle
∴∠AED=15°
It can be calculated that △ BOF is a right triangle with ∠ FBO = 30 ° and ∠ BOF = 60 °
Then, according to the fact that the opposite side of 30 ° in a right triangle is half of the hypotenuse (that is, sin30 ° = 0.5)
It can be concluded that BF = 2 * of = 2
According to Pythagorean theorem, we can get Bo = root 3
｜ AB = BC = radical 6

Quadratic radical 3 (17:16:56:17)
&#If y = (3x-6) + (6-3x) + + x ^ 3 & # 160; &# 160; then the square root of 10Y + 2x is
& 160; & 160; (& 160; & 160; & 160;) is the root sign

(3x-6) ≥ 0 and (6-3x) ≥ 0
Ψ 3x-6 ≥ 0 and 6-3x ≥ 0
∴3x-6=0
The solution is x = 2
∴y=(3x-6)+(6-3x)+x^3
=0+0+8
=8
｜ positive and negative (10Y + 2x) = positive and negative (80 + 4) = positive and negative 2 (21)
() is the root sign

In parallelogram ABCD, ab ⊥ AC, ab = 1, BC =, diagonal lines AC and BD intersect at point O. rotate straight line AC clockwise around point O and intersect BC and ad at points E and f respectively. What is correct in the following conclusions--------
1. When the rotation angle is 90 degrees, the quadrilateral abef is a parallelogram
2. During rotation, the segments AF and EC are always equal
3. When the rotation angle is 45 degrees, the quadrilateral BEDF is a diamond

1.2.1. When the rotation angle is 90 degrees, the angle AOF = angle OAB, AB is parallel to Fe, and AF is parallel to be, so the quadrilateral abef is a parallelogram. 2. The parallelogram ABCD is a centrosymmetric figure with O symmetry center, so a and C, F and E are symmetry points respectively

Inverse scale function (1 19:56:19)
It is known that y = Y1 + Y2, Yi is positively proportional to √ x, Y2 is inversely proportional to X2, and when x = 1, y = 11; when x = 4, y = 6.5
(1) Find the relationship between Y and X (2) when x = 9, find the value of Y

First, let Y1 = K1 * x, y2 = K2 / X
So y = Y1 + y2 = K1 * x + K2 / X
Then, when x = 1, y = 11; and when x = 4, y = 6.5
That is, K1 + K2 = 11, 4k1 + K2 / 4 = 6.5
The solution is K1 = 1, K2 = 10
So y = x + 10 / X
(2)y=9+10/9=100/9

Univariate quadratic equation (19 21:56:0)
If the equation x2 + 2 (a + 1) x - (b-2) 2 = 0 about X has two equal real roots, find: (1) the value of A.B; (2) the value of a2007 + B2

Because the equation x2 + 2 (a + 1) x - (b-2) 2 = 0 has two equal real roots
So 4 (a + 1) ^ 2 + 4 (b-2) ^ 2 = 0
So a = - 1, B = 2
a^2007+b^2
=-1+4
=3

Fill in the blanks (25 19:56:50)
1. In the right triangle ABC, the ∠ B is 90 °, AB is 3cm and AC is 5cm. Fold the triangle so that point C coincides with point a, and the crease De is obtained. Then the perimeter of the triangle Abe is (&# 160; &# 160; &# 160; &# 160; &# 160; &# 160;)
2. The points with the same distance from the inside of the triangle to each side of the triangle must be in the triangle ( )

Question 1: 8 cm.ABE=AB +AE+BE=AB+AC
Question 2: the intersection of the vertical bisectors of each side

It is known that the equations of X and y are 1.3x + 2Y = m + 1 & # 160; 2.2x + y = m-2 & # 160; &# 160; when m is what value, x > y?

x＝m-5,y＝-m+8.x＞y↔m-5＞-m+8↔m＞13/2

A simple algorithm of 56 + 78-46 + 12