Calculation: LG25 + LG2 · LG50 + (LG2) 2

Calculation: LG25 + LG2 · LG50 + (LG2) 2

The original formula is: 2lg5 + LG2 · (1 + lg5) + (LG2) 2 = 2lg5 + LG2 (1 + lg5 + LG2) = 2lg5 + 2lg2 = 2

Calculation: (1) LG25 + lg2lg50 + (LG2) 2 (2) 813 + (12) − 2 + (27-1 + 16-2) 0 + 416

(1) The original formula = 2lg5 + LG2 (1 + lg5) + lg22 = 2lg5 + LG2 + LG2 (lg5 + LG2) = 2 (lg5 + LG2) = 2. (2) the original formula = (23) 13 + (2-1) - 2 + 1 + 2 = 2 + 2 + 3 = 7

Calculation: LG25 + LG2 · LG50 + (LG2) 2

The original formula is: 2lg5 + LG2 · (1 + lg5) + (LG2) 2 = 2lg5 + LG2 (1 + lg5 + LG2) = 2lg5 + 2lg2 = 2

（8x^3+8x^2+4x+1)(8x^3-8x^2+4x-1)

Solution
（8x^3+8x^2+4x+1)(8x^3-8x^2+4x-1)
=[（8x^3+4x）+（8x^2+1)][(8x^3+4x)-(8x^2+1)]
=（8x^3+4x)^2-(8x^2+1)^2
=[(8x^3)^2+(4x)^2+2*32x^4]-[64x^4+1+2*8x^2]
=64x^6+16x^2+64x^4-64x^4-1-2*8x^2
=64x^6-1

ABCD is a square, e is a point on the edge of BC, AF bisecting angle DAE

Prove: lengthen CB, make BM = DF, connect am, lengthen the extension line of BC and AF intersect n, because the quadrilateral ABCD is square, so AB = ad angle Abe = angle ADF = angle bad = 90 degree ad parallel BC, so DAF = angle ane, because AF bisects the angle DAE, so EAF = angle DAF, so EAF = angle ane, because the angle ABM + angle Abe = 180 degree, so the angle

Number reasoning 1,2,1,6,9,10
A.13 B.12 C.19 D.17

The sum of every three terms is 4, 9, 16 and 25, which is the complete square
So 9 + 10 + () = 36 means 17

Completion
1. The sum of the volume of a cylinder and a cone with the same base and height is 48 cubic decimeters. The volume of the cylinder is () cubic decimeters, and that of the cone is () cubic decimeters
2. The area of the bottom of a cone and a cylinder is equal. The volume ratio of the cylinder to the cone is 3:2. If the height of the cylinder is 12 decimeters, the height of the cone is () decimeters
The volume of a cylindrical wood block is 3.6 cubic decimeters. Cut it into the largest cone, and the volume of the cut part is () cubic decimeters
A.1.2 B.2.4 C.3.6
The ratio of the diameter of the bottom to the height of the cylinder is ()
A.2π:1 B.1:1 C.1:π D.π:1
Practical questions
1. There is a cylindrical wood, which is 6 meters long. If it is sawn into three sections of cylindrical wood, the surface area will be increased by 12.56 square decimeters, which is related to the volume of the wood
2. The diameter of the bottom of a cylindrical tank filled with water is 10 meters. After using part of the water, the water surface drops by 40 cm, and the remaining water is exactly 5 / 6 of the water in the tank. Calculate the volume of the tank

The volume of a cone with equal base and height is one third of that of a cylinder
What is the volume of a cylinder
48 ÷ (1 ＋ 1 / 3) = 36 (cubic decimeter)
What is the volume of a cone
36 × 1 / 3 = 12 (cubic decimeter)
Suppose the area of the base of the cone and the cylinder is s, and the height of the cone is h, then the height of the cylinder is 2 / 3H
Volume of cone v = 1 / 3 * s * H = 12 = > s * H = 36
Then the volume of cylinder v = s * (2 / 3H) = 2 / 3 * 36 = 24
So the volume of the cylinder is 24 cubic decimeters
The total surface area of a 2-meter-long cylinder is increased by 12.56 square decimeters after it is sawed into three small cylinders. The volume of the cylinder is (62.8) cubic decimeters
After sawing, four bottom areas are increased, each of which is 12.56/4 = 3.14,
Volume = 3.14 * 20 = 62.8 cubic decimeter
6 bottom radius: 3.14 / (2 * 3.14) = 5m
Bottom area: 3.14 * 5 * 5 = 78.5 square meters
The water used is: 78.5 * 0.4 = 31.4 cubic meters
The volume is: 31.4 / (1-7 / 8) = 251.2 cubic meters

If a trapezoid is reduced by 6cm, it becomes a parallelogram. If a trapezoid is reduced by 4cm, it becomes a triangle. How many centimeters was the trapezoid before

10 cm
Reduce 4 cm from the bottom into a triangle, know the bottom is 4 cm
If the lower base is reduced by 6cm and becomes a parallelogram, the lower base of the parallelogram = upper base = 4cm, plus the reduced 6cm, the total length is 10cm

Skillfully calculated 87.4 - (43.27-22.6) - 9.73

87.4-43.27+22.6-9.73=87.4+22.6-（43.27+9.73）=110-53=57

The classical problem of Pythagorean theorem in the second grade of junior high school
Seek some classic, often test
Good questions give high marks

Method 1: A & amp; sup2; + B & amp; sup2; + C & amp; sup2; - 6a-8b-10c + 50 = 0 & nbsp; (A & amp; sup2; - 6a-8a + 9) + (B & amp; sup2; - 8b + 16) + (C & amp; sup2; - 10C + 25) = 0 & nbsp; (A-3) {% &