Mathematics second year questions, help solve, help solve a second year of mathematics problems

1)(x+y 3) (1 4x-2y)=(12x 2+4xy)(3-24xy) The numerator and denominator are multiplied by 12x

2)1/(3-m)=(m+3)/(9-m^2)=-(m+3)/(m^2-9)

3) (20b+30a) (12b-15a) multiplication 60ab4) x y=3, i.e. x=3y.

x^2+2xy-3y^2)/(x2-xy+y^2)=(x-y)(x+3y)/(9y^2-3y^2+y^2)=12y^2/7y^2=12/7

1.Multiply the numerator denominator by 12 to get (12x+4y) (3x-24y)3（1/3a+1/2b）=(3a+2b)/6ab （1/5a-1/4b）=(4b-5a)/20ab

1/3a+1/2b）/（1/5a-1/4b）=(20b+30a)/(12b-15a)

Push out x=3y into the original formula (x +2xy-3y)=12y (x -xy+y)=7y

x²+2xy-3y²）/（x²-xy+y²）=12/7

1, the numerator and denominator are multiplied by 12 to become (12x+4y) (3x-24y)2, and the second formula is multiplied by a —1, and then multiply the numerator and denominator by (m+3) at the same time, becoming (m+3) (m -9).

3, the numerator and denominator are multiplied by 60 to become (20a+30b) (12a-30b)4, (x +2xy-3y) (x -xy+y) the numerator and denominator are divided by y, and x y=3 is brought in, and 12 7 is obtained

1. Multiply 12 up and down

Get: 12x+4y 3x-24y

2. The second up and down multiplication (-m-3).

Get: -m-3 (m2-9).

3. Multiply 60 up and down

Get: (20a+30b) (12a-15b)4, divide by y2

Gotcha: (X y is too hard to play, I use Z).

z2+2z-3）/（z2-z+1）=12/7

1. Multiply the numerator and denominator by 12x at the same time to obtain (12x 2+4xy) (3-23xy).

2. m (m -9) = m (m+3) (m-3), so the numerator denominator of 1 (3-m) is multiplied by (m+3) at the same time to obtain - (m+3) (m+3) (m-3), and 3. the numerator denominator is multiplied by 60ab at the same time to obtain (20b+30a) (12b-15a).

4. x y=3, y x=1 3, the numerator denominator is divided by xy at the same time, (x +2xy-3y) (x -xy+y) = (x y+2-3y x) (x y-1+y x) = (3+2-3 3) (3-1+1 3) = 12 7

The fourth answer is wrong.

Let bg=x(x+2) 2=16+(6-x) 2(right-triangle egc, Pythagorean theorem) get x=3

The CEG area is 6

The CGF area is 6*(3 5)=

1. Set bg=x

Then (x+2) 2=16+(6-x) 2 (right triangle EGC, Pythagorean theorem) gives x=3

The CEG area is 6

The area of CGF is 6*(3 5)=3.

2 It has been proved that BG=3 then BG=CG=GF=33 because it is an isosceles triangle so GFC= GCF, because congruence, then AGF= AGB, because AGF+ AGB+ FGC=180 degrees, GFC+ GCF+ FGC=180 degrees.

So agf= agb= gfc= gcf, so fc ag

4. With the elimination method, the first three are correct, so the fourth answer is wrong.

The symmetry point of the first straight line is denoted as A1, which connects A1B and intersects the second line at point C, and the point C makes CD perpendicular to the river, and Cd is the required bridge. Reason: The width of the river is unchanged, so the length of the bridge is unchanged, that is, CD is unchanged, so as long as AD+BC is required to be the shortest, after crossing the point A as a symmetry point, A1B is AD+BC, that is, the shortest distance liquid draft (the shortest line segment between two points) can also use the distance formula between two points:

In the in-plane states and:

Let a(x1,y1) and b(x2,y2), then ab = x1-x2)2+(y1-y2).

1+k2)x1x2, or ab = x1

x2∣secα=∣y1

y2 sin, where is the angle of inclination of the straight line ab and k is the slope of the straight line ab.

In space: let a(x1,y1,z1),b(x2,y2,z2)ab|=√x1-x2)^2

y1-y2)^2

z1-z2)^2)]

Because the number of cars must be greater than or equal to 0, it is obtained from the formula of the column.

The second question combines the results of the first question, first of all, do you understand the column form of the second question? This is a practical problem, so can you understand that (8-x) and [7-(10-x)] in the column must be 0? If you don't understand, read the question 10 more times.

(1) The sum of the inner angles of the pentagon = 180° 3 = 540°, and the sum of the adjacent outer angles of an inner angle and other internal angles is 600 degrees, so this outer angle = 60° + the corresponding inner angle cannot be determined.

2) The sum of the inner angles of the quadrilateral = 360°, so this outer angle = 240° + the corresponding inner angles, it is impossible.

The sum of the inner angles of the hexagon = 720°, so this outer angle = the corresponding internal angle -120°, uncertain.

(1) The pentagonal shape can be decomposed into 3 triangles, so the sum of the inner angles is 540 degrees, and the sum of the outer angles and the inner angles is 600 degrees, so the degrees of the outer angles = 600-540 = 60 (degrees).

2) does not exist, the sum of the inner angles of the quadrilateral is 360 degrees, then the number of outer angles is 240 degrees, and because the number of outer angles cannot be greater than 180 degrees, there is no change in the quadrilateral and the following polygons; The sum of the inner angles of a hexagon is 720 degrees, which is greater than 600, so there is no such situation for polygons above the hexagon.

1: Solution: Let the two-digit number be x and the single digit number be y, then this number is (10x+y)10x+y-3(x+y)=23 (1).

10x+y=5(x+y)+1 (2)

Obtained from (1), (2).

7x-2y=23

5x-4y=1

Solution: x=5

y=6A: This two-digit number is 56

2.Let the guess be split: an addition is xThe other plus is y

x+10y=341

10x+y=242

x=21 y=32

3: Set: It took x minutes to go uphill and y minutes to go downhill.

x+y=16

4800*x 60+12000*y 60=1880 solution: x=11

y=5Answer: It took 11 minutes for the uphill slope and 5 minutes for the downhill, for a total of 16 minutes and the bell was closed.

4.Solution: Suppose you need two kinds of candy x and y kg each.

x+y=100 (1)

18x+10y=15×100 (2)

1) 18-(2) De Zhaokai.

8y=300

y=substitute y= into (1).

x+x=A: Two kinds of candy are needed in kilograms, kilograms each.