Mathematics in the second year of junior high school is one urgent and one in the second mathemati

2a(a-5) (a-2) 2 divided by [(1-a)(2-a)-12] (2-a).

2a(a-5) (a-2) 2 divided by (a2-3a+2-10) (2-a).

2a(a-5) (a-2) 2 divided by (a+2)(a-5) (2-a)2a(a-5) (a-2) 2 divided by (a+2)(a-5) [-(a-2)].

Then remove (a-5) and (a-2).

2a/(a-2)(a+2)]

2a/(a^2-4)

And because a 2 = 6a + 4

So the variant gets: a 2-4 = 6a

So the original formula = -2a (a 2-4).

2a 6a haha, I haven't done the question for a long time, and the format is not right.

The problem is to put a 2 = 6a + 4

Replace a 2 in an algebraic formula

2a^2-10a)/(a^2-4a+4)]=[(2*6a+2*4-10a)/(6a+4-4a+4)]=[(2a+8)/(2a+8)]=1

1-a-12 (2-a)] general score = [(1-a)*(2-a)-12] (2-a)=[2-3a+a 2-12] (2-a)=[2-3a+6a+4-12] (2-a)=(3a-6) (2-a)=-3

Substituting a 2 = 6A + 4 into the equation sought, substituting the complex fraction into the simplest fraction, and then substituting the above equation into the final result is -1 3

The front is 1 and the back is -3

The result is a third of -1 3 minus.

The volume of the cuboid is 48*6=288 (cubic centimeters), and the edge lengths of the three cubes are x, 2x, and 3x; Then the volumes of the three cubes are x, 8x, and 27x.

So there is x +8x +27x =288. That is, there is 36x = 288 to get x = 2.

So the edge length of the maximum cube is 3x=6.

The surface area of the maximum cube is 6*6*6=216

1.Find the known volume 48*6=288 cubic centimeters.

2.The edge-to-length ratio is.

Corresponding volume ratio.

The largest cube has a volume of 288*(27 36)=216 and its edge length is 6cm

The surface area is 6*6*6=216 square centimeters.

The maximum cube surface area is 216 (cm3).

Answer: The volume of steel: 48 6 = 288 (cm3) The ratio of the edge length of the three cubes is 1:

2:3, then the volume ratio is 1:(2 2 2):

3 3 3) = 1:8:27 (the volume of the cube is to the 3rd power of the edge length).

The largest box has a volume of 288 [27 (1+8+27)] = 216 (cm3).

Then calculate the edge length from his volume:

The maximum cube volume is 216 = 6 6 6

The maximum cube edge length is 6.

Its surface area is: 6 (6 6) = 216.

Note: Although the surface area and volume results are the same, both are 216, but the meaning expressed in the calculation is different. ）

bc=be, proof as follows:

In RT Triangle ADC vs. RT Triangle AFE.

ad=af (known).

ac=ae (known).

So two triangles congruence (hl).

So the angle ACD angle AEF (congruent triangle corresponds to the same angle) so the angle ACB angle AEB (flat angle definition of the equation property) and because AD is perpendicular to DB, af is perpendicular to be, and ad af so the angle CBA angle ABE (the distance from any point on the angle bisector to both sides of the angle is equal Angle bisector definition).

In the triangle ACB vs. triangle AEB.

Angular ACB Angular AEB (verified).

Horn CBA Angle ABE (verified).

ab=ab (common edge).

So two triangles are congruent (AAS).

So bc=be (congruent triangles correspond to equal sides).

Answer: Solution: (1) If the newsstand orders x copies of the evening newspaper from the newspaper every day, then x should meet the requirements of 60 x 100, and x is a positive integer then a total of (20x 10 60) copies will be sold every month, and 10 (x-60) copies will be returned to the newspaper Because the newspaper sold makes a profit of 0 3 yuan per copy, and the returned newspaper loses 0 5 yuan per copy, the profit obtained per month is, y = 0 3 (20 x + 10 60) one.

The value range of the independent variable x is 60 x 100, and x is a positive integer (2) When 60 x 100, y increases with the increase of x, and when x=100, y has a maximum value

Y max = 100 480 = 580 (yuan).

The newsstand should order 100 newspapers from the newspaper office in order to make the maximum profit per month, and the maximum profit is 580 yuan.

1) x should be greater than or equal to 60. y ≤ 100 = (20x+10*60)*1 - 30x* +10(x-60)* =x+480 (60<=x<=100).

2) y=x+480 when x takes the maximum, y is the maximum. Therefore, x takes 100 y=580 yuan, and earns 580 yuan a month by making 100 copies a day.

Solution: Early slag (1) proves: the quadrilateral abcd is a parallelogram, ab df (parallelogram is parallel to the opposite side), abe= fce (two straight lines are parallel, the inner wrong angle is equal), e is the midpoint of bc, be=ce, in abe and fce, abe= fce (verified); be=ce (verified); AEB = CEF (to the vertex angle); abe fce(, ab=fc (congruent triangles correspond to equal sides).

When doing (2) BC=AF, the quadrilateral ABFC is rectangular. Rationale: ab=fc (proven), ab cf (proven), quadrilateral abfc is a parallelogram (a set of quadrilaterals with opposite sides parallel and equal is a pure parallelogram), bc=af, quadrilateral abfc is rectangular (parallelogram rectangular with equal diagonals).

Analysis: (1) BPC= BCP is inferred from the equality of the two base angles of an isosceles triangle; Then, the conclusion of bpq= bqp is obtained by the vertical definition and equivalent substitution.

2) As ph bc, the vertical foot is the point h by solving the right angle abc to know abc=60°, bc=2 then according to the diagram and the pythagorean theorem, ph 2 + qh 2 = cq 2-(ph 2 + ch 2), that is, 2ph 2 + qh 2 = cq 2-ch 2 So substituting the length of the relevant line segment into it, you can get the relationship between y and x

Hope it helps.