Junior High School Math Problems Help Urgent!! Thank you, dear friends

One. Fill-in-the-blank questions.

a*b=b*a

Yuan. Rice. Year. 10,000 yuan

7.①5a+4②（2x/3）-2③3*（x+a）/4④2n-2，2n，2n+2

Yuan. Yuan. 10.Explain the meaning of the following algebraic expressions in words.

The cube of the pier can be interpreted as:

The volume of a cylinder with a circle at the base and a radius of height.

If the ** of a teddy bear is x yuan, then 1 2x can be expressed as the teddy bear after a 5% discount.

5a+4b can be interpreted as:

The sum of 5 A's and 4 B's.

11.Fill-in-the-blank questions.

If algebraic equation 3 a-2 makes sense, then a is not equal to 2

a=4, b=3, then 2a-5b=

When a=3, b=2, c=-2, ab-2 c= if x+y=3, xy=-2, then 2xy-(x+y)= if x=, then the algebraic formula x-2 x=

If a=-1, then the exponent of algebraic formula a2-a 3+2=-1 and the exponent 2-6= of algebraic equation 1 and 4x when x=4

a+b=b+a

ab=ba2,,x+3

m-4+35,a(1+20%)

6,10a+b

7,5a+4

2x/3-2

3/4(a+x)

2n-22n

2n+28,9,1000m+300n

10, 1 of the volume of the circle 3

Half the price.

The price of a book is a pen.

The price of B5 books and 4 pens costs a total of 5A+4B

11, a is not equal to 2

a+b=b+a

a*b=b*a

x+3m-4+3=m-1

a*(1+20%)=

10a+b5a+42/3x-2

x+a)3/4

2n-2,2n,2n+2

n*80%=

9\1000m+300n

The area of the circle. The price of the teddy bear after the discount.

5 times for A plus 4 times for B.

11. (1) A is not equal to 2

2a-5b=-7

2)2a-5b=-7

One. Fill-in-the-blank questions.

ab = meta. x+3m.

m-4 years old. 120% A million yuan.

10a+b7.Expressed in algebraic formula:

5a+4②（2x/3）-2

3*（x+a）/4

2n-2，2n，2n+2

80% N.

1000m+300n

Yuan. 10.Explain the meaning of the following algebraic expressions in words.

The cube of the pier can be interpreted as:

r 3 volume of a cylinder.

If the ** of a teddy bear is x yuan, then 1 2x can be expressed as?

50% off the price.

5a+4b can be interpreted as:

The sum of five A and four B.

11.Fill-in-the-blank questions.

Not equal to 2

Analyzing the problem, because it contains a and b, I thought of converting it into a completely flat way.

The formula contains -2a, so (a-1) is obtained, and (b-2) has a -2a+1+b -4b+4=a +b -2a-4b+5 and the difference between the original formula and 1,6=5+1

From this the original formula = (a-1) +b-2) +1

a-1) +b-2) +1 is evergreen at zero, regardless of any real number a and b, the value of a +b -2a-4b+6 is always positive.

Solution: a + b -2a - 4b + 6

a²-2a+1+b²-4b+4+1

a-1)²+b-2)²+1

Because (a-1) +b-2) +1 is greater than or equal to 0, a +b -2a-4b+6 is a positive number.

Solution: a + b -2a - 4b + 6

a²﹣2a＋1＋b²﹣4b＋4＋1

a-1)²＋b-2)²＋1

1 proves that no matter a or b is any real number, the value of a b 2a 4b 6 is always positive.

The key is to convert 6 to 1+4+1 so that you can merge with the A and B recipes, and you can add anything you don't understand.

a²+b²-2a-4b+6

a²-2a+1+b²-4b+4+1

a-1)²+b-2)²+1

Because (a-1) 0

b-2) 0 (the square is non-negative).

So (a-1) +b-2) +1 1

a b 2a 4b 6=(a-1) 2+(b-2) 2+1 because (a-1) 2 is greater than or equal to 0 and (b-2) 2 is greater than or equal to 0, so (a-1) 2+(b-2) 2+1 is greater than 0 In summary:

Regardless of any real number a and b, the value of a b 2a 4b 6 is always positive.

Solution: Original formula = a + b -2a-4b + 6

a²-2a+1+b²-4b+4+1

a-1)²+b-2)²+1＞0

So the original formula is always positive.

The recipe is simple.

a²+b²-2a-4b+6

a²-2a+1+b²-4b+4+1

a-1)²+b-2)²+1

Because (a-1) is greater than or equal to 0

b-2) greater than or equal to 0

So a + b -2a - 4b + 6 is greater than or equal to 1

Regardless of any real number a and b, the value of a b 2a 4b 6 is always positive.

A squared 2a+1 b squared 4b+4+1=(a-1) square (b-4) square 1

Because (a-1) squared (b-4) squared or 0 then a squared b square 2a-4b+6 or 1 is always positive.

Simplification: (A-1) 2+(B-2) 2+1

So it is positive for any real number.

The original formula = the square of (a-1) + the square of (b-2) + 1, which is a positive number.

Solution: Because the two square roots of a positive number are opposite to each other, 3x-2+5x+6=0

Solution: x=-1 2

Then the square root is: -7 2 or 7 2

So this positive number is (7 2) 2 = 49 4

If the positive number is 49 4, then x=1 2 The positive number is 196, x=-4 Is there a problem with the question?

(1) When the two numbers are unequal, the square root of the positive number is the opposite number, then there is (3x-2)+(5x+6)=0, and the solution is x=, so this number is (3x-2) 2=

2) When the two numbers are equal, 3x-2=5x+6, and x=-4 is solved

So the number is 196

The square root of a positive number is equal to 3x-2 and 5x+6, then (3x-2) 2=(5x+6) 2, i.e. 9x 2-12x+4=25x 2+60x+36So 16x 2 + 72x + 32 = 0, i.e. 2x 2 + 9x + 4 = 0

The solution yields x=-1 2 or x=-4. When x=-1 2, 3x-2=-7 2, 5x+6=7 2. The positive number is 49 4.

When x = -4, 3x-2 = -14, 5x+6 = -14. The positive number is 196.

I don't know if there are 2 equal real roots under x 2-2 * root number 2 * x * sin +1 = 0, find the degree of .

If there are two equal real roots, then =b 2-4ac=0, and then find the angle.

The line l2: y=2x+8 when y=0, x= -4, so the coordinates of g (or b) are (-4,0).

The straight line l1: y=-x+2 when x= -4, y=6, so the coordinates of point c are (-4,6), so bc=6

Therefore, the longitudinal coordinates of point d are equal to 6, that is, the line l2: y=2x+8, y=6, and find x= -1, that is, the coordinates of point d are (-1,6).

1.Therefore, cd is equal to the absolute value of the abscissa difference between the two points, that is, cd = -1 - (-4) = 32, and the second question should be the intersection of the line L2 and the y axis at the point m.

1) g is the intersection point of L2 and the X axis, so Y=0 is substituted into L2: Y=2X+8 to get X=-4, so the coordinates of point C are (-4, Y), and point C is on L1, and point C is substituted into L1: Y=-X+2 to get :

y=6, so the coordinates of point c are (-4,6);

At this time, the coordinates of point D are set to (x,6) and it is on the line L2, so it satisfies the equation of L2 y=2x+8, and the coordinates of point D are substituted to obtain x=-1, so the coordinates of point D are (-1,6);

So: dc length is 3, bc length is c and d the ordinate is 62) bcme can not reach the quadrilateral, you take a good look is not bad is required is the area of the triangle bce.

Solution: e(2,0)b(-4,0) is known

So c(-4,6)d(-1,6)a(-1,0) so dc=|-1-(-4)|=3 bc=|0-6|=6 If the L1 and Y axes are compared to the M point, the CME is on the same straight line, and the BCE is a right triangle.

If the L2 and Y axes are compared to the M point, the extension of Cd is compared to the M axis, and the extension of Cd is compared to H, and the Y axis is compared to Ne(2,0) M(0,8).

Straight line em:y=-4x+8, so h(,6).

be=|-4-2|=6 ， ch=| ，mn=|8-6|=2s quadrilateral, bcme=s, trapezoidal bche+s, triangle, mchs, trapezoidal bche=bc*be 2=18

S triangle mch=ch*mn2=

So the s quadrilateral bcme = 18+

1) First of all, the equations of L1 and L2 should be solved to obtain point F as (-2,4), and then substituted Y=0 into L1 and L2 respectively to obtain E(2,0), G(-4,0), because B coincides with point G, that is, x=-4 of point C, substitute X=-4 into L1, obtain Y=6, that is, point C is (-4,6), according to the y of point D of the rectangle is 6, substitute L2 to obtain X=-1, that is, the coordinates of point D are (-1,6). The above can be obtained with a CD length of 3 and a BC of 6.

2) Since the intersection of the straight line L1 and the Y axis at the point M, and the extension intersection at the point of the X axis is E, finding BCME is actually just finding the area of the BCE triangle, and the area (if there is no mistake in the question) is 6*6 2=18.

Solution: Since the line l2:y=2x+8 intersects the x-axis at the point g, the point g(-4,0).

And because the point b coincides with the point g.

So point b(-4,0).

Let the point c(-4,y), because the point c is on the straight line l1:y=-x+2, so y=4+2=6

So point c(-4,6).

So bc=6

Let the point d(x,6), because the point d is on the line l2:y=2x+8, so 6=2x+8, x=-1

So point d(-1,6).

So dc=3

1) G(-4,0) is obtained from the straight line L1, and X=-4 is brought into the straight line Y=-X+2 by the BC X-axis to obtain C(-4,6), so BC=6

Substituting y=6 into y=2x+8 from the cd y-axis yields d(-1,6), so cd=3

2, Question 2 I feel you are typing wrong, it should be L2 and Y axis intersect with mThen find the coordinates of each point and divide it into triangles.

1. There is the intersection point of the line L2 and the x-axis, b(-4,0), and c(-4,6)d(-1,6)a(-1,0).

So dc=ab=3, bc=6

2.Which point is required for the second question? Is there anything wrong with that?

The area of the quadrilateral BCME, point m at **?

Can't find it!

If A, B, and C cars are arranged with X, Y, and Z respectively, then 6X+5Y+4Z=80

x+y+z=15

The minimum value of the function t=100x+150y+100z is obtained by substituting z=15-x-y into the function.

t=50y+1500 (y takes at least 4, so the minimum value is y, takes 4) and then solves x=8, z=3

The minimum shipping fee is 1700 yuan.

6 cars of Type A.

5 cars of type B. 4 cars of type C.

That's it!!!

Let A have x cars, B has Y cars, and C has Z cars.

6x+5y+4z＝80

x+y+z=15

Find the minimum value of 100*6x+150*5y+100*4z, because a>=5, b>=4So.

6x+5*4+4z=80

x+4+z=15

x=8,y=4,z=3.

100*6x+150*5y+100*4z=100*6*8+150*5*4+100*4*3=9000.

6 cars of Type A.

5 cars of type B. 4 cars of type C.

A: 6 cars for Type A, 5 cars for Type B, 4 cars for Type C.

First of all, let the A car be x, B be y, and C be z, so we get 6x+5y+4z=80

Inequality x is greater than or equal to 5 y is greater than or equal to 4 x + y + z = 15 Three inequalities find the minimum value of 100x+150y+100z First eliminate the element Remove z to get the formula of two unknowns.

Get x greater than or equal to 5 y greater than or equal to 4 80-2x-y less than or equal to 15 find the minimum value of 2000-50x+25y from the feasible field The minimum value is calculated by yourself Hehe.