In the first year of junior high school, students should do math problems with inequalities and ineq

1) The area of the square is not more than 25cm, then the side length of the square is not more than 5cm, so the circumference of the square, that is, a 4 25, that is, a 20cm2) The area of the circle is greater than 100cm, that is, r 100, r 10, r 10

So the circumference of the circle, that is, a 2 10, i.e. a 20 3) when a = 8, the square side length = 8 4 = 2, so the square area = 2 2 = 4 The radius of the circle = 8 2 = 4 , so the area of the circle = (4) = 16

The area of the circle is large.

When a=12, the side length of the square = 12 4 = 3, so the area of the square = 3 3 = 9 The radius of the circle = 12 2 = 6, so the area of the circle = (6) = 36

The area of the circle is large.

4) When the circumference is equal, the area of the circle is greater than the area of the square.

If you don't know, ask again; Satisfied! May you open and good luck!! 】

1)(a/4)²≤25，a/4≤5，a≤202）π×a/2π）²100

3) Both are circles with a large area.

4) In squares and circles with equal circumference, the area of the circle is large.

The solution is as follows: 1) Establish the inequality: (a 4) 2 25, and the solution is:

a 202), establish an inequality: a 2 ) 2>100, solution: a>20 3), when a=8, the area of the square is (8 4) 2=4, and the area of the circle is (8 2) 2, so the area of the circle is large.

When a=12, the area of the square is (12 4) 2=9 and the area of the circle is (12 2) 2, so the area of the circle is large.

4) Guess that when the circumference of the square and the circumference of the circle are equal, the area of the circle is the largest.

1) (a 4) 25, a 4 5, a 202) a 2 ) 100, so a 20 3) both are large in the area of the circle (I suggest you derive this yourself) 4) in a square with equal circumference and in a circle, the area of the circle is large.

1. 5x-12≤2(4x-3)

2. 5x-1＞2x+5

x-4＜3x+1

3. 4x+5／3 ≥ 5／6 - 1-x ／2

4.The inequality group x+a 0 , 1-2x x-2 has a solution. Find the range of values for a.

lx-2l+lx-3l≥m

The whole solution is divided into three cases:

When x 2, lx-2l = 2-x, lx-3l = 3-x. The original form becomes:

2-x+3-x≥m

x≤(5-m)/2

When 2 x 3, lx-2l=x-2, lx-3l=3-x. The original form becomes:

x-2+3-x≥m

m 1 when x 3, lx-2l = x-2, lx-3l = x-3. The original form becomes:

x-2+x-3≥m

x≥(m+5)/2

Since m 1, (5-m) 2 2, the result of the formula is x 2. (m+5) 2 3, the result is x 3.

The final answer is: x 2 or x 3

Note: The result of the above three equations is the preceding "when x......The intersection of the x range and the solved x range, the final x result is the union of the results of the above three formulas, because I don't know any other news about m, just use the formula to solve the result of m 1, but this result is also limited to x, and it is not fully convincing. In other words, the intersection of x 2 and x (5-m) 2, and the intersection of x 3 and x (m+5) 2 cannot be determined, and the answer cannot be given.

Categorical discussions.

x-2>=0

x-3<0

x-2+3-x>=m, i.e., when m<=1, 2<=x<3 x-2>=0

x-3>=0

x-2+x-3>=m gives x>=3 or x>=(m=5) 2

If (m+5) 2>3, i.e., m>1 at 3<=x<=(m+5) 2if(m+5) 2<3, i.e. m<1 (m+5) 2<=x<=3 if(m+5) 2=3, i.e. m=1 x=3

x-2<0

x-3>=0

2-x+x-3>=m When m<=1, x<2 or x>=3

x-2<0

x-3<0

2-x+3-x>=m gives x<2 or x<(5-m) 2

If (5-m) 2<2 i.e. m>1 x<(5-m) 2if(5-m) 2>2 i.e. m<1 x<2 if(5-m) 2=2 i.e. m=1 x=2, I hope I am not mistaken.

You can consider three cases: x is greater than or equal to 3, x is less than or equal to 2, and x is greater than 2 and less than 3.

The image of LX-2L+LX-3L is a frying pan type, and the minimum value is taken within [2,3], m<=1

Do you have a mobile phone? This one is not good for typing, I will send it to you by text message.

From 2x-a<0 we get 1:x2x-1 to get 2:x<4 And because the solution set of the inequality group is all negative numbers, then: a 2 should be less than or equal to 0, i.e. a is less than or equal to 0

Solution: 2x-a 0 = x2x-1 = x<4 } a 0

The solution set is all negative.

The known system of equations about x and y is 3x+y=2+3m; x+3y=2-3m is a positive number for solving the cavity trap x and y, and the value range of m is obtained.

Solve a system of equations about x and y.

3x+y=2+3m

x+3y=2-3m

Get: x=(1+3m) 2

y=(1-3m)/2

The solution is positive. x=(1+3m) Bu Gai 2>0,1+3m>0,m>-1 Wu Xiannian 3

y=(1-3m) 2>0,1-3m>0,m<1 3m value range: -1 3x-1 The solution set is 1 x<2, find the value of a.

x-3(x-2)≤4

x-3x+6≤4

3x-x>=6-4

x>=1

a+2x/3>x-1

a+2x>3x-3

x, so the 1 x solution set is 1 x<2

So a+3=2

a=-1

Solution: If the price of 300 is 50% of the higher purchase price, then the purchase price of this year is 300 (1 + 50%) 200 yuan, if the price of 300 is 100% of the high purchase price, then the purchase price of hunger is 300 (1 + 100%) 150 yuan, so the purchase price of the clothing is between 150 yuan and 200 yuan. Let our counteroffer be $x, then the inequality group can be listed as follows:

150 (1+20%) x 200 (1+20%) solves: 180 x 240

So you should make a counteroffer between 180 yuan and 240 yuan.