Detailed answers to math problems in junior high school Hope master speed Thank you I give points I

1) Let the growth rate x , 64(1+x) 2=100 , and the solution is x =1 4

End of 2010: 100(1+x)=100x5 4=1252) Set up indoor x, outdoor y pcs.

5000x+1000y=150000 Simplification yields 5x +y=150 , y=150-5x

and 2x<=y<= So 2x<=150-5x<=Inequality group: Eq. 1: 2x<=150-5x

Equation 2: 150-5x<=

Equation 1 gives : x<=150 7, 150 7 is approximately equal to Equation 2 : x>=20, so 20<=x<=when x=20, y=50, and when x=21, y=45

1) Let the growth rate be x, then 64(1+x) = 100 x= 100(1+, so there were 125 cars at the end of 2010.

2) Set up x parking spaces indoors and Y parking spaces outdoors, then 2x y, 5000x+1000y 150000

That is, 5x+y 150 when y takes the maximum value, x=20, then y=50, when y takes the minimum value of 2x, 7x 150, x 21 at this time y=40, so you can build up to 20 indoors, 50 outdoors, or 21 indoor and 42 outdoor.

1.Let the annual growth rate be a, then 64x(1+a)(1+a)=100If a is obtained, then in 2010 it will reach 100x(1+a)=125 vehicles.

2。B indoors and C in the open air

150000=5000b+1000c

c is greater than 2b and less than.

When c is equal to 2b, then b = 21 and c = 45

When c=, then b=20, c=50

You can build up to 70 of them.

1) Let the increase rate be x 64 (1+x)2=100x1= x2=rounding) 100 (1+

Answer... 2) Set, indoor y open-air z

5000y+1000z<=150000

2y<=z<=

Solution. y z

1.A: 1 [(1 20)*5 (4+5)] = 36 hours B:

1 [(1 20)*4 (4+5)] = 45 hours2 a^2+1/a^2=(a+1/a)^2 -2=2^2 -2=2

3.1) x<-1 or 13

4. 5/v - 5/(v+a)

5. 1-1/(1-x)=1/(1-x)1/(1-x)=1/2

1-x=2x=-1

6. v+(a-b)/v=b+1

v^2-(b+1)v+a-b=0

v=[b+1 (b+1) 2-4(a-b)] 2=[b+1 *b 2+6b-4a+1)] 25,The expression of question 6 is really not agreeable, and I really can't understand it without parentheses.

1.A 20 hours to complete the total work of 5 9. B completes 4 9 then A alone takes 20 (5 9) = 36 hours.

B alone takes 20 (4 9) = 45 hours.

2。Solution: a=1

a2+1/a2=1

3.Solve the equation 3-x x2-1=0 and calculate it yourself. (Note that x cannot be equal to +1 and -1).

1。A: 20*(5+4) 4=45B: 20*(5+4) 4=36.

1 or 13, it is negative.

6。If both sides are multiplied by v, then v 2-(b+1)v+a-b=0Solving. v=(b+1)/2+ -b+1)^2-4a+4b

By o as od ac in d, it is easy to know ao=5, od=4, so ad=3, ac=6;

2) Let apc be an isosceles triangle after t seconds, then ap=10-t, if ac=pc, and the point c is ch ab in h, then ahc ado, ac:ah=oa:ad, i.e., ac:

5:3, after passing apc is an isosceles triangle;

If ap=ac, from pb=x, ab=10, ap=x, and ac=6, then 10-t=6, the solution is t=4s, after 4s, apc is an isosceles triangle;

If ap=cp, p and o coincide, then ap=bp=5, after 5s, apc is an isosceles triangle

This one is relatively simple, and you can have the cosine theorem.

Solution: Derived from the question:

0∴（m+1)²－4×1×(2m－1)≥0m²+2m＋1－4×2m＋4≥0

m²－6m＋5≥0

m－1）（m－5）≥0

Solution: 1 m 5

The integer m has the following values: 1,2,3,4,5

Solution: Derived from the question:

0∴（m+1)2－4×1×(2m－1)≥0m2+2m＋1－4×2m＋4≥0

m2－6m＋5≥0

m－1）（m－5）≥0;m≥5，m≤1;

x（1,2)=[-(m+1)±√m2－6m＋5)]/2;

The two are integers and hold when m=5 or m=1.

Deformed:

x＋2)m＝－x^2－x＋1

m＝(－x^2－x＋1)/(x＋2)

－x^2－x＋2－1)/(x＋2)

x＋1－1/(x＋2)

x 2 is a factor of 1, thus x 1 or 3

When x 1, m 1

When x 3, m 5

3. All the middle line is not written, and the first line below is not written.

Which question in the picture needs to be answered?

FYI. In the first problem, l2 is converted to y=a(x-m) 2+2m+1, and substituting a=1 to obtain the vertex coordinates (m,2m+1) because the center symmetries the point (1,3).And because the vertex of L1 is on the Y axis, the vertex of L1 is (0,n); And because of the central symmetry point, m+0 2=1, m=2

The second question, the beginning of the lack of time is similar to the above, that is, there is no need to substitute a=1I won't play it, and substitute m=2 into the equation to get the vertex coordinates of l2 (2,5), and from the symmetry center formula, answer the vertex of l1 (0,1), from the meaning of the question, the analytical formula of l1 is y=x 2+n, and substituting the vertex to get that n is equal to 1, so the analytical formula of l1 is y=x 2+1When y=0, x=1 or -1, because point c is on the positive x-axis, so c(1,0)

From the point of central symmetry, 1+x 2=x=x=1, y=6Then we know that the point c(1,0) is about the center point of symmetry, and the point of Qingchawang on l2 is (1,6).Finally, substitute the point (1,6) into y=a(x-2) 2+5Derive a