Extended math for junior high school, extended math for junior high school

One (1).It does not exist, because the line segment between two points is the shortest, the shortest distance is 10cm, and it cannot be 8cm.

3）.Point c can be on a straight line ab, but not absolutely, because there are many such points, two on the straight line ab, and more than one outside the straight line ab. As shown in the figure below.

Two. You're wrong.

Solution, 1, can not be the shortest straight line between two points.

If point c is to hold, then the sum of its distances from ab should be greater than 10 cm, and it is equal to 8 cm

does not exist. 2. In the same way, point c does not exist.

3. Set point C on the right side of AB.

The sum of the distances from point C to points A and B is equal to 20 cm

If it exists, the distance to AB should be 15 cm and 5 cm respectively, and the same goes for two.

But I don't know if it can not be on the straight line ab, if it is on the plane, it is an infinite number) 1, on the same floor, I also think you wrote it wrong, if aob+ eof=135°, then you can know.

OE divides the AOC

aoc=∠aob+∠boc=150°

eoc=1/2∠aoc=75°

eof=∠aof-∠aoe=25°

of boc

bof=1/2∠boc=30°

eof=∠eof+∠bof=55°

Of course, the first question does not exist, and the sum of the two sides of the triangle is greater than the third time.

The reason for the second question is the same as above. The third question is possible on a straight line AB, the distance A or B5cm is OK, so that AC15cm, BC5cm, or vice versa, the question stem is satisfied. Of course, it is also possible to no longer ab, and the simplest example is an equilateral triangle.

If I'm not mistaken, according to the conditions, aob+ eof=135° is... Did you write it wrong? The angular bisector is very distinct, and it cannot be 156°.

According to the first formula, the root number a=(7-5|b|3, and then substitute the second formula to get: s=(14-10|b|Let Zen 3-3|b|=(14-19|b|)/3;

Gentan car dust data|b|>=0, the value of the sail elimination to s;

The same is according to the first formula: |b|=。Then it is also the second equation that is substituted: we get s=..Then according to the root number a is also "=0", and then get a value of s:

Combining the two ranges of values gives the range of values for s.

Correct, the two triangles are congruent, and the angle a is equal to the angle acf, according to the triangle inner angle and 180, so it is correct.

1)(1/2+……1/2006)*(1/2+……1/2005)+(1/2+……1/2006) -1/2+……1/2006)*(1/2+……1/2005)-(1/2+……1 2005) = 1 20062) abc in 3 positive, original formula = 3; 2 positive 1 negative, original = 1; 1 positive 2 negative, original formula = -1; 3 negative manuscript finch, original formula = -3 synthesized upper state Jing spine, the possible values of the original sail infiltration type are 3,1,-1,-33)2 55=(2 5) 11=32 11, 3 44=(3 4) 11=81 11, 4 33=(4 3) 11=64 11 32<64<81 2 55<4 33<3 444)abc<0, 3 negative in abc, or 1 negative 2 positive a+b+c>0, 1 negative 2 positive in abc, x=1x 19-92x+2=1-92+2=-89

According to the first article, let the Zen formula obtain: the root number a=(7-5|b|3, and then substitute the second formula to get: s=(14-10|b|）/3-3|b|=(14-19|b|)/3;According to |b|>=0 to get the value of s; The same is according to the first formula: |

b|=。Then it is also the second equation that is substituted: we get s=..

Then according to the root sail elimination number a is also tan car dust is "=0, and then get a value of s: combine the range of two values to get the value range of s.

Solution: a1=-1 2

a2=1/(1-a1)=1/[1-(-1/2)]=2/3a3=1/(1-a2)=1/(1-2/3)=3a4=1/(1-a3)=1/(1-3)=-1/2a5=1/(1-a4)=1/[1-(-1/2)]=2/3……It follows that a1 = -1 2, and from the second number onwards, each number is equal to "the reciprocal of the difference between 1 and the number in front of it", which is the cycle of .

2012 3 = 670 more than 2

a2012=2/3

Try to count a few groups.

a1 a2 a3 ..Derived.

1/2， 2/3， 3， -1/2， 2/3， 3， 。Cycles are every 3 cycles.

2012 divided by 3 remainder 2 , so a2012 = 2 3 (see a2 a5 a8 a11 are all remainder 2, the value is 2 3) qed

Solution: a1=-1 2

a2=1/(1+1/2)=2/3

a3=1/(1-2/3)=3

a4=1/(1-3)=-1/2;

The distribution of this number is regular; Its period is 4-1=3;

then 2012 3=67....2;

then a2012=a2=2 3;

a2012=2 3, for the following reasons: because a1=-1 2, and from the second number onwards, each number is equal to the reciprocal of the difference between "1 and the number in front of it, so a2=1 [1-(-1 2)]=2 3,a3=1 (1-2 3)=3,a4=1 (3-1)=1 2,In summary, it can be seen that every 3 numbers are in a cycle, 2012 3=670....2, so a2012=2 3

Let's start with the first few numbers of this sequence: the first number is 1 2;The second number is the reciprocal of 1 - 1 2 = 1 2, which is 2; The third number is the reciprocal of 1-2=-1, which is -1; The fourth number is the reciprocal of 1-(-1)=2, which is 1 2.

We see that the first number and the fourth number are repeated, and we can judge that this number series is a cycle of three numbers in a section, that is, three numbers are in a group, and the number at the beginning is 1 2

a2012 is the 2012th number, 2012 divided by 3 = 670 more than 2, we can see that the 2012th number is the second number of group 671. The first number in a group of three is 1 2 and the second number is 2So, the value of a2012 is 2

Column: a1=1 2

The reciprocal of a2=(1-1 2) = 2

The reciprocal of a3 = (1-2) = -1

The reciprocal of a4=[1-(-1)] = 1 2

The number column is: 1 2 2 -1 1 2 2 -1....

i.e. a2012=2

a1=-1/2，a2=2/3, a3=3, a4=-1/2

These numbers are cyclic in groups of 3, so divide 2012 by 3, and the remainder is the number from a1 to a3.

According to the above law, it is found that these numbers are cyclic in a group of 3, 2012 3=670....2，∴a2012=a2=2/3

a1=-1/2，a2=2/3, a3=3, a4=-1/2

Every three is a cycle, so a2012 = 2 3