Math Problems I don t know what to do anymore

If AC is connected, then S(ABC) = (1 2) S(ABCD) = 35 2

Thus, be ab=s(bce) s(abc)=14 (35 2)=4 5

bf/bc= s(abf)/s(abc)=5/(35/2)=2/7

Then there are: BC CF=7 5, AE BE=1 4

Thus Menelaus' theorem knows: (bc cf) (fg ag) (ae be)=1

i.e.: 7 5 (fg ag) 1 4=1

So fg ag=20 7, i.e., fg af=20 27

and s(acf) = s(abc) - s(abf) = 35 2-5 = 25 2

So s(cfg) s(acf) = fg af, i.e. s(cfg) = s(acf) (fg af) = (25 2) (20 27) = 250 27

So s(begf) = s(bce)-s(cfg) = 14-250 27 = 128 27

First find the side length of the square, then the length of the BF and BE sides, and subtract the area of the triangle CGF from the area of the triangle BCE.

x-1=0, x=1, x-squared to the fourth power + mx to the third power + nx-16=1 to the fourth power + m-16=m to the third power + n-16=m to the third power + n-15=0 (1).

x-2 = 0, x = 2, x to the fourth power + mx to the third power + nx-16 = 2 to the fourth power + m2 to the third power + n2-16 = 16 + 2m to the third power + 2n-16 = 2m to the third power + 2n = 0 (2).

Synact 2(1)=(2).

2 (m to the third power) + 2n-30 = 2m to the third power + 2n6 (m to the third power) = -30

m=- 5 under the cubic root number

n=20

m=-5

n=20, the specific process is to bring x=1 and x=2 into the equation respectively, and then follow the 2-element system of equations.

It's the same as the first one.

There are common factors (x-1) and (x-2), which means that the quadratic equation can be converted into the form f(x) = (x-1) (x-2) (x-a) (x-b).

is a natural number. i.e., when x=1, and x=2 are the solutions to this equation (x-1)(x-2)(x-a)(x-b)=0.

Bring x=1, and x=2 into the original equation.

1+m+n-16=0

16+8m+2n-16=0

Solve the equation m = -5

n=20

Altogether available.

5 20 = 100 (root).

Each one should be sawn 5-1 times.

So a total saw.

5-1) 20=80 (times).

We're happy to answer for you!

A total of 5*20=100 short logs can be obtained.

A total of 4 * 20 = 80 times to saw.

an+1=2an+λ

an+1+λ=2(an+λ)

an+1+ an+ =2, a1+ =1+ is a proportional series with 1+ as the first term and the common ratio of 2.

an+λ=(1+λ)2^(n-1)

an=(1+λ)2^(n-1)-λ

In 1, the upstairs solution should be noted that when = 1, it is not a proportional series.

Null 1: 12 with 7, 6 with 11, 9 with 8, 5 with 10 Null 2: 10 with 7, 11 with 8

Null 3: 7 vs. 11, 8 vs. 10

Null 4: 2 vs. 10, 4 vs. 11, 1 vs. 9, 3 vs. 12 Null 5: 3 vs. 10, 4 vs. 9

Null 6: 4 with 10, 3 with 9

Since the multiplication of the three is less than 0, there are two cases, all three numbers are negative, but the sum of the three equals 2, this situation is excluded, and the second case is that two numbers are positive and one number is negative.

Suppose a<0, b>0, c>0

Then, the absolute value of ab is -ab, and the absolute value of ca is -ca, then x=-1+1-1=-1

The required equation can be reduced to a+b+c-5=2-5=-3

From ABC < 0, it can be seen that there are 1 or 3 negative numbers in ABC (a+b+c=2 excludes the second case); In this case, 2 of AB, BC, AC are negative and 1 is positive (verifiable); This yields: x = 1;Substitution is available:

Equation = 3;