(x2-5) / (x-1) + (10x-10) / (x2-5) = 7 solutions X2 is the square of X

(x2-5) / (x-1) + (10x-10) / (x2-5) = 7 solutions X2 is the square of X

Let y = (X & # 178; - 5) / (x-1)
The original equation can be reduced to
y+10/y=7
y²-7y+10=0
y=2 y=5
When y = 2, (X & # 178; - 5) / (x-1) = 2
x²-2x-3=0
X = - 1 or x = 3
When y = 5, (X & # 178; - 5) / (x-1) = 5
x²-5x=0
X = 0 or x = 5
Test: x = - 1, x = 0, x = 3, x = 5 are the solutions of the equation

Solve the equation (10 / x ^ 2 + X-6) + (2 / 2-x) = 1

10/(x^2+x-6)+2/(2-x)=10/(x+3)(x-2)-2/(x-2)=1 10-2(x+3)=(x+3)(x-2)=x^2+x-6
X ^ 2 + X-6 = 10-2x-6 x ^ 2 + 3x-10 = 0 (x + 5) (X-2) = 0 x + 5 = 0 x = - 5 X-2 = 0 x = 2 (rounding) x = - 5

Solution equation (6 / 4-x2) + (1 / x + 2) = (1 / 2-x)

(6/4-x^2)+(1/x+2)=(1/2-x)
6+2-x=2+x
2x=6
x=3
It is proved that x = 3 is the solution of the equation

Given the first n terms and Sn of the arithmetic sequence {an} with tolerance greater than zero, and satisfying A1 * A6 = 21, S6 = 66, find the general term formula an of the sequence {an}

a1*a6=a1*(a1+5d)=21
S6=6a1+(6+5)/2*d=66
The results show that A1 = 1, d = 31 / 5 or A1 = 21, d = - 9 / 5
The latter group is removed because D is required to be greater than 0
an=1+(n-1)*31/5=(31n-26)/5

In the sequence {an}, an = 3n-1, the definition is used to prove that {an} is an arithmetic sequence, and its tolerance is obtained

an=3n-1
a（n+1）=3（n+1）-1=3n+2
a（n+1）-an=（3n+2）-（3n-1）=3
So {an} is an arithmetic sequence with a tolerance of 3

In the sequence {an}, A1 = 1, an + 1 = 3an + (n + 1) multiplied by 3N (n belongs to n *)
Let BN = an / N, find the general formula of BN

A (n + 1) = 3an + (3N (n + 1))
Let a (n + 1) + X (n + 1) ^ 2 + y (n + 1) + Z = 3 (an + xn + yn + Z)
We get x = 3 / 2, y = 3, z = 9 / 4
So this sequence is an equal ratio sequence with 31 / 4 as the first term
an+3/2n+3n+9/4=31/4×3^(n-1)
Then BN is also obtained

How many natural numbers from 1 to 100 are multiples of 2, 3 and 5

2*3*5=30
30*2=60 30*3=90
So there are three, 30, 60, 90

Find out all the multiples of 8 in the natural numbers from 1 to 100, which is the smallest of the multiples of 9

8、16、24、32、40、48、56、64、72、80、88、96.
72 meet the requirements

Divide 100 into the sum of two natural numbers, one is a multiple of 7, the other is a multiple of 11. These two natural numbers are______ And______ ．

Because 100 = 7 × 8 + 11 × 4, the two natural numbers are 56 and 44, so the answer is 56, 44

In natural numbers 1-100, how many numbers are not multiples of 7 or 13

100 △ 7 = 14,2
100 △ 13 = 7,9
The common multiple of 7 and 13 is 91
So:
A number that is not a multiple of 7 or 13
=100-14-7+1
=80