Set a six digit. 1abcde multiplied by 3, it becomes. Abcde1______ ．

Set a six digit. 1abcde multiplied by 3, it becomes. Abcde1______ ．

According to the above analysis: 142857 × 3 = 428571

If we know a six digit 1abcde and multiply it by 3, the result will be abcde1. Let five digit ABCDE be x, then the equation is 3 × (100000 + x) = ()
The solution is x = (), so the original six digit is ()

The equation is as follows
3×(100000+x)=10x+1
300000+3x=10x+1
10x-3x=300000-1
The solution is x = 42857, so the original six digit number is 142857

125 and 8 points, 7 times 8. Is there a simple way to calculate it

125 and eight, seven times eight
=125*8+7/8*8
=1000+7
=1007

Given that a, B, C and D are four different rational numbers, and (a + C) (a + D) = 1, (B + C) (B + D) = 1, then the value of (a + C) (B + C) is______ ．

From (a + C) (a + D) = 1, A2 + (c + D) a + CD = 1 is obtained. ① from (B + C) (B + D) = 1, B2 + (c + D) B + CD = 1 is obtained. ② according to ① and ②, a and B are two unequal real roots of the equation x2 + (c + D) x + CD-1 = 0. By Weida's theorem, ab = CD-1, a + B = - C-D, and (a + C) (B + C) = C2 + (a + b) C + ab = c2-cd + CD-1 = - 1

The average number of M is a, the average number of n is B, the average number of M + n is a______ ．

So the answer is: am + BNM + n

For three numbers a, B, C, m {a, B, C} is used to represent the average of the three numbers
If M {a, B, C} = min {a, B, C}, then (fill in the size relationship of a, B, c) ". Prove the conclusion you found
I want to know how to prove a = b = C

Let a > = b > = C, then m {a, B, C} = (a + B + C) / 3, min {a, B, C} = C;
And m {a, B, C} = (a + B + C) / 3 > = C, if and only if a = b = C, m {a, B, C} = C = min

Given that a, B, C and D are four different rational numbers, and (a + C) (a + D) = 1, (B + C) (B + D) = 1, find the value of (a + C) + (B + D)
I believe you must be smart

According to the first equation, (a + C) = 1 / (a + D)
According to the first equation, (B + D) = 1 / (B + C)
(a+c)+(b+d)=1/(a+d)+1/(b+c)=(a+b+c+d)/(a+d)(b+c)
The leftmost a + B + C + D of this equation can be eliminated from the rightmost a + B + C + D
1 / (a + D) (B + C) = 1, that is, (a + D) (B + C) = 1
And because (a + C) (a + D) = 1, so a + C = B + C, a = B, similarly, a = b = C = D can be obtained. Substituting any equation in the condition, a = b = C = D = 1 / 2, so (a + C) + (B + D) = 2

Given that a, B, C are rational numbers, and | a + 1 | + | B-2 | + | C-5 | = 0, find the value of a + 2B + 3C

The sum of three nonnegative numbers is 0, so all three numbers are 0
So a = - 1, B = 2, C = 5
a+2b+3c=-1+4+15=18

Let a, B, C be rational numbers, satisfying (3a-2b + C-2) &# 178; + (a + 2b-3c) &# 178; + │ 2a-b + 2c-3 │ ≤ 0, and find the value of a, B, C

Given that a, B and C are rational numbers, then the opposite number of - A + 2B + 3C is
A.a-2b+3c B.a-2b-3c C.a+2b-3c D.a+2b+3c

-(-a+2b+3c)
=a-2b-3c
Choose B