There are two kinds of sulfuric acid solutions with different mass fractions. The mass fraction of type A is 90% and that of type B is 75%. Now we need to prepare 12 kg of 85% sulfuric acid solution. How many kg of each of the two sulfuric acid solutions?

X+Y=12
90%X+75%Y=12×85%
Combine the above two equations and find out
X=8,Y=4
Unit: kg

(√2+√5-√3)/(2√30-6√2+4√3)

(√2+√5-√3)/(2√30-6√2+4√3)
=The steps of √ 6 (√ 2 + √ 5 - √ 3) / √ 6 (2 √ 30-6 √ 2 + 4 √ 3) are multiplied by √ 6
=√6(√2+√5-√3)/(12√5-12√3+12√2)
=√6/12*(√2+√5-√3)/(√2+√5-√3)
=√6/12
Hope to adopt

Junior high school mathematics will not keep up, is that so

No, mathematics is a step-by-step discipline, and you can't learn without more than ten years of basic skills. There is still a gap between primary school mathematics and junior high school mathematics. It's not the same. Primary School Mathematics Olympiad can teach you some difficult mathematics at most, and it's not related to junior high school mathematics to expand your thinking!
Mathematics is a serious science, and so is Mathematical Olympiad. Now "Olympics" is used by some people as a tool to make money, thinking that children are easy to cheat!

Divide the square of 20012000 by the square of 20011999 + the square of 20012001 - 2
(the original question is expressed as a fraction)

Denominator = square of 20011999 + square of 20012001 - 2 = (20012000-1) square + (20012000 + 1) square - 2 = square of 20012000 - 2x20012000 + 1 + square of 20012000 + 2x20012000 + 1-2 = square of 2x20012000

Calculation: (- 2x) 2 - (- 2x + 3) (- 2x-3)

（-2x）2-（-2x+3）（-2x-3），=4x2-[（-2x）2-32]，=4x2-（4x2-9），=9．

In 1-100 integers, so that any number is not three times of another number, how many numbers are there at most in this number set?
My solution is 78 with 100 minus can be divided by 3 and not times the number of 9!

Yes
From 1 to 100, there are 33 numbers that can be divided by 3
There are 11 numbers divisible by 8
So the answer is 100-33 + 11 = 78
Your method is right

A. B, C three micro robots around a circular orbit high-speed movement, they start clockwise at the same time, a in 2 seconds to catch up with B, 2.5 seconds to catch up with C, when C catch up with B, C and B movement distance ratio is 3:2, ask the first minute, a around this Circular orbit movement how many circles?

42 laps
Let the length of the track be l and the velocity of A.B.C be x.y.z,
Let C overtake B in T seconds
Then TZ: ty = 3:2 = = > Z: y = 3:2
From the meaning of the title
L = 2x-2y and L = 2.5x-2.5z
Substituting Z: y = 3:2 gives X: y = 7:2 and L = 5Y
Then the time required for each lap of a is L / x = 10 / 7
The number of laps per minute is 60 / (10 / 7) = 42

ABCDE is a five digit natural number, where a, B, C, D, e are Arabic numerals, and a

When a

1. Given that the sum of - 2 / 3A ^ XB ^ y + 8 and 4A ^ 2yb ^ 3x-y is a monomial, find the value of x ^ y
2. Let X and y be rational numbers, and M = 2x ^ 2 + 9y ^ 2 + 8x-12y + 12, then what kind of number is m?

And is monomial, so
-2/3a^xb^y+8
4a^2yb^3x-y
Must be of the same kind
The exponents of the same letters must be equal
x=2y
y+8=3x-y
Explanation
x=4
y=2
So x ^ y = 16
M=2x^2+9y^2+8x-12y+12
=(2x^2+8x+4)+(9y^2-12y+4)+4
=(2x+2)^2+(3y-2)^2+4
It's a positive number

(x+1)(x+1)=x^2-1
(x-1)(x^2+x+1)=x^3-1
(x-1)(x^3+x^2+x+1)=x^4-1
(x-1)(x^4+x^3+x^2+x+1)=x^5-1
① Observe and try to find 2 ^ 6 + 2 ^ 5 + 2 ^ 4 + 2 ^ 3 + 2 ^ 2 + 2 + 1
② Judgment 2 ^ 2008 + 2 ^ 2007 + 2 ^ 2006 + +What are the digits of 2 ^ 2 + 2 + 1

2^6+2^5+2^4+2^3+2^2+2+1=（2-1）（2^6+2^5+2^4+2^3+2^2+2+1）=2^7-12^2008+2^2007+2^2006+… +2^2+2+1=(2-1)(2^2008+2^2007+2^2006+… +2 ^ 2 + 2 + 1) = 2 ^ 2009-1 = 2x (2 ^ 4) ^ 502-1 (2 ^ 4) ^ 502 is 6, 2x (2 ^ 4) ^ 502 is 2, 2x (...)

There are two kinds of sulfuric acid solutions with different mass fractions. The mass fraction of type A is 90% and that of type B is 75%. Now we need to prepare 12 kg of 85% sulfuric acid solution. How many kg of each of the two sulfuric acid solutions?