The sum of three consecutive natural numbers is less than 15. There are several groups in such an array

The sum of three consecutive natural numbers is less than 15. There are several groups in such an array

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The sum of three continuous natural numbers is 15. Find the three natural numbers

15÷3=5,5-1=4,5+1=6
So it's four, five, six

The sum of three consecutive natural numbers is less than 13. How many groups of such natural numbers are there? Write them separately?

123 234 345 three groups

In ABC, if log (2) Sina log (2) CoSb loog (2) sinc = 1 (the base in brackets), what triangle is this triangle

Sina > 0, CoSb > 0, sinc > 0 log (2) Sina log (2) CoSb loog (2) sinc = 1, that is, Sina / (cosbsinccosb) = 2sina = 2cosbsinccosb = Sina / (2sinc) a / Sina = C / sinc, that is, CoSb = A / (2C) = (A & # 178; + C & # 178; - B & # 178;) / 2acc (A & # 178; + C & # 178; - B & # 178;) = a

a> B > 0, then a square + 1 / AB + 1 / A * (a-b) is the minimum
A square + 1 / AB + 1 / A * (a-b)

a>b>0
a^2+1/ab+1a(a-b)
=a(a-b)+ab+1/ab+1/a*(a-b)
Under the root sign of 4 times or more (a (a-b) * AB * 1 / AB * 1 / A * (a-b)) = 4
This is a generalization of the mean inequality, focusing on the construction

If the square of | a + 2 | + (B-3) is 0, then AB is square=

|The square of a + 2 | + (B-3) = 0
a+2=0 b-3=0
a=-2 b=3
ab^2=-2*9=-18
(ab)^2=(-2*3)^6=36

(a + b) of (ab-b's Square) / [(a-b's Square] of (AB + B's Square)

The original formula = (a + b) / b (a-b) × (a-b) & # / b (a + b)
=(a-b)/b²

If a squared - B squared = AB, find a squared part of B squared + B squared part of a squared part
Be quick, I'll use it

∵a²-b²=ab
∴(a²-b²)²=a²b²
∴a²/b²+b²/a²=(a^4+b^4)/a²b²=(a^4+b^4)/ (a²-b²)²=(a^4+b^4)/ (a^4-2a²b²+b^4)= (a^4+b^4-2a²b²+2a²b²)/(a^4-2a²b²+b^4)=1+2a²b²/(a²-b²)²=1+2=3

AB a + B = a square B?

On the left side of the equation: denominator AB, numerator a + B
On the right side of the equation: denominator a * a * B,
If you multiply the denominator by a, you naturally have to multiply the numerator by a to ensure that both sides of the equation are equal
So the molecule is a (a + b)
A (a + b)

The square of a in ab - the square of B △ the square of (a-b)
Division of fractions
3Y squared / 2x squared x6x / 5Y △ 21x squared / 10Y

1、[(a^2-b^2)/ab]÷(a-b)^2
=[(a+b)(a-b)/ab]÷(a-b)^2
=(a+b)/[ab(a-b)]；
2、(3y^2/2x^2)×(6x/5y)÷(21x^2/10y)
=(9y/5x)÷(21x^2/10y)
=6y^2/7x^3.