(3m²-n²)²-(m²-3n²)²=

(3m²-n²)²-(m²-3n²)²=

(3m 2 - n ^ 2) 2 - (M? - 3N? 2) = (3m ^ 2-N ^ 2 + m ^ 2-3n ^ 2) * (3m ^ 2-N ^ 2-m ^ 2 + 3N ^ 2) = (4m ^ 2-4n ^ 2) * (2m ^ 2 + 2n ^ 2) = 8 * (m ^ 2-N ^ 2) * (m ^ 2 + n ^ 2) (1) = 8 * (m ^ 4-N ^ 4) = 8m ^ 4-8n ^ 4 (2) if factorized, then (1) if simplified, then (2

Fraction: M 2 - 3 M / 9-m 2

m²-3m/9-m²
=m(m-3)/(3-m)(3+m)
=-m(3-m)/(3-m)(3+m)
=-M / 3 + m same term cancellation
Answer: the original formula is - M / M + 3

If the cubic root a 3 = root a 2, then the value range of a is

Known (a 3) ˆ (1 / 3) = (a 2) ˆ (1 / 2)
∵ when a

If M + 2 = - | 2m-n |, and N = 3M + X, find the value of X under the root sign

√(m+2)=-|2m-n|,
√(m+2)+|2m-n|=0
√(m+2)=0,|2m-n|=0
m+2=0,2m-n=0
m=-2,n=-4
n=3m+x
-4=(-2)*3+x
-4=-6+x
X=2
√x
=√2

Let a and B be rational numbers and satisfy the equation a + B root 3 = root 6 × root (2 + root 3), then what is a + B equal to?

Root 6 × root (2 + root 3) = √ 6 * √ ((1 / 2) (3 + 2 √ 3 + 1)) = √ 6 * √ ((1 / 2) (1 + √ 3) ^ 2)
=√6*√（1/2）*√（1+√3）^2=√3*（1+√3）
=3+√3=a+b√3
Since a and B are rational numbers, a = 3, B = 1

Is there a positive integer a, B (a less than B) such that it satisfies the root a + root B = root 1404? Thank you. Thank you. Thank you

√a+√b=√1404=6√39=√39+5√39=2√39+4√39
A

If the sequence {an} is positive and a1 + A2 +. + the square of an = n + 3N (n belongs to positive integer), then A1 / 2 + A2 / 3 + What is an / N + 1 equal to

√a1+√a2+… +√a(n-1)+√an=n²+3n
√a1+√a2+… +√a(n-1) =(n-1)²+3(n-1)
Subtract the two formulas to get √ an = 2n + 2
∴an=4(n+1)²
∴an/(n+1)=4(n+1)=4n+4
∴a1/2+a2/3+...+an/(n+1)
=4(1+2+...+n)+4n
=4(1+n)n/2+4n
=2n²+6n

First simplify, then evaluate, (x-2-x-2 / 5) divided by 2x + 4 / x-3, where x = radical 2-3

The first denominator is x + 2
The original formula = [(x + 2) (X-2) - 5] / (x + 2) × (2x + 4) (x-3)
=(x+3)(x-3)/(x+2)×2(x+2)/(x-3)
=(x+3)/2
=√2/2

First simplify and then evaluate: (1 part of A-B + 1 part of a) divided by a + B part AB, where a = root 2 + 1, B = root 2

(1 in A-B + 1 in a) divided by ab in a + B
=(b+a+a-b) /[ (a+b)(a-b) ]× (a+b)/ab
=2/[b(a-b)]
=2/(ab-b²)
=2 / (2 + radical 2 - 2)
=Radical 2

First simplify, then evaluate: (a + radical 3) (a-radical 3) - A (a-6) + 6, where a = 1 / 2 + radical 1 / 2 First simplify, then evaluate: (a + radical 3) (a-radical 3) - A (a-6), do not need to add 6

The original formula = a 2 - 3-A 2 + 6A + 6
=6a+3
=6×(1/2+√2/2)+3
=6+3√2