Multiply this problem by (4/4), take the 4 in the denominator into the square root (as 16) and divide each of the two terms by that 16. For the first term, split the 16 into 2x2x2x2 and divide each of 500, 502, 504, 506 by 2; for the second term, +16 divided by 16 leaves you with +1. All of that leaves you with 4x the classic problem of the square root of four consecutive integers multiplied plus one. Sqrt(a x (a+1) x (a+2) x (a+3) +1) = a^2 + 3a +1 Sqrt (500 x 502 x 504 x 506 +16) = 4 x Sqrt (250 x 251 x 252 x 253 + 1) = 4 x (250^2 + 3 x 250 +1) However, recognizing this and reducing the problem to one you already know doesn't actually make the final computation easier. :)
(a²-9)(a²-1) = a²(a²-1) - 9(a²-1) (a + b)(c +d) = a(c+d) + b(c+d), each one of the first terms is multiplied to the second tem as a whole (a + b)(c +d) also equals (ac + ad) + (bc + bd), this is probably what ur used to, now factor out a and b u get a(c+d) + b(c+d) , now the comon factor is (c+d) and can be factored out to (c+d)(a+b) .. wich is what u started with.. when doing mutliplication it doesent matter if we do 'ab' or 'ba' .. one imprtant note is that either term can be negative ex a( b + (-c) ) = ab + (-ac) = ab - ac
Multiply this problem by (4/4), take the 4 in the denominator into the square root (as 16) and divide each of the two terms by that 16. For the first term, split the 16 into 2x2x2x2 and divide each of 500, 502, 504, 506 by 2; for the second term, +16 divided by 16 leaves you with +1. All of that leaves you with 4x the classic problem of the square root of four consecutive integers multiplied plus one. Sqrt(a x (a+1) x (a+2) x (a+3) +1) = a^2 + 3a +1
Sqrt (500 x 502 x 504 x 506 +16) = 4 x Sqrt (250 x 251 x 252 x 253 + 1) = 4 x (250^2 + 3 x 250 +1)
However, recognizing this and reducing the problem to one you already know doesn't actually make the final computation easier. :)
500＝Ｘと置くと，与式＝Ｘ×（Ｘ＋2）×（Ｘ＋4）×（Ｘ＋6）＋16＝［（Ｘの2乗）＋6Ｘ＋4］の2乗……Ｘ＞0より，（Ｘの2乗）＋6Ｘ＋4＞0…従って求める答えは…（500の2乗）＋3000＋4＝253004…となる。
502=a √[(a-2)a(a+2)(a+4)+16] = √[(a²+2a-8)(a²+2a)+16]
a²+2a=U √[(U-8)U+16] = √[U²-8U+16] = √(U-4)² = U-4
U-4 = a²+2a-4 = a²+2a+1-5 = (a+1)²-5 = 503²-5 = 253004
Super thanks
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short trick for this question 500×506+4
3:15???
(a²-9)(a²-1) = a²(a²-1) - 9(a²-1)
(a + b)(c +d) = a(c+d) + b(c+d),
each one of the first terms is multiplied to the second tem as a whole
(a + b)(c +d) also equals (ac + ad) + (bc + bd), this is probably what ur used to, now factor out a and b u get a(c+d) + b(c+d) , now the comon factor is (c+d) and can be factored out to (c+d)(a+b) .. wich is what u started with.. when doing mutliplication it doesent matter if we do 'ab' or 'ba' .. one imprtant note is that either term can be negative
ex a( b + (-c) ) = ab + (-ac) = ab - ac
?²=(503²-1)(503²-9)+16
?=503²-5
Roundabout 503², I dont care 😅