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1) x4y2 + x2y4 + x4y3 + x2y5 = (x4y2 + x2y4) + (x4y3 + x2y5) = x2y2.(x2 + y2) + x2y3.(x2 + y2) = x2y2.(x2+ y2) (1 + y) = [xy.(x2 + y2)].[xy(1+y)]
=> x4y2 + x2y4 + x4y3 + x2y5 chia cho xy.(x2 + y2) bằng xy.(1+ y)
2) A = (n2 - 8)2 + 36 = n4 - 16n2 + 100 = (n4 + 20n2 + 100) - 36n2 = (n2 + 10)2 - (6n)2 = (n2 - 6n+ 10).(n2 + 6n+ 10)
Vậy để A là số nguyên tố thì n2 - 6n + 10 = 1 hoặc n2 + 6n + 10 = 1
Mà n là số tự nhiên nên n2+ 6n + 10 > 1
=> n2 - 6n + 10 = 1 => n2 - 6n + 9 = 0 => (n -3)2 = 0 => n = 3
Vậy....
3) a) = xy(x - y) - xz(x + z) + yz.[(x+ z) + (x - y)] = xy(x - y) - xz(x + z) + yz.(x + z) + yz(x - y)
= [xy(x - y) + yz.(x - y)] + [(yz.(x+ z) - xz(x+z)] = y(x - y)(x+ z) + z(x + z).(y - x) = (x+ z)(x- y).(y - z)
b) = (x2 + x)2 - (2x)2 - 4(x+3) = (x2 + x + 2x).(x2 + x- 2x) - 4(x+3) = (x2 + 3x).(x2 - x) - 4(x+3)
= (x+3).[x.(x2 - x) - 4] = (x+3).(x3 - x2 - 4) = (x+3).(x3 - 8 + 4 - x2) = (x+3).[(x - 2)(x2 + 2x + 4) - (x - 2).(x+2)]
= (x + 3).(x - 2).(x2 + 2x + 4 - x- 2) = (x + 3).(x - 2).(x2 + x + 2)
4) a) n4 + 1/4 = (n4 + n2 + 1/4) - n2 = (n2 + 1/2)2 - n2 = (n2 - n + 1/2).(n2 + n + 1/2) = [n(n - 1) + 1/2].[n.(n+1) + 1/2]
Áp dụng công thức ta có:
A = \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right).\left(4^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}=\frac{\frac{1}{2}.\left(1.2+\frac{1}{2}\right).\left(2.3+\frac{1}{2}\right).\left(3.4+\frac{1}{2}\right)...\left(18.19+\frac{1}{2}\right).\left(19.20+\frac{1}{2}\right)}{\left(1.2+\frac{1}{2}\right).\left(2.3+\frac{1}{2}\right).\left(3.4+\frac{1}{2}\right).\left(4.5+\frac{1}{2}\right)...\left(19.20+\frac{1}{2}\right).\left(20.21+\frac{1}{2}\right)}\)
A = \(\frac{\frac{1}{2}}{20.21+\frac{1}{2}}=\frac{1}{841}\)
\(1,x+y+z=0=>x=-\left(y+z\right)\)
\(=>x^2=\left(y+z\right)^2=y^2+2yz+z^2\)
\(=>x^2-y^2-z^2=2yz\)
\(=>\left(x^2-y^2-z^2\right)^2=\left(2yz\right)^2=4y^2z^2\)
\(=>x^4+y^4+z^4-2x^2y^2-2x^2z^2+2y^2z^2=4y^2z^2\)
\(=>x^4+y^4+z^4=4y^2z^2-2y^2z^2+2x^2z^2+2x^2y^2=2x^2y^2+2y^2z^2+2x^2z^2\)
\(=>2\left(x^4+y^4+z^4\right)=\left(x^2+y^2+z^2\right)^2\left(đpcm\right)\)
\(2,A=2\left(x^6-y^6\right)-3\left(x^4+y^4\right)\)
\(=2\left[\left(x^2\right)^3-\left(y^2\right)^3\right]-3\left(x^4+y^4\right)\)
\(=2\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)-3\left(x^4+y^4\right)\)
\(=2\left(x^4+x^2y^2+y^4\right)-3\left(x^4+y^4\right)\)
\(=2x^4+2x^2y^2+2y^4-3x^4-3y^4=-x^4+2x^2y^2-y^4\)
\(=-\left(x^4-2x^2y^2+z^4\right)=-\left[\left(x^2-y^2\right)^2\right]=-1\) (do x2-y2=1)
\(3,\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)+15\)
\(=\left(x-3\right)\left(x+3\right)\left(x-1\right)\left(x+1\right)+15=\left(x^2-9\right)\left(x^2-1\right)+15\left(1\right)\)
Đặt \(x^2-5=t\),khi đó (1) trở thành :
\(\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t-1\right)\left(t+1\right)\)
\(=\left(x^2-6\right)\left(x^2-4\right)=\left(x^2-6\right)\left(x-2\right)\left(x+2\right)\)
\(4,a,20^n-1=20^n-1^n=\left(20-1\right)\left(20^{n-1}+20^{n-1}+...+1^{n-1}\right)\)
chia hết cho (20-1)=19
=>20n-1 là hợp số vì có nhiều hơn 2 ước
b) đang kẹt,vấn đề nằm ở đề