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a, 7x^3 + 5 ( x - y )^2 v- 7y^3
= 7 ( x^3 - y^3 ) + 5 ( x-y )^2
= 7 ( x - y )^3 + 5 ( x-y ) ^2
= [ 7 ( x- y ) + 5 ] ( x-y) ^2
\(1,x+y+z=0=>x=-\left(y+z\right)\)
\(=>x^2=\left(y+z\right)^2=y^2+2yz+z^2\)
\(=>x^2-y^2-z^2=2yz\)
\(=>\left(x^2-y^2-z^2\right)^2=\left(2yz\right)^2=4y^2z^2\)
\(=>x^4+y^4+z^4-2x^2y^2-2x^2z^2+2y^2z^2=4y^2z^2\)
\(=>x^4+y^4+z^4=4y^2z^2-2y^2z^2+2x^2z^2+2x^2y^2=2x^2y^2+2y^2z^2+2x^2z^2\)
\(=>2\left(x^4+y^4+z^4\right)=\left(x^2+y^2+z^2\right)^2\left(đpcm\right)\)
\(2,A=2\left(x^6-y^6\right)-3\left(x^4+y^4\right)\)
\(=2\left[\left(x^2\right)^3-\left(y^2\right)^3\right]-3\left(x^4+y^4\right)\)
\(=2\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)-3\left(x^4+y^4\right)\)
\(=2\left(x^4+x^2y^2+y^4\right)-3\left(x^4+y^4\right)\)
\(=2x^4+2x^2y^2+2y^4-3x^4-3y^4=-x^4+2x^2y^2-y^4\)
\(=-\left(x^4-2x^2y^2+z^4\right)=-\left[\left(x^2-y^2\right)^2\right]=-1\) (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
x4 - 5x2 +4
=x4 -4x -x +4
=(x4 - 4x) - (x-4)
=x3(x-4)-(x -4)
=(x3 -1).(x-4)
3, x3 - x + y3 - y
=( x3 + y3) - (x + y)
=(x +y)(x2 - xy +y2) -(x +y)
= x2 - xy + y2
4, x3 - 3x2 -4x +12
= (x3 - 3x2)-(4x -12)
=x2(x - 3) - 4(x - 3)
= (x2 - 4)(x -3)
=(x-2)(x+2)(x-3)
7, 45 +x3 - 5x2 - 9x
=( x3 - 5x2 )-( 9x - 45)
= x2(x-5)-9(x-5)
=( x2- 9)(x-5)
= (x-3)(x+3)(x-5)
Bài 1
1) 4x - x2 - 4 = 0
⇔ -( x2 - 4x + 4 ) = 0
⇔ -( x - 2 )2 = 0
⇔ x - 2 = 0
⇔ x = 2
2) 4( x - 1 )2 - ( 5 - 2x )2 = 0
⇔ 22( x - 1 )2 - ( 5 - 2x )2 = 0
⇔ ( 2x - 2 )2 - ( 5 - 2x ) = 0
⇔ ( 2x - 2 - 5 + 2x )( 2x - 2 + 5 - 2x ) = 0
⇔ ( 4x - 7 ).3 = 0
⇔ 4x - 7 = 0
⇔ x = 7/4
3) 9( x - 2 )2 - 4( 3 - x )2 = 0
⇔ 32( x - 2 )2 - 22( x - 3 )2 = 0
⇔ ( 3x - 6 )2 - ( 2x - 6 )2 = 0
⇔ ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 ) = 0
⇔ x( 5x - 12 ) = 0
⇔ x = 0 hoặc 5x - 12 = 0
⇔ x = 0 hoặc x = 12/5
4) x2 - 6x + 5 = 0
⇔ x2 - 5x - x + 5 = 0
⇔ x( x - 5 ) - ( x - 5 ) = 0
⇔ ( x - 5 )( x - 1 ) = 0
⇔ x - 5 = 0 hoặc x - 1 = 0
⇔ x = 5 hoặc x = 1
Bài 2.
1) x2 - z2 + y2 - 2xy
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - z2
= ( x - y - z )( x - y + z )
2) a3 - ay - a2x + xy
= ( a3 - a2x ) - ( ay - xy )
= a2( a - x ) - y( a - x )
= ( a - x )( a2 - y )
3) 2xy + 3z + 6y + xz
= ( 2xy + 6y ) + ( xz + 3z )
= 2y( x + 3 ) + z( x + 3 )
= ( x + 3 )( 2y + z )
4) x2 + 2xz + 2xy + 4yz
= ( x2 + 2xy ) + ( 2xz + 4yz )
= x( x + 2y ) + 2z( x + 2y )
= ( x + 2y )( x + 2z )
5) ( x + y + z )3 - x3 - y3 - z3
= x3 + y3 + z3 + 3( x + y )( y + z )( x + z ) - x3 - y3 - z3
= 3( x + y )( y + z )( x + z )
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
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 = \(\frac{\frac{1}{2}}{20.21+\frac{1}{2}}=\frac{1}{841}\)