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(x+1)(x-4)(x+2)(x-8)+4x^2
=[(x+1)(x-8)][(x-4)(x+2)]+4x2
=(x2-7x-8)(x2-2x-8)+4x2
Đặt t=x2-2x-8 ta được:
(t-5x).t+4x2
=t2-5xt+4x2
=t2-xt-4xt+4x2
=t.(t-x)-4x.(t-x)
=(t-x)(t-4x)
thay t=x2-2x-8 ta được:
(x2-3x-8)(x2-6x-8)
Vậy (x+1)(x-4)(x+2)(x-8)+4x^2=(x2-3x-8)(x2-6x-8)
1. (x-1)(x-3)(x-5)(x-7)-20=0
<=> (x-1)(x-7)(x-3)(x-5)-20=0
<=> (x^2-8x+7)(x^2-8x+15)-20=0
Đặt x^2-8x+7=a => x^2-8x+15= a+8
=> a(a+8)-20=0
<=> a^2+8a-20=0
<=>(a^2+8a+16)-36=0
<=> (a+4)^2=36
=> {a+4=6a+4=−6{a+4=6a+4=−6
<=>{a=2a=−10{a=2a=−10
*a=2 => x^2-8x+7=2
<=> x^2-8x+5=0
<=>(x^2-8x+16)-11=0
<=>(x-4)^2=11
<=>x-4=√11
<=> x=√11 +4
*a=-10 => x^2-8x+7=-10
<=> x^2-8x+17=0
<=> (x^2-8x+16)+1=0
<=> (x-4)^2=-1 (PT vô nghiệm)
Vậy pt có nghiệm x=√11 +4
mk chỉ biết vậy thôi
3, \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)-3=\left(x^2+x\right)\left(x^2+x-2\right)-3\)
Đặt \(x^2+x=t\)
\(t\left(t-2\right)-3=t^2-2t-3=\left(t-3\right)\left(t+1\right)\)
Theo cách đặt \(\left(x^2+x-3\right)\left(x^2+x+1\right)\)
1. ( x2 - x + 2 )4 - 3x2 ( x2 - x + 2 )2 + 2x4
Đặt t = x2 - x + 2 , ta có :
t4 - 3x2t2 + 2x4
= t4 - 2x2t2 - x2t2 + 2x4
= t2 ( t2 - 2x2 ) - x2 ( t2 - 2x2 )
= ( t2 - x2 ) ( t2 - 2x2 )
= ( t - x ) ( t + x ) ( t2 - 2x2 )
= ( x2 - x + 2 - x ) ( x2 - x + 2 + x ) [ ( x2 - x + 2 )2 - 2x2 ]
= ( x2 - 2x + 2 ) ( x2 + 2x ) ( x2 - 3x + 2 ) ( x2 + x + 2 )
2. 3 ( - x2 + 2x + 3 )4 - 26x2 ( - x2 + 2x + 3 )2 - 9x4
Đặt y = - x2 + 2x + 3 , ta có :
3y4 - 26x2y2 - 9x4
= x2y2 + 3y4 - 9x4 - 27x2y2
= y2 ( x2 + 3y2 ) - 9x2 ( x2 + 3y2 )
= ( y2 - 9x2 ) ( x2 + 3y2 )
= ( y - 3x ) ( y + 3x ) ( x2 + 3y2 )
= ( - x2 + 2x + 3 - 3x ) ( - x2 + 2x + 3 + 3x ) [ x2 + 3 ( - x2 + 2x + 3 )2 ]
= ( - x2 - x + 3 ) ( - x2 + 5x + 3 ) ( 3x4 - 12x3 - 5x2 + 36x + 27 )
(x+1)(x+2)(x+3)(x+4)-24
= (x^2 + 5x + 4)(x^2 + 5x + 6) - 24
đặt x^2 + 5x + 5 = a
ta có : (a - 1)(a + 1) - 24 = a^2 - 1 - 24
= a^2 - 25
= (a - 5)(a+5)
= (x^2 + 5x + 5 - 5)(x^2 + 5x + 5 + 5)
= (x^2 + 5x)(x^2 + 5x + 10)
= x(x + 5)(x^2 + 5x + 10)
x(x+1)(x+2)(x+3)+1
= (x^2 + 3x)(x^2 + 3x + 2) + 1
đặt x^2 + 3x + 1 = a
ta có : (a - 1)(a+1) + 1 = a^2 - 1 + 1 = a^2
= (x^2 + 3x + 1)^2
(x+1)(x+2)(x+3)(x+4)−24f,(x+1)(x+2)(x+3)(x+4)−24
=(x+1)(x+4)(x+2)(x+3)−24=(x+1)(x+4)(x+2)(x+3)−24
=(x2+5x+4)(x2+5x+6)−24=(x2+5x+4)(x2+5x+6)−24
Đặt t=x2+5x+4t=x2+5x+4 , ta có
t(t+2)−24t(t+2)−24
=t2+2t−24=t2+2t−24
=(t2+2t+1)−25=(t2+2t+1)−25
=(t+1)2−52=(t+1)2−52
=(t+1−5)(t+1+5)=(t+1−5)(t+1+5)
=(t−4)(t+6)=(t−4)(t+6)
=(x2+5x+4−4)(x2+5x+4+6)=(x2+5x+4−4)(x2+5x+4+6)
=(x2+5x)(x2+5x+10)
x(x+1)(x+2)(x+3)+1x(x+1)(x+2)(x+3)+1
=[x(x+3)][(x+1)(x+2)]+1=[x(x+3)][(x+1)(x+2)]+1
=(x2+3x)(x2+2x+x+2)+1=(x2+3x)(x2+2x+x+2)+1
=(x2+3x)(x2+3x+2)+1=(x2+3x)(x2+3x+2)+1(1)
Đặt x2+3x=t⇒x2+3x+2=t+2x2+3x=t⇒x2+3x+2=t+2
Do đó (1)=t(t+2)+1=t2+2t+1=(t+1)2(1)=t(t+2)+1=t2+2t+1=(t+1)2(*)
Vì t=x2+3xt=x2+3x nên
(*)=(x2+3x+1)2
1, (x-1)(x+2)(x+3)(x-6)+32x^2
= (x^2 - 7x + 6)(x^2 + 5x + 6) + 32x^2
đặt x^2 - x + 6 = a ta có
(a - 6x)(a + 6x) + 32x^2
= a^2 - 36x^2 + 32x^2
= a^2 - 4x^2
= (a - 2x)(a + 2x)
= (x^2 - x + 6 - 2x)(x^2 - x + 6 + 2x)
= (x^2 - 3x + 6)(x^2 + x + 6)
2, (x+1)(x-4)(x+2)(x-8)+4x^2
= (x^2 + 7x - 8)(x^2 - 2x - 8) + 4x^2
đặt x^2 + 2,5x - 8 = a ta có
(a + 4,5x)(a - 4,5x) + 4x^2
= a^2 - 81/4x^2 + 4x^2
= a^2 - 65/4x^2
\(=\left(a-\sqrt{\frac{65}{4}}x\right)\left(a+\sqrt{\frac{65}{4}}x\right)=\left(x^2+\frac{5}{2}x-8+\sqrt{\frac{65}{4}}x\right)\left(x^2+\frac{5}{2}x-8-\sqrt{\frac{65}{4}x}\right)\)