Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 )
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
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
\(=\left(a-\sqrt{\frac{65}{4}}x\right)\left(a+\sqrt{\frac{65}{4}}x\right)=\left(x^2+\frac{5}{2}x-8+\sqrt{\frac{65}{4}}x\right)\left(x^2+\frac{5}{2}x-8-\sqrt{\frac{65}{4}x}\right)\)
a) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)-4=\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\)
Đặt \(t=x^2+6x+5\)
\(PT=t\left(t+3\right)-4=t^2+3t-4=\left(t-1\right)\left(t+4\right)\)
Thay t: \(PT=\left(x^2+6x+5-1\right)\left(x^2+6x+5+4\right)=\left(x^2+6x+4\right)\left(x^2+6x+9\right)=\left(x^2+6x+4\right)\left(x+3\right)^2\)
b) Đặt \(t=\left(2x+1\right)^2\)
\(PT=t^2-3t+2=\left(t^2-3t+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(t+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(t+1\right)\left(t+2\right)\)
Thay t:
\(PT=\left[\left(2x+1\right)^2+1\right]\left[\left(2x+1\right)^2+2\right]=\left[4x^2+4x+2\right]\left[4x^2+4x+3\right]=2\left[2x^2+2x+1\right]\left[4x^2+4x+3\right]\)
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 \(\left(x^2+x-3\right)\left(x^2+x+1\right)\)
\(\left(x^2+x\right)^2-2x^2-2x-15\)
\(=\left(x^2+x\right)^2-\left(2x^2+2x+15\right)\)
\(=\left(x^2+x\right)^2-\left[\left(2x^2+2x\right)+15\right]\)
\(=\left(x^2+x\right)^2-\left[2.\left(x^2+x\right)+15\right]\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\) \(\left(1\right)\)
đặt \(x^2+x=t\)
\(\left(1\right)\)\(=\) \(t^2-2t-15\)
\(=\left(t-1\right)^2-16\)
\(=\left(t-1-4\right)\left(t-1+4\right)\)
\(=\left(t-5\right)\left(t+3\right)\)
thay \(t=x^2+x\) ta có
\(\left(1\right)=\left(x^2+x-5\right)\left(x^2+x+3\right)\)
các câu còn lại tương tự nha
học tốt