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21 tháng 8 2017

\(f,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt \(t=x^2+5x+4\) , ta có
\(t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=\left(t^2+2t+1\right)-25\)

\(=\left(t+1\right)^2-5^2\)

\(=\left(t+1-5\right)\left(t+1+5\right)\)

\(=\left(t-4\right)\left(t+6\right)\)

\(=\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(g,\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)

\(=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(t=x^2-8x+7\), ta có:

\(t\left(t+8\right)-20\)

\(=t^2+8t-20\)

\(=\left(t^2+8t+16\right)-36\)

\(=\left(t+4\right)^2-6^2\)

\(=\left(t+4+6\right)\left(t+4-6\right)\)

\(=\left(t+10\right)\left(t-2\right)\)

\(=\left(x^2-8x+7+10\right)\left(x^2-8x+7-2\right)\)

\(=\left(x^2-8x+17\right)\left(x^2-8x+5\right)\)

21 tháng 8 2017

f ) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt \(x^2+5x+5=t\), ta có :

\(\left(t-1\right)\left(t+1\right)-24\)

\(=t^2-1-24=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)\)

Thay và ta có :

\(\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)

21 tháng 7 2021

Bài 10:

a) (x+2)2 -x(x+3) + 5x = -20

=> x2 + 4x + 4 - x2 - 3x + 5x = -20

=> 6x = -20 + (-4)

=> 6x = -24

=> x = -4

b) 5x3-10x2+5x=0   

=>5x(x2-2x+1)=0

=>5x(x-1)2 =0

=> 5x=0 hoặc (x-1)2=0

=>x=0 hoặc x=1

c) (x- 1)- (x+ x+ 1)(x- 1) = 0

=> (x2 - 1)[(x- 1)2 -  (x+ x+ 1)] = 0

<=> (x2 - 1)(x4 - 2x2 + 1 - x- x- 1) = 0

<=>  (x2 - 1)(-3x2) = 0

<=> (x2 - 1)=0 hoặc (-3x2) =0

<=> x2=1 hoặc x2=0

<=> x=−1;1 hoặc x=0

d)

(x+1)3−(x−1)3−6(x−1)2=-19

⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0

⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0

⇔12x+13=0⇔12x+13=0

⇔12x=-13

⇔x=-23/12

Học tốt nhé:333banhqua

 

 

 

17 tháng 5 2018

a) ( x 2  – 4x + 1)( x 2  – 2x + 3).

b) ( x 2  + 5x – 1)( x 2  + x – 1).

5 tháng 10 2021

\(3,=\left(x-y\right)^3+\left(y-x+x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3+\left(y-x\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-x+x-z\right)+\left(x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3-\left(x-y\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-z\right)-\left(z-x\right)^3+\left(z-x\right)^3\\ =3\left(y-x\right)\left(x-z\right)\left(y-z\right)\)

\(4,=\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-3x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)

2 tháng 9 2021

Bài 2:

a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)

b) \(x^2+6x+9-4y^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)

c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)

d) \(4x^2-y^2-6x+3y=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)

e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)

f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)

g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)

i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)

 

a: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=\left(x+1\right)\left(3x-10\right)\)

b: \(x^2+6x+9-4y^2\)

\(=\left(x+3\right)^2-4y^2\)

\(=\left(x+3-2y\right)\left(x+3+2y\right)\)

c: \(x^2-2xy+y^2-5x+5y\)

\(=\left(x-y\right)^2-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-5\right)\)

a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)

b: \(=25-\left(x-2y\right)^2\)

\(=\left(5-x+2y\right)\left(5+x-2y\right)\)

24 tháng 4 2022

a) Thu gọn:

P(x) = x4+(-7x2+4x2)+(x+6x)-2x3-2

P(x) = x4-3x2+7x-2x3-2

Sắp xếp: P(x) = x4-2x3-3x2+7x-2

Thu gọn:

Q(x) = x4+(-3x+x)+(-5x3+6x3)+1

Q(x) = x4-2x+x3+1

Sắp xếp: Q(x)= x4x3-2x+1

b/ Nếu x=2, ta có:

P(2) = 24-2.23-3.22+7.2-2

        = 16 - 2.8 - 3.4 + 14 -2

        = 16-16-12+14-2

        = -12+14-2 

        = 0

=> x=0 là nghiệm của P(x)

Q(2)= 24+ 23-2.2+1

= 16+8-4+1

= 24-4+1

=21

mà 21≠0

Vậy: x=2 không phải là nghiệm của Q(x)

=>