x³ + 3x²+3x+1=(x+1)³

Ta có
x3+3x2+3x+1=0⇔(x+1)3=0x3+3x2+3x+1=0⇔(x+1)3=0
⇔⇔ x + 1 = 0
⇔⇔ x = -1
Vậy x = -1

\(\left(x-3\right)\left(2x+5\right)+8x\left(1-x\right)-\left(2x+1\right)\left(5-3x\right)\)

\(=2x^2-x-15+8x-8x^2-\left(-6x^2+5+7x\right)\)

\(=-6x^2+7x-15+6x^2-5-7x=-20\)

Trả lời:

( 5x - 1 ) + 2 ( 1 - 5x ) ( 4 + 5x ) + ( 5x + 4 )²

= ( 5x - 1 )² - 2 ( 5x - 1 ) ( 5x + 4 ) + ( 5x + 4 )²

= [ ( 5x - 1 ) - ( 5x + 4 ) ]²

= ( 5x - 1 - 5x - 4 )² 

= ( - 5 )²

= 25

Trả lời:

a, ( 4xy )² + 8xy + 1 = ( 4xy + 1 )²

b, 25x² - 36 = ( 5x + 6 ) ( 5x - 6 )

c, ( 1/2 - x )² = 1/4 - x + x²

=_= ......bt chết liền (hoa mắt,chóng mặt,ù tai)

Trả lời:

8a³ - 12a²b + 6ab² - b³

= ( 2a )³ - 3.( 2a )²b + 3.2a.b² - b³

= ( 2a - b )³

thu gọn

nhầm là n+89

=(2a-b)^3 nhé, k cho mình đấy

\(8a^3-12a^2b+6ab^2-b^3\)

\(=\left(2a\right)^3-3.\left(2a\right)^2b+3.2a.b^2-b^3\)

\(=\left(2a-b\right)^3\)

Ta có:

\(x+y+z=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\).

\(x+y+z=\frac{\left(a-b\right)\left(b+c\right)\left(c+a\right)+\left(b-c\right)\left(a+b\right)\left(c+a\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Ta có:

\(\left(a-b\right)\left(b+c\right)\left(c+a\right)+\left(b-c\right)\left(c+a\right)\left(a+b\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)\).

\(=\left(c+a\right)\left[\left(a-b\right)\left(b+c\right)+\left(b-c\right)\left(a+b\right)\right]+\left(c-a\right)\left(a+b\right)\left(b+c\right)\).

\(=\left(c+a\right)\left(ab+ac-b^2-bc+ab+b^2-ac-bc\right)\)\(+\left(c-a\right)\left(ab+ac+b^2+bc\right)\).

\(=\left(c+a\right)\left(2ab-2bc\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).

\(=2b\left(c+a\right)\left(a-c\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).

\(=\left(2bc+2ab\right)\left(a-c\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).

\(=\left(a-c\right)\left(2ab+2bc-ab-ac-b^2-bc\right)\).

\(=\left(a-c\right)\left(ab+bc-b^2-ac\right)=\left(a-c\right)\left[\left(ab-b^2\right)-\left(ac-bc\right)\right]\).

\(=\left(a-c\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]=\left(a-c\right)\left(a-b\right)\left(b-c\right)\).
Do đó\(x+y+z=\frac{\left(a-c\right)\left(a-b\right)\left(b-c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\).

Mà \(xyz=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)nên:

\(x+y+z=-xyz\).

\(\Rightarrow x+y+z+xyz=0\)(điều phải chứng minh).

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

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

Thay x = 101 vào biểu thức trên ta được : 

\(\left(101-1\right)^3=100.100.100=1000000\)

b, \(x^3+9x^2+27x\Leftrightarrow x\left(x^2+9x+27\right)\)

Thay x = 97 vào biểu thức trên ta được : 

\(97\left[\left(97\right)^2+9.97+27\right]=97.10309=999973\)

bạn xem lại đề ý b nhé 

Trả lời:

b, x³ - 3x² + 3x - 1 = ( x - 1 )³ 

Thay x = 101 vào biểu thức trên, ta có:

( 101 - 1 )³ = 100³ = 1000000

c, x³ + 9x² + 27x 

Thay x = 97 vào biểu thức trên, ta có:

97³ + 9.97² + 27.97 = 999973