làm bừa thui,ai tích mình mình tích lại

Số số hạng là : 

Có số cặp là :

50 : 2 = 25 ( cặp )

Mỗi cặp có giá trị là :

99 - 97 = 2 

Tổng dãy trên là :

25 x 2 = 50

Đáp số : 50

\(3x^2+3x+1+x^3\)

\(=x^3+3x^2+3x+1\)

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

Giải:

a) \(-x^3+3x^2-3x+1\)

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

\(=-\left(x^3-3.x^2.1+3.x.1^2-1^3\right)\)

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

Vậy ...

Chúc bạn học tốt!

a) ta có : \(-x^3+3x^2-3x+1=1-3.1^2.x+3.1.x^2-x^3=\left(1-x\right)^3\)

b) ta có : \(64-48x+12x^2-x^3=4^3-3.4^2x+3.4.x^2-x^3=\left(4-x\right)^3\)

Ta có: a/ -x³+3x²-3x+1 = -(x³-3x²+3x-1)

= -(x-1)³

b/ 64-48x+12x²-x³ = 4³-3.4².x+3.a.x²-x³

= (4-x)³

a) -x³ + 3x² - 3x + 1

= -(x³ - 3x² + 3x - 1)

=-(x - 1)³

b) 8 - 12x + 6x² - x³

= 2³ - 3.2².x + 3.2.x² - x³

= (2 - x)³

Ta có  x 3 + 3 x 2 + 3 x + 1 = x 3 + 3 x 2 . 1 + 3 x . 1 2 + 1 3 = ( x + 1 ) 3 .

Bài 5:

a: \(x^3-1-\left(x^2+2x\right)\left(x-2\right)=5\)

=>\(x^3-1-\left(x^3-2x^2+2x^2-4\right)=5\)

=>\(x^3-1-x^3+4=5\)

=>3=5(vô lý)

b: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

=>\(x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\)

=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)

=>12x-4=-10

=>12x=-6

=>x=-6/12=-1/2

BÀi 3:

a: \(A=\left(a+b\right)^3-\left(a-b\right)^3\)

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

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

b:Sửa đề: \(A=\left(u-v\right)^3+3uv\left(u-v\right)\)

\(=u^3-3u^2v+3uv^2-v^3+3u^2v-3uv^2=u^3-v^3\)

c: \(C=6\left(c-d\right)\left(c+d\right)+2\left(c-d\right)^2-\left(c-d\right)^3\)

\(=6\left(c^2-d^2\right)+2\left(c^2-2cd+d^2\right)-c^3+3c^2d-3cd^2+d^3\)

\(=6c^2-6d^2+2c^2-4cd+2d_{}^2-c^3+3c^2d-3cd^2+d^3\)

\(=-c^3+3c^2d-3cd^2+d^3+8c^2-4cd-4d^2\)

Bài 2:

a: \(x^3+3x^2+3x+1=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x+1\right)^3\)

b: \(m^3+9m^2n+27mn^2+27n^3\)

\(=m^3+3\cdot m^2\cdot3n+3\cdot m\cdot\left(3n\right)^2+\left(3n\right)^3\)

\(=\left(m+3n\right)^3\)

a. (3a + 1)³

= 27a³ + 27a² + 9a + 1

Bài 3: 

b: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)+10=0\)

\(\Leftrightarrow6x^2+12-6x^2+12x-6=0\)

hay \(x=-\dfrac{1}{2}\)

Bài 2: 

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

b: \(m^3+9m^2n+27mn^2+27n^3=\left(m+3n\right)^3\)

a) \(A=8x^3+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)

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

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

d)  \(D=27+27y^2+9y^4+y^6=\left(3+y^2\right)^3\)