2x2=16
3x3=21
4x4=4
vậy:1x1=?
S1=1x1
S2=2x2-1x1
S3=3x3-(2x2-1x1)
S4=4x4-(3x3-(2x2-1x1))
..................................
a)Hãy viết dãy số S5
b)Nếu dãy số này cứ tiếp tục như thế thì S2011 có giá trị là bao nhiêu?
S5=5x5-(4x4-(3x3-(2x2-1x1)))
S2011=2001x2001-(2000x2000-(1999x1999-(....)))
Bạn ơi tính như vậy thì phần b tính kiểu jr bao nhiêu dấu ngoạc làm sao tính được
Biến đổi mỗi biểu thức sau thành một phân thức đại số:
a) 1+1x1−1x1+1x1−1x; b) 1−2x+11−x2−2x2−11−2x+11−x2−2x2−1.
a) 1+1x1−1x1+1x1−1x =(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1=(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1
b) 1−2x+11−x2−2x2−11−2x+11−x2−2x2−1 =(1−2x+1):(1−x2−2x2−1)=(1−2x+1):(1−x2−2x2−1)
=x+1−2x+1:x2−1−(x2−2)x2−1=x+1−2x+1:x2−1−(x2−2)x2−1
=x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1)=x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1)
=x−1x+1.(x−1)(x+1)1=(x−1)2=x−1x+1.(x−1)(x+1)1=(x−1)2.
a) 1+1x1−1x1+1x1−1x =(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1=(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1 b) 1−2x+11−x2−2x2−11−2x+11−x2−2x2−1 =(1−2x+1):(1−x2−2x2−1)=(1−2x+1):(1−x2−2x2−1) =x+1−2x+1:x2−1−(x2−2)x2−1=x+1−2x+1:x2−1−(x2−2)x2−1 =x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1)=x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1) =x−1x+1.(x−1)(x+1)1=(x−1)2=x−1x+1.(x−1)(x+1)1=(x−1)2.
1x1+2x2+3x3+...+19x19
1x1! + 2x2! + 3x3!+... + 16x16!=?
1x1=2x2=4=1=5x11898324789358=???
1x1!+2x2!+3x3!+4x4!+...+100x100!
1x1!+2x2!+3x3!+4x4!+...+100x100! = ?
A=1x1+2x2+3x3+...+100x100
Rút gọn 1x1!+2x2!+......+15x15!
= 1.1! + (3 - 1).2! + (4 - 1).3! + ...+ (16 - 1).15!
= 1! + 3.2! - 2! + 4.3! - 3! + ...+16.15! - 15!
= 1! + 3! - 2! + 4! - 3! + ...+16! - 15!
= 1! + (3! + 4! + ...+ 16!) - (2! + 3! + ...+ 15!)
= 1 + 16! - 2! = ....
Ta có: 1x1!+2x2!+......+15x15!
= 1.1! + (3 - 1).2! + (4 - 1).3! + ...+ (16 - 1).15!
= 1! + 3.2! - 2! + 4.3! - 3! + ...+16.15! - 15!
= 1! + 3! - 2! + 4! - 3! + ...+16! - 15!
= 1! + (3! + 4! + ...+ 16!) - (2! + 3! + ...+ 15!)
= 1 + 16! - 2!
Vậy ...............
hok tốt
a=1x1^2+2x2^3+3x4^2+......+98x99^2