E = 1 x 1 + 2 x2 + 3 x 3 + . . . + 15 x 15

E = 1( 2 -1 ) + 2 ( 3 -1 ) + 3 ( 4-1 ) + . . . + 15 ( 16 -1 )

E = 1.2 -1 + 2.3 - 2 + 3.4 - 3 + . . . + 15 . 16 - 15

E = ( 1.2 + 2.3 + 3.4+ . . . + 15 .16 ) - ( 1 + 2 + 3 + . . . + 15 )

E = 1360- 120

E  = 1240

Tk mk nha

S5=5x5-(4x4-(3x3-2x2)

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

\(E=1.1+2.2+3.3+4.4+...+99.99\)

\(\Rightarrow E=1^2+2^2+3^2+4^2+...+99^2\)

\(\Rightarrow E=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}\)

\(\Rightarrow E=\dfrac{99.100.199}{6}\)

\(\Rightarrow E=33.50.199=328350\)

E = 1 x 1 + 2 x 2 + 3 x 3 + 4 x 4 +...+ 99 x 99

E = 1x(2-1) + 2 x (3-1)+...+ 99 x (100 -1)

D = 1 x 2 - 1 + 2 x 3 - 2 +...+ 99 x 100 - 99

D = 1x2 + 2 x 3 +...+ 99 x 100 - ( 1 + 2 +...+ 99)

Đặt A = 1x2 + 2 x 3 +...+ 99 x 100 

      B =  1 + 2 + ...+ 99

     1x2 x 3 = 1x2x3

     2x3x3  = 2x 3 x (4-1) = 2x3x4 - 1x2x3

     3 x 4 x 3 = 3 x 4 x ( 5 - 2) = 3 x 4 x 5 - 2 x 3 x 4

     ................................................

     99 x 100 x 3 = 99 x 100 x (101 - 98) = 99x100x101 - 98 x 99 x 100

Cộng vế với vế ta có: 3A = 99 x 100 x 101

                                   A = 99 x 100 x 101 : 3 = 333300

B = 1 + 2 + 3 + ...+ 99

B = (99 + 1).[(99 -1):1 +1]:2 = 4950

E = 33300 - 4950 = 328350

a là cocaiconcac nhé e

b là cocaidaccau ok nhé

a: \(=60\cdot\dfrac{17}{20}=51\)

b: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}=\dfrac{1}{5}\)

A=1x2+2x3+3x4+...+49x50

3A= 3(1.2+2.3+3.4+...+49.50)

3A= 1.2.3+2.3.3+3.4.3+...+49.50.3

3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)

3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51

3A= 49.50.51

A= 49.50.51/3=41650

B=1x3+3x5+5x7+...+99x101

B=1/1.3 +1/3.5 +...+1/99.101

2B=2/1.3 + 2/3.5 +...+2/99.101

2B=1-1/3+1/3-1/5+...+1/99-1/101

2B=1-1/101

2B=100/101

B=100/101:2=100/202

C=1x2x3+2x3x4+3x4x5+...+50x51x52
Nhân C với 4 ta được:
C x 4 = 1x2x3x4 + 2x3x4x 4 + 3x4x5x4 +…+50x51x52x4
C x 4 = 1x2x3x4 + 2x3x4x(5-1) + 3x4x5x(6-2) + ... + 50x51x52x(53-49)
C x 4 = 1x2x3x4 + 2x3x4x5 - 1x2x3x4 + 3x4x5x6 - 2x3x4x5 + ... +49x 50x51x52 - 50x51x52x53
Sau khi cộng - trừ giản ước ta có : C x 4 = 50x51x52x53
C = 50x51x52x53 : 4 = 1756950

Phương trình −2 x 2 − 6x − 1 = 0 có  = ( − 6 ) 2 – 4.(− 2).(−1) = 28 > 0 nên phương trình có hai nghiệm x 1 ;   x 2

Theo hệ thức Vi-ét ta có  x 1 + x 2 = − b a x 1 . x 2 = c a ⇔ x 1 + x 2 = − 3 x 1 . x 2 = 1 2

Ta có

N = 1 x 1 + 3 + 1 x 2 + 3 = x 1 + x 2 + 6 x 1 . x 2 + 3 x 1 + x 2 + 9 = − 3 + 6 1 2 + 3. − 3 + 9 = 6

Đáp án: A

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.