b. \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2003}\)= \(\frac{1}{2013}\)

c. \(5.\left(\frac{1}{14}+\frac{1}{84}+\frac{1}{204}+\frac{1}{374}\right)\)= 5. \(\frac{1}{11}\)= \(\frac{5}{11}\)

Mình biết 2 câu này thôi, thông cảm nhá...!!!

d) Ta có: 100-(1+1/2+1/3+1/4+...+1/100)

=1x100-(1+1/2+1/3+1/4+...+1/100)

=(1-1)+(1-1/2)+(1-1/3)+(1-1/4)+....+(1-1/100)

=1/2+2/3+3/4+...+99/100

a)10,11+11,12+12,13+....+98,99+99,10

=10+0,11+11+0,12+12+0,13+...+98+0,99+99+0,10

=(10+11+12+...+98+99)+(0,10+0,11+0,12+...+0,99)

Bạn biết cách tính tổng dãy số cách đều rồi thì tính lần lượt 2 cái tổng trong ngoặc.

Còn nếu chưa thì tính số số hạng thế này:  (số lớn nhất - số bé nhất):2+1

Tính tổng thì: (số lớn nhất + số bé nhất) x (số số hạng):2

A = 4.113.25 - 5.112.20

  = 100.113 - 100.112

  =  100. ( 113-112)

  =100 . 1 = 100

D=673015

E=2550

C=...

G=781875

lấy lời giải ra hộ  tui

D=1+4+7+...+2008
D=(2008-1):3+1
D=670.
VẬY D=670.
E=2+4+6+...+100
E=(100-2):2+1
E=49.
VẬY E=49.
C=2+4+6+...+1024
C=(1024-2):2+1
C=512.
VẬY C=512.
 

\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)

\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)

\(\Rightarrow x=-\dfrac{81}{100}\)

\(a)\)\(\left(50-6.x\right).18=2^3.3^2.5\)

\(\Leftrightarrow\)\(\left(50-6.x\right).18=8.9.5\)

\(\Leftrightarrow\)\(\left(50-6.x\right).18=360\)

\(\Leftrightarrow\)\(\left(50-6.x\right)=360\div18\)

\(\Leftrightarrow\)\(50-6.x=20\)

\(\Leftrightarrow\)\(6.x=50-20\)

\(\Leftrightarrow\)\(6.x=30\)

\(\Leftrightarrow\)\(x=5\)

\(b)\)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=7450\)

\(\Leftrightarrow\)\(100x+\left(1+2+3+...+100\right)=7450\)

\(\Leftrightarrow\)\(100x+5050=7450\)

\(\Leftrightarrow\)\(100x=7450-5050\)

\(\Leftrightarrow\)\(100x=2400\)

\(\Leftrightarrow\)\(x=24\)

b.

(x+1)+(x+2)+...+(x+100)=7450

=> 100x + (1+2+3+...+100)=7450

=>100x + (100+1).50=7450

=>100x=2400

=>x=24

(x+1)+(x+2)+(x+3)+....(x+100)=7450

=> 100x + (1+2+3...+100)=7450

=> 100x + 5050=7450

=> 100x =7450-5050

=> 100x = 2400

=> x= 24

bạn ghi lại đề được ko

Bài 2:

100! = 1.2.3.4.   ...................   .100

E = 1.2+2.3+3.4+......+99.100
Gấp E lên 3 lần ta có:
E . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
E . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)
E . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100                                                                                       E . 3 = 99.100.101
E = 99.100.101 : 3
E = 33.100.101
E = 333 300

k mik nha

E = 1.2 + 2.3 + 3.4 + ... + 99.100

=> 3E = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3

=> 3E = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ 99.100.(101-98)

=> 3E = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

=> 3E = 99.100.101

=> E = 333300

Ta có : A = 1/1.2 + 1/2.3 + .... + 1/98.99 + 1/99.100 .

=>       A = 1 - 1/2 + 1/2 - 1/3 + .... + 1/98 - 1/99 + 1/99 - 1/100 .

=>       A = 1 - 1/100 .

=>       A = 99/100 .

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=1-\frac{1}{100}\)

\(\Rightarrow A=\frac{99}{100}\)

Cảm ơn nhìu nha