S = 1 x 2 + 2 x 3 + ... + 99 x 100

3S = 1 x 2 x 3 + 2 x 3 x (4 - 1) + ..... + 99 x 100 x (101 - 98)

3S = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + .... + 99 x 100 x 101 - 98 x 99 x 100

3S = 99 x 100 x 101 = 999900

S = 999900 : 3 = 333300 

\(1\times\left(1+1\right)+2\times\left(2+1\right)+3\times\left(3+1\right)\)

\(=1\times2+2\times3+3\times4\)

\(=2+6+12\)

\(=20\)

\(a=215\times62+42-52\times215\)

\(a=215\times\left(62-52\right)+42\)

\(a=215\times10+42\)

\(a=2150+42\)

\(a=2192\)

\(b=14\times29+14\times71+\left(1+2+3+...+99\right)\times\left(199199\times198-198198\times199\right)\)

\(b=14\times\left(29+71\right)+\left(1+2+3+...+99\right)\times\left(199\times1001\times198-198\times1001\times199\right)\)

\(b=14\times100+0\)

\(b=1400\)

1: Quá dễ

1 . (1 + 1) + 2 . (2 + 1) + 3 . (3 + 1)

= 1 . 2 + 2 . 3 + 3 . 4

= 2 + 6 + 12

= 20

2:

a = 215 . 62 + 42 - 52 . 215

= 215 . (62 - 52) + 42

= 215 . 10 + 42

= 2150 + 42

= 2192

b = 14 . 29 + 14 . 71 + (1 + 2 + 3 + ... + 99) . (199199 . 198 - 198198 . 199)

= 14 . (29 + 71) + (1 + 2 + 3 + ... + 99) . (199 . 1001 . 198 - 198 . 1001 . 199)

= 14 . 100 + (1 + 2 + 3 + ... + 99) . 0

= 1400 + 0 = 1400

a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )

Vậy x = 1

b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)

=> x + 100 = 0

=> x           = -100

c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)

=> x - 100 = 0

=> x           = 100

Chúc bạn học tốt

có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai

\(=\dfrac{x^2-25-x^2-10}{x\left(x-5\right)}=\dfrac{-35}{x\left(x-5\right)}\)

viết có chắc chữ giải mà cũng đúng thật vô lý

phương trình này nhìn từ đầu cũng bik vô nghiệm ko có x

\(\frac{x+1}{99}+\frac{x+2}{99}=\frac{x+10}{99}+\frac{x+20}{99}\)

Nhân 2 vế cho 99 ta được:

\(99.\left(\frac{x+1}{99}+\frac{x+2}{99}\right)=99.\left(\frac{x+10}{99}+\frac{x+20}{99}\right)\)

=>x+1+x+2=x+10+x+20

=>2x+3=2x+30

=>0x=27 (vô lí)

Vậy ko tìm dc x

\(\frac{x+1}{99}+\frac{x+2}{99}+\frac{x+3}{99}+\frac{x+4}{99}=-4\)

=>\(\frac{\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)}{99}=-4\)

=> (x+1)+(x+2)+(x+3)+(x+4)=-4.99=-396

=>4x+10=-396

4x=-406

x=-406:4=-101,5