\(a,A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-..-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{100}-\left(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\right)\)

\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

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

\(A=\frac{2}{100}-1\)

\(A=\frac{1}{50}-1\)

\(A=\frac{-49}{50}\)

b,\(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n=2^{n+34}\)        (1)

Đặt \(B=2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\)

\(\Rightarrow2B=2.\left(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\right)\)

             \(=2.2^3+3.2^4+4.2^5+...+\left(n-1\right).2^n+n.2^{n+1}\)

\(2B-B=\left(2.2^3+3.2^4+4.2^5+..+\left(n-1\right).2^n+n.2^{n+1}\right)\)

                 \(=(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n)\)

             \(B=-2^3-2^4-2^5-...-2^{n+1}-2.2^2\)

                 \(=-\left(2^3+2^4+2^5+...+2^n\right)+n.2^{n+1}-2^3\)

Đặt \(C=2^3+2^4+2^5+2^n\)

\(\Rightarrow2C=2.(2^3+2^4+2^5+...+2^n)\)

         \(C=2^4+2^5+2^6+...+2^{n+1}\)

\(2C-C=\left(2^4+2^5+2^6+...+2^{n+1}\right)-\left(2^3+2^4+2^5+...+2^n\right)\)

\(C=2^{n+1}-2^3\)

Khi đó :  \(B=-(2^{n+1}-2^3)+n.2^{n+1}-2^3\)

                  \(=-2^{n+1}+2^3+n.2^{n+1}-2^3\)

                   =\(=-2^{n+1}+n.2^{n+1}=\left(n-1\right).2^{n-1}\)

Vậy từ (1) ta có:\(\left(n-1\right),2^{n+1}=2^{n+34}\)

                           \(2^{n+34}-\left(n-1\right).2^{n+1}=0\)

                          \(2^{n+1}.[2^{33}-\left(n-1\right)]=0\)

Do đó \(2^{33}-n+1=0\)( Vì \(2^{n+1}\ne0\)với mọi \(n\))

\(n=2^{33}+1\)

Vậy \(n=2^{33}+1\)

\(H=\frac{1}{100}-\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-...-\frac{1}{2\cdot1}\)

\(U=\frac{1}{100}-\left(\frac{1}{100\cdot99}+\frac{1}{99\cdot98}+...+\frac{1}{2\cdot1}\right)\)

\(U=\frac{1}{100}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\right)\)

\(H=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(HU=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(UH=\frac{1}{100}-1+\frac{1}{100}\)

\(HU=\frac{2}{100}-1=-\frac{49}{50}\)

Chậc =)))

chac j` vay` cu'ng

C = 1/100 - ( 1/2.1 + 1/3.2 + ... + 1/98.97 + 1/99.98 + 1/100.99

C = 1/100 - (  1- 1/2+ 1/2 - 1/3 + ... + 1/97 - 1/98 + 1/98 - 1/99 + 1/99 - 1/100 )

C = 1/100 - ( 1 - 1/100 )

C = 1/100 - 99/100

C = \(\frac{-49}{50}\)

\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-1+\frac{1}{100}\)

\(C=\frac{-49}{50}\)

C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 - 1/2.1

C = 1/100 - (1/100.99 + 1/99.98 + 1/98.97 + ... + 1/3.2 + 1/2.1)

C = 1/100 - (1/1.2 + 1/2.3 + ... + 1/98.99 + 1/99.100)

C = 1/100 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/98 - 1/99 + 1/99 - 1/100)

C = 1/100 - (1 - 1/100)

C = 1/100 - 99/100

C = -98/100 = -49/50

\(c=\frac{1}{100}-\frac{1}{100.98}\frac{1}{99.98}\frac{1}{98.97}-......-\frac{1}{3.2}-\frac{1}{2.1}\)=\(\frac{1}{100}-\left[\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right]\) =\(\frac{1}{100}-\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\right]\)=\(\frac{1}{100}-\left[1-\frac{1}{100}\right]=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{49}{50}\)

\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)

Ta xét 2 trường hợp 

\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)

tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian 

Bài 1: Tìm x, biết:

\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}}\)

a) \(\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{99}-\left(\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

đặt \(A=\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

\(A=1-\frac{1}{99}\)

\(A=\frac{98}{99}\)

thay A vào, ta được :

\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)

b) \(\frac{2}{100.99}-\frac{2}{99.98}-...-\frac{2}{3.2}-\frac{2}{2.1}\)

\(=\frac{2}{100.99}-\left(\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\right)\)

đặt \(A=\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\)

\(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{98.99}\)

\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)

\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)

\(A=2.\left(1-\frac{1}{99}\right)\)

\(A=2.\frac{98}{99}\)

\(A=\frac{196}{99}\)

Thay A vào, ta được :

\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)

Ta có:

\(A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}\right)-\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{97}-\frac{1}{98}\right)-...-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(1-\frac{1}{2}\right)\)

\(=\frac{1}{100}-\frac{1}{99}+\frac{1}{100}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)

\(=\frac{1}{100}+\frac{1}{100}-1\)

\(=\frac{1}{50}-\frac{50}{50}\)

\(=-\frac{49}{50}\)

Câu này khó quá ta mình suy nghĩ này giờ mà vẫn chưa ra