A=1/1+2+1/1+2+3+1/1+2+3+4+.....+1/1+2+3+4+...+50

Ta có 1/1+2+3+...n=1/[n*(n+1)/2]=2*[1/n(n+1)]=2*[1/n-1/n+1]

Thay n=1;2;3;4;5;6;...;50 Ta có A=2*[1/2-1/51]=49/51

vậy.......................................................

a ) 13/20

B)

C..........................................................

minh dang tính

lấy máy tính mà bấm

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

\(\Rightarrow\left(2x-\frac{4}{3}\right)^2=\frac{1}{2}-\frac{1}{4}\)

\(\Rightarrow\left(2x-\frac{4}{3}\right)^2=\frac{1}{4}\)

\(\Rightarrow\left(2x-\frac{4}{3}\right)=\sqrt{\frac{1}{4}}\)

\(\Rightarrow\left(2x-\frac{4}{3}\right)=\frac{1}{2}\)

\(\Rightarrow2x=\frac{1}{2}+\frac{4}{3}\)

\(\Rightarrow2x=\frac{11}{6}\)

\(\Rightarrow x=\frac{11}{6}\div2\)

\(\Rightarrow x=\frac{11}{6}\times\frac{1}{2}\)

\(\Rightarrow x=\frac{11}{12}\)

câu 1 :1/100

câu 2 :1

b)ta đặt A:  \(A=\frac{1}{99}+\frac{2}{98}+..+\frac{99}{1}\)

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

                  \(A=\frac{100}{99}+\frac{100}{98}+..+\frac{100}{2}+\frac{100}{100}\)

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

A=1 đoán :))

S = 1+1/2.(1+2)+1/3.(1+2+3)+...+1/100.(1+2+3+...+100)

   = 1+1/3.(1+2+3)+1/5.(1+2+3+4+5)+...+1/99(1+2+3+...+99) +  1/2.(1+2)+1/4.(1+2+3+4)+...+1/100.(1+2+3+...+100)

   =  (1+2+3+...+50)+(3/2+5/2+7/2+...+101/2)

   =  1275+1300

   =       2575

làm giùm bn í đi mọi người ........ mk cx k cho ......

\(\dfrac{2}{5}+\dfrac{1}{4}=\dfrac{8}{20}+\dfrac{5}{20}=\dfrac{13}{20}\)
\(\dfrac{1}{3}+\dfrac{3}{5}=\dfrac{5}{15}+\dfrac{9}{15}=\dfrac{14}{15}\)
\(\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)
\(\dfrac{1}{8}+\dfrac{5}{6}=\dfrac{3}{24}+\dfrac{20}{24}=\dfrac{23}{24}\)

ĐÂSDWADAS