\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)

\(2S=\frac{2.5}{3.13}+\frac{2.5}{13.23}+....+\frac{2.5}{83.93}\)

\(2S=\frac{10}{3.13}+\frac{10}{13.23}+.....+\frac{10}{83.93}\)

\(2S=\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\)

\(2S=\frac{1}{3}-\frac{1}{93}=\frac{30}{93}\)

\(S=\frac{30}{93}.\frac{1}{2}=\frac{15}{93}\)

Sửa đề:

\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)

\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\left(\frac{31}{93}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\frac{10}{31}\)

\(S=\frac{5}{31}\)

bn lớp 5 ak

Bạn làm theo cách này nhé:

\(\frac{7}{5}\div\frac{4}{5}=\frac{7}{4}\)(5 ở trên tử và 5 ở dưới mẫu triệt tiêu còn 1)

Ta có: \(\frac{7}{4}>1>\frac{2005}{2006}\Rightarrow\frac{7}{4}\div\frac{4}{5}>\frac{2005}{2006}\)

7/5:4/5=35/20 nhẩm ra =7/4.Mà 7/4>1,2005/2006<1neen 7/5:4/5>2005/2006  Mik nghi ko đến 4s đâu

Ta có

\(S=\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}\right)\)

\(S>\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{1}{2}.4=2\)

Vậy S>2

a,<=>\(\frac{\left(2x+1\right)^2}{4}\)+\(\frac{2\left(2x-1\right)^2}{4}\)≥\(\frac{12\left(x+5\right)^2}{4}\)

<=>4x²+4x+1+2(4x²-4x+1)≥12(x²+10x+25)

<=>4x²+4x+1+8x²-8x+2≥12x²+120x+300

<=>4x²+4x+1+8x²-8x+2-12x²-120x-300≥0

<=>-124x-297≥0

<=>124x+297≤0

<=>124x≤-297

<=>x≤\(\frac{-297}{124}\)

b, Tương tự câu a

c,

TH1: 5-3x=2+x

<=> -3x - x = 2 - 5

<=> -4x = -3

<=> x = 3/4

TH2: 5-3x = -2 - x

<=> -3x + x = -2 - 5

<=> -2x = -7

<=> x = 7/2

\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(B=\frac{5}{3}.\left(1-\frac{1}{2017}\right)\)

\(B=\frac{5}{3}.\frac{2016}{2017}=\frac{10080}{6051}\)

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{2014.2017}\)

\(3M=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\right)\)

\(3M=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(3M=5\left(1-\frac{1}{2017}\right)\)

\(3M=5.\frac{2016}{2017}\)

\(3M=\frac{10080}{2017}\)

\(\Rightarrow M=\frac{3360}{2017}\)

Ta có: 

\(B=\frac{5}{1.4}+\frac{5}{4.7}+.....+\frac{5}{2014.2017}\)

\(\Rightarrow B.\frac{3}{5}=\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{2014.2017}\)

\(=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+......+\left(\frac{1}{2014}-\frac{1}{2017}\right)\)

\(=1-\frac{1}{2017}\)

\(=\frac{2016}{2017}\)

\(\Rightarrow B=\frac{2016}{2017}:\frac{3}{5}\)

\(=\frac{3360}{2017}\)

Vậy B \(=\frac{3360}{2017}\)~~~~~

ĐKXĐ: \(x\notin\left\{0;\frac{3}{2}\right\}\)

Ta có: \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)

\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)

Suy ra: \(x-3=10x-15\)

\(\Leftrightarrow x-3-10x+15=0\)

\(\Leftrightarrow-9x+12=0\)

\(\Leftrightarrow-9x=-12\)

hay \(x=\frac{4}{3}\)(tm)

Vậy: \(S=\left\{\frac{4}{3}\right\}\)