Bài 1: Ta có:

\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)

\(=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\)

\(=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\)

\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+...+\dfrac{1}{26}-\dfrac{1}{32}\right)\)

\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{32}\right)=\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)

Vậy \(M=\dfrac{5}{16}\)

Bài 2: Ta có:

\(A=\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+...+\dfrac{1}{210}\)

\(=\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+...+\dfrac{1}{14.15}\)

\(=\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{15}\)

\(=\dfrac{1}{6}-\dfrac{1}{15}=\dfrac{1}{10}\)

Vậy \(A=\dfrac{1}{10}\)

Giải:

\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}.\)

\(M=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+\dfrac{5}{208}.\)

\(M=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+\dfrac{5}{13.16}.\)

\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right).\)

\(M=\dfrac{5}{3}\left[\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+\left(\dfrac{1}{10}-\dfrac{1}{10}\right)+\left(\dfrac{1}{13}-\dfrac{1}{13}\right)+\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\right].\)

\(M=\dfrac{5}{3}\left[0+0+0+\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\right]\)

\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\)

\(M=\dfrac{5}{3}\left(\dfrac{4}{16}-\dfrac{1}{16}\right).\)

\(M=\dfrac{5}{3}.\dfrac{3}{16}.\)

\(M=\dfrac{15}{48}=\dfrac{5}{16}.\)

M=\(\dfrac{5}{28}+\dfrac{1}{14}+\dfrac{1}{26}+\dfrac{5}{208}\)

M=\((\dfrac{5}{28}+\dfrac{1}{14})+\left(\dfrac{1}{26}+\dfrac{5}{208}\right)\)

M=\(\dfrac{1}{4}+\dfrac{1}{16}\)

M=\(\dfrac{5}{16}\)

A=\(\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+...+\dfrac{1}{13\times14}+\dfrac{1}{14\times15}\)

A=\(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{15}\)

Sau khi giản ước các phân số cho nhau Ta có:

A=\(\dfrac{1}{6}-\dfrac{1}{15}\)

A=\(\dfrac{1}{10}\)

\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}+\dfrac{1}{156}+\dfrac{1}{182}+\dfrac{1}{210}\\ =\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\\ =\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{13}-\dfrac{1}{14}\\ =\dfrac{1}{5}-\dfrac{1}{14}\\ =\dfrac{14}{70}-\dfrac{5}{70}=\dfrac{9}{70}\)

\(\dfrac{7}{110}+\dfrac{7}{132}+\dfrac{7}{156}+...+\dfrac{7}{4830}\)

\(=\dfrac{7}{10\cdot11}+\dfrac{7}{11\cdot12}+\dfrac{7}{12\cdot13}+...+\dfrac{7}{69\cdot70}\)

\(=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)

\(=7\cdot\dfrac{3}{35}=\dfrac{3}{5}\)

Giải:

a)A=1/56+1/72+1/90+1/110+1/132+1/156

   A=1/7.8+1/8.9+1/9.10+1/10.11+1/11.12+1/12.13

   A=1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12+1/12-1/13

   A=1/7-1/13

  A=6/91

b)B=4/21+4/77+4/165+4/285+4/437+4/621

   B=4/3.7+4/7.11+4/11.15+4/15.19+4/19.23+4/23.27

   B=1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27

   B=1/3-1/27

   B=8/27

c) C=1/21+1/77+1/165+1/285+1/437+1/621

    C=1/3.7+1/7.11+1/11.15+1/15.19+1/19.23+1/23.27

    C=1/4.(4/3.7+4/7.11+4/11.15+4/15.19+4/19.23+4/23.27)

    C=1/4.(1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27)

    C=1/4.(1/3-1/27)

    C=1/4.8/27

    C=2/27

d) D=1/1.6+1/6.11+1/11.16+1/16.21+1/21.26+1/26.31

    D=1/5.(5/1.6+5/6.11+5/11.16+5/16.21+5/21.26+5/26.31)

    D=1/5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21+1/21-1/26+1/26-1/31)

    D=1/5.(1/1-1/31)

    D=1/5.30/31

    D=6/31

Nếu câu d cậu viết thiếu thì làm như vầy nhé!

Chúc bạn học tốt!

Nếu như câu d ko chép sai thì làm thế này nha:

d) D=1/1.6+1/6.11+1/11.16+1/16.21+1/26.31

    D=1/5.(5/1.6+5/6.11+5/11.16+5/16.21)+1/806

    D=1/5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21)+1/806

    D=1/5.(1/1-1/21)+1/806

    D=1/5.20/21+1/806

    D=4/21+1/806

    D=3245/16926

Chúc bạn học tốt!

a)  >

b)  <

ôi bn ơi

a: \(\Leftrightarrow\dfrac{32}{x}=\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{99}\)

=>32/x=1/3-1/5+1/5-1/7+...+1/9-1/11

=>32/x=1/3-1/11=8/33

=>x=32:8/33=132

b: \(\Leftrightarrow1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{56}=\dfrac{x}{16}\)
\(\Leftrightarrow6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{x}{16}\)

=>x/16=6-1/2+1/8=11/2+1/8=45/8=90/16

=>x=90

c: \(\Leftrightarrow\dfrac{22}{x}=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\)

=>22/x=1/2*2/3*...*9/10*3/2*4/3*...*11/10

=>22/x=1/10*11/2=11/20=22/40

=>x=40

a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)

\(\Leftrightarrow3x-9=90-5+10x\)

\(\Leftrightarrow3x-9=10x+85\)

\(\Leftrightarrow3x-10x=85+9\)

\(\Leftrightarrow-7x=94\)

hay \(x=-\dfrac{94}{7}\)

Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)

b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)

\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)

\(\Leftrightarrow6x-4-60=9-6x-42\)

\(\Leftrightarrow6x-64=-6x-33\)

\(\Leftrightarrow6x+6x=-33+64\)

\(\Leftrightarrow12x=31\)

hay \(x=\dfrac{31}{12}\)

Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)

c) Ta có: \(3\left(x-1\right)+3=5x\)

\(\Leftrightarrow3x-3+3=5x\)

\(\Leftrightarrow3x-5x=0\)

\(\Leftrightarrow-2x=0\)

hay x=0

Vậy: S={0}

d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)

\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)

\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)

\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)

\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)

mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)

nên x+101=0

hay x=-101

Vậy: S={-101}

a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)

Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt

b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)

Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt

c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)

Vậy x = 3 là nghiệm của pt

d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)

Vậy x = -101 là nghiệm của pt

e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)

Vậy x = 100 là nghiệm của pt

f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)

Vậy x = 100 là nghiệm của pt

e) Ta có: \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\)

\(\Leftrightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1=0\)

\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\)

\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\)

mà \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}>0\)

nên 100-x=0

hay x=100

Vậy: S={100}

f) Ta có: \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)

\(\Leftrightarrow\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)

\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)

mà \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}>0\)

nên x-100=0

hay x=100

Vậy: S={100}