Vì \(\dfrac{1}{6}=\dfrac{4}{24},\dfrac{1}{7}=\dfrac{4}{28}\) nên các phân số đó là: \(\dfrac{4}{25},\dfrac{4}{26},\dfrac{4}{27}\)

4/20

Từ 50 đến 99 có 50 số; ta cho tất cả các phân số đó về 1/100; ta có 50/100 = 1/2; còn dư một số phần chênh giữa 1/100 va các phân số đó.

A = \(-1\dfrac{1}{5}\). \(\dfrac{4.\left(\dfrac{1}{3}-\dfrac{3}{7}-\dfrac{3}{53}+3\right)}{\dfrac{1}{3}-\dfrac{3}{7}-\dfrac{3}{53}+3}\): \(\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{21}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{21}}\)

A = \(-1\dfrac{1}{5}\). \(4\) : \(\dfrac{4.\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}\right)}{5.\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}\right)}\)

A = \(-4\dfrac{4}{5}\): \(\dfrac{4}{5}\)

A = \(-4\dfrac{4}{5}\) . \(\dfrac{5}{4}\)

A = \(-6\)

\(\dfrac{x+7}{x+2}=1+\dfrac{5}{x+2}\)

Để \((x+7) \vdots (x+2)\) thì \(\dfrac{5}{x+2}\) là số nguyên

     \(=>x+2 \in Ư_{5}\)

  Mà \(Ư_{5}=\){\(\pm 1 ;\pm 5\)}

 \(=>x=\){`-3;-1;-7;3`}

`(2345-45)+2345`

`=2300+2345`

`=4645`

__________________

`(-2010)-(119-2010)`

`=-2010-119+2010`

`=(-2010+2010)-119`

`=0-119`

`=-119`

$(2345-45)+2345$

$=2300+2345$

$=4645$

$(-2010)-(119-2010)$

$=-2010-119+2010$

$=(-2010+2010)-119$

$=0-119$

$=-119$

Lời giải:
\(\frac{11}{2}.\frac{12}{2}.\frac{13}{2}....\frac{20}{2}=\frac{(12.14.16.18.20)(11.13.15.17.19)}{2^{10}}\)

\(=\frac{12}{2^2}.\frac{14}{2}.\frac{16}{2^4}.\frac{18}{2}.\frac{20}{2^2}(11.13.15.17.19)\)

\(=3.7.1.9.5.11.13.15.17.19=1.3.5.7.9.11.13.15.17.19\)

bang 50000000

Ban đầu cô Yên phải gửi: 3 400 000 : 6,8 x 100

Lưu ý: Đây là dạng toán lớp 5

B=1+3+\(3^2\)+\(3^3\)+....+\(3^{1991}\)

B=1+3+\(3^2\)+\(3^3\)+....+\(3^{1991}\)

=(1+3+\(3^2\)+\(3^3\))+(\(3^4\)+\(3^5\)+\(3^6\)+\(3^7\))+.....+(\(3^{1988}\)+\(3^{1989}\)+\(3^{1990}\)+\(3^{1991}\))

=(1+\(3^4\))(1+3+\(3^2\)+\(3^3\))(\(3^8\)+....+\(3^{1988}\))

=82.(1+3+\(3^2\)+\(3^3\))(\(3^8\)+....+\(3^{1988}\))

Vì 82⋮41

→E⋮41

→B⋮41(đpcm)

Bạn tham khảo nha: 

B=1+3+32+33+....+31991B=1+3+32+33+....+31991

=(1+3+32+33)+(34+35+36+37)+.....+(31988+31989+31990+31991)=(1+3+32+33)+(34+35+36+37)+.....+(31988+31989+31990+31991)

=(1+3+32+33)+34(1+3+32+33)+....+31988(1+3+32+33)=(1+3+32+33)+34(1+3+32+33)+....+31988(1+3+32+33)

=(1+3+32+33)+(1+34+....+31988)=(1+3+32+33)+(1+34+....+31988)

=(1+34)(1+3+32+33)(38+....+31988)=(1+34)(1+3+32+33)(38+....+31988)

=82.(1+3+32+33)(38+....+31988)=82.(1+3+32+33)(38+....+31988)

Vì 82⋮4182⋮41

→82.(1+3+32+33)(38+....+31988)⋮41→82.(1+3+32+33)(38+....+31988)⋮41

→B⋮41(đpcm)

4/9

giải các bước ra giúp mình