Bắc lấy số quả cam là:

\(120\times\dfrac35=72\) (quả cam)

Tổng số cam của Ninh và Quế là:

\(120-72=48\) (quả cam)

Ta có sơ đồ:

Ninh: .____.____.____.

Quế : .____.____.____.____.____.

(Tổng = 48)

Theo sơ đồ, tổng số phần bằng nhau là:

\(3+5=8\) (phần)

Ninh lấy số quả cam là:

\(48:8\times3=18\) (quả cam)

Quế lấy số quả cam là:

\(48-18=30\) (quả cam)

Đáp số:...

\(\frac23+\frac34+\frac45+\frac{12}{9}+\frac{28}{16}\)

\(=\frac23+\frac34+\frac45+\frac43+\frac74\)

\(=\left(\frac23+\frac43\right)+\left(\frac34+\frac74\right)+\frac45\)

\(=2+\frac52+\frac45\)

\(=\frac{20}{10}+\frac{25}{10}+\frac{8}{10}\)

\(=\frac{20+25+8}{10}\)

\(=\frac{53}{10}=5,3\)

Ta có:

\(36=2^2\cdot3^2\)

\(60=2^2\cdot3\cdot5\)

\(72=2^3\cdot3^2\)

\(\rArrƯCLN(36,60,72)=2^2\cdot3=4\cdot3=12\)

\(\frac{49}{5}-\frac12-\frac14-\frac18-\frac{1}{16}-\frac{1}{32}\)

\(=\frac{49}{5}-\left(\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}\right)\)

\(=\frac{49}{5}-\frac{31}{32}\)

\(=\frac{1568}{160}-\frac{155}{160}\)

\(=\frac{1413}{160}=8\)\(\frac{133}{160}\)

Ta có: \(1\) phút \(20\) giây \(=80\) giây

Vậy vận tốc của người đó là \(\dfrac{420}{80}=5,25\) \(m\)/\(s\)

a)

\(\dfrac{3}{3x-2}=\dfrac{x}{x+2}-1\)

\(\dfrac{3}{3x-2}=\dfrac{x-\left(x+2\right)}{x+2}\)

\(\dfrac{3}{3x-2}=\dfrac{x-x-2}{x+2}\)

\(\dfrac{3}{3x-2}=\dfrac{-2}{x+2}\)

\(\rArr3(x+2)=-2(3x-2)\)

\(3x+6=-6x+4\)

\(3x+6x=4-6\)

\(9x=-2\)

\(x=-\dfrac29\)

Vậy \(x=-\dfrac29\)

b)

\(\dfrac{x+6}{x+5}+\dfrac32=2\)

\(\dfrac{x+6}{x+5}=\dfrac12\)

\(\rArr2(x+6)=1(x+5)\)

\(2x+12=x+5\)

\(2x-x=5-12\)

\(x=-7\)

Vậy \(x=-7\)

c)

\(\dfrac{2}{x-2}-\dfrac{3}{x-3}=\dfrac{3x-20}{\left(x-3\right)\left(x-2\right)}\)

\(\dfrac{2(x-3)-3(x-2)}{\left(x-3\right)\left(x-2\right)}=\dfrac{3x-20}{\left(x-3\right)\left(x-2\right)}\)

\(\dfrac{\left(2x-6\right)-\left(3x-6\right)}{\left(x-3\right)\left(x-2\right)}=\dfrac{3x-20}{\left(x-3\right)\left(x-2\right)}\)

\(\dfrac{2x-6-3x+6}{\left(x-3\right)\left(x-2\right)}=\dfrac{3x-20}{\left(x-3\right)\left(x-2\right)}\)

\(\dfrac{-x}{\left(x-3\right)\left(x-2\right)}=\dfrac{3x-20}{\left(x-3\right)\left(x-2\right)}\)

\(\rArr-x=3x-20\)

\(\rArr-4x=-20\)

\(\rArr x=5\)

Vậy \(x=5\)

\(\dfrac67+\dfrac54\)

\(=\dfrac{24}{28}+\dfrac{35}{28}\)

\(=\dfrac{24+35}{28}\)

\(=\dfrac{59}{28}=2\dfrac{3}{28}\)

\(\dfrac56+\dfrac98\)

\(=\dfrac{20}{24}+\dfrac{27}{24}\)

\(=\dfrac{20+27}{24}\)

\(=\dfrac{47}{24}=1\dfrac{23}{24}\)

\(\dfrac14+\dfrac56\)

\(=\dfrac{3}{12}+\dfrac{10}{12}\)

\(=\dfrac{3+10}{12}\)

\(=\dfrac{13}{12}\)

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