\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\)và \(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)
=> \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Ta có :
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
=> \(\left\{{}\begin{matrix}\dfrac{x}{8}=2\Rightarrow x=2.8=16\\\dfrac{y}{12}=2\Rightarrow x=2.12=24\\\dfrac{z}{15}\Rightarrow2.15=30\end{matrix}\right.\)
@Kfkfj

B = {1; 5; 9}
C = {1; 6; 8}
D = {2; 4; 9}
E = {2; 5; 8}
F = {2; 6; 7}
G = {3; 4; 8}
H = {3; 5; 7}
I = {4; 5; 6}
@Hoàng Thu Ngân

b, \(\dfrac{\left(3\dfrac{1}{5}.1,4-2,5.\dfrac{7}{108}\right):2\dfrac{7}{18}+4\dfrac{1}{2}.0,1}{70,5-5,28:7\dfrac{1}{2}}\)

= \(\dfrac{\left(\dfrac{16}{5}.\dfrac{7}{5}-\dfrac{5}{2}.\dfrac{7}{108}\right):\dfrac{43}{18}+\dfrac{9}{2}.\dfrac{1}{10}}{\dfrac{141}{2}-\dfrac{132}{25}:\dfrac{15}{2}}\)

= \(\dfrac{\left(\dfrac{112}{25}-\dfrac{35}{216}\right):\dfrac{43}{18}+\dfrac{9}{20}}{\dfrac{141}{2}-\dfrac{88}{125}}\)

= \(\dfrac{\dfrac{112}{25}:\dfrac{43}{18}-\dfrac{25}{216}:\dfrac{43}{18}+\dfrac{9}{20}}{\dfrac{17449}{250}}\)

= \(\dfrac{\dfrac{2016}{1075}-\dfrac{25}{516}+\dfrac{9}{20}}{\dfrac{17449}{250}}\)

= \(\dfrac{\dfrac{2016}{1075}+\dfrac{259}{645}}{\dfrac{17449}{250}}\)
= \(\dfrac{\dfrac{7343}{3225}}{\dfrac{17449}{250}}\)

@Christina

a, (3\(\dfrac{1}{5}-2\dfrac{1}{3}+\dfrac{5}{6}\)) - (1,5 - \(\dfrac{2}{3}+\dfrac{1}{6}\))

= \(\dfrac{16}{5}-\dfrac{7}{3}+\dfrac{5}{6}-\dfrac{3}{2}+\dfrac{2}{3}-\dfrac{1}{6}\)

= (\(\dfrac{16}{5}-\dfrac{3}{2}\)) - (\(\dfrac{7}{3}-\dfrac{2}{3}\)) + (\(\dfrac{5}{6}-\dfrac{1}{6}\))

= \(\dfrac{17}{10}-\dfrac{5}{3}+\dfrac{2}{3}\)
= \(\dfrac{17}{10}-1\)

= \(\dfrac{7}{10}\)

@Christina

2n + 1 \(⋮6-n\)
Có -2(6 - n) \(⋮6-n\)
=> 2n + 1 + 12 - 2n \(⋮6-n\)
=> 13 \(⋮6-n\)
=> 6 - n \(\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
=> n = 5; 7; -7; 19 (thỏa mãn)
@Nguyễn Thị Kiều Trang

Ta có : a : 24 dư 10
=> a = 24k + 10 = 2.12k + 2.5 = 2.(12k +5) \(⋮2\)
Do 24k chia hết cho 4 mà 10 không chia hết cho 4
=> a không chia hết cho 4
@Nguyễn Thị Kiều Trang

Số học sinh lớp 6C chiếm :
\(\dfrac{8}{9}.\dfrac{17}{16}=\dfrac{136}{144}\) (số học sinh lớp 6B)
Số học sinh cả 3 lớp chiếm :
\(\dfrac{144}{144}+\dfrac{8}{9}+\dfrac{136}{144}=\dfrac{408}{144}\) (số học sinh lớp 6B)
Số học sinh lớp 6B là :
102 : \(\dfrac{408}{144}=36\) (học sinh)
Số học sinh lớp 6A là :
36 x \(\dfrac{8}{9}=32\) (học sinh)
Số học sinh lớp 6C là :
32 x \(\dfrac{17}{16}=34\) (học sinh)
@Thùy Nguyễn Thị Bích

Ta có :
\(\dfrac{a}{b}=\dfrac{18}{27}=\dfrac{2}{3}\)
ƯCLN(a; b) = 17
<=> \(\left\{{}\begin{matrix}a=2k\\b=3k\end{matrix}\right.\) (k \(\in\) N)
<=> 2k \(⋮17;3k⋮17\)<=> 3k - 2k = k \(⋮17\)
<=> k = 17
<=> \(\left\{{}\begin{matrix}a=17.2=34\\a=17.3=51\end{matrix}\right.\)
@Gin Tuấn Kiệt

\(\dfrac{1}{5}3-2x=\dfrac{1}{10}1\)

=> \(\dfrac{3}{5}-2x=\dfrac{1}{10}\)

=> 2x = \(\dfrac{1}{2}\)

=> x = \(\dfrac{1}{4}\)

@Bùi Thị Phương Anh

3\(\dfrac{1}{5}-2x=1\dfrac{1}{10}\)

=> \(\dfrac{16}{5}-2x=\dfrac{11}{10}\)

=> 2x = \(\dfrac{21}{10}\)

=> x = \(\dfrac{21}{20}\)

@Bùi Thị Phương Anh