Có: 

Có: 

Ta có:

\(a.\frac{Y}{3}=\frac{27}{Y}\Leftrightarrow\frac{Y^2}{3Y}=\frac{3.27}{3Y}\)

\(\Rightarrow Y^2=3.27=81\)

Do \(\orbr{\begin{cases}81=\left(-9\right)^2\\81=9^2\end{cases}\Rightarrow}\orbr{\begin{cases}Y=-9\\Y=9\end{cases}}\)

Vậy \(Y\in\left\{-9;9\right\}.\)

\(b.\frac{15}{Y}=\frac{Y}{9}\Leftrightarrow\frac{15.9}{Y.9}=\frac{Y^2}{9.Y}\)

\(\Rightarrow15.9=135=Y^2\)

Do \(\orbr{\begin{cases}135=\left(-\sqrt{135}\right)^2\\135=\sqrt{135}^2\end{cases}\Rightarrow}\orbr{\begin{cases}Y=-\sqrt{135}\\Y=\sqrt{135}\end{cases}}\)

Vậy \(Y\in\left\{-\sqrt{135};\sqrt{135}\right\}.\)

\(c.\frac{15}{48}:y=\frac{10}{4}\)

\(\Rightarrow y=\frac{15}{48}:\frac{10}{4}=\frac{1}{8}\)

Vậy \(y=\frac{1}{8}.\)

\(d.\frac{20}{y}.9=\frac{60}{135}\)

\(\Rightarrow\frac{20}{y}=\frac{60}{135}:9=\frac{20}{135}\)

\(\Leftrightarrow y=135\)

Vậy \(y=135.\)

\(e.\frac{16}{Y}=\frac{Y}{9}\Leftrightarrow\frac{16.9}{Y.9}=\frac{Y^2}{Y.9}\)

\(\Rightarrow16.9=144=Y^2\)

Do \(\orbr{\begin{cases}144=\left(-12\right)^2\\144=12^2\end{cases}\Rightarrow\orbr{\begin{cases}Y=-12\\Y=12\end{cases}}}\)

Vậy \(Y\in\left\{-12;12\right\}\)

\(g.\frac{25}{Y}=\frac{Y}{9}\Leftrightarrow\frac{25.9}{Y.9}=\frac{Y^2}{9.Y}\)

\(\Rightarrow25.9=225=Y^2\)

Do \(\orbr{\begin{cases}225=\left(-15\right)^2\\225=15^2\end{cases}\Rightarrow\orbr{\begin{cases}Y=-15\\Y=15\end{cases}}}\)

Vậy \(Y\in\left\{-15;15\right\}.\)

\(h.y-\frac{12}{42}=\frac{10}{12}\)

\(\Rightarrow y=\frac{10}{12}+\frac{12}{42}=\frac{47}{42}\)

Vậy \(Y=\frac{47}{42}.\)

\(i.y+\frac{5}{30}=\frac{18}{2}\)

\(\Rightarrow y=\frac{18}{2}-\frac{5}{30}=\frac{53}{54}\)

Vậy \(y=\frac{53}{54}.\)

mấy câu này gần như là giống nhau nên mik giải 1 câu rồi bạn dựa theo mà giải tiếp nha

y/3 = 27/y => y² = 3 . 27

               => y² = 81 => y = 9 hoặc -9

vậy .....

100 nha ban

2x . 3 = 15 => x = \(\frac{15}{6}\)= 2,5

=> \(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\)

= \(\frac{\frac{15}{6}+16}{9}=\frac{18,5}{9}=\frac{37}{18}\)

=> y-25/16=37/18 => y-25=296/9 => y=521/9=57,(8)

=> z+9/25=37/18 => z+9=925/18 => z=925/18=51,3(8)

Tick mk nha Đứcc Mạnhh

Đứcc Manhjh ak, bn có sai ko, 2x^3=16 hay 15, nếu bằng 16 thì tổng của 3 số trên = 100 đấy !

Ta có: \(2x^3-1=15\Leftrightarrow x^3=8\Rightarrow x=2\)

\(\Rightarrow\dfrac{18}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\Rightarrow\left\{{}\begin{matrix}\dfrac{y-25}{16}=2\Rightarrow y=57\\\dfrac{z+9}{25}=2\Rightarrow z=41\end{matrix}\right.\)

Vậy \(B=x+y+z=2+57+41=100\)

`2x^3-1=15=>2x^3=16=>x^3=8=>x=2`

Có:`[x+16]/9=[y-25]/16`

`=>[2+16]/9=[y-25]/16=>y=57`

Có:`[x+16]/9=[z+9]/25`

`=>[2+16]/9=[z+9]/25=>z=41`

Ta có:`B=x+y+z=2+57+41=100`

a: =>x/15=-3/5

=>x=-9

b: =>36/y=4/7

=>y=36:4/7=63

c: =>xy=-12

=>(x,y) thuộc {(-1;12); (12;-1); (1;-12); (-12;1); (2;-6); (-6;2); (6;-2); (-2;6); (3;-4); (-4;3); (-3;4); (4;-3)}

d: =>xy=-18

=>(x,y) thuộc {(1;-18); (-18;1); (-1;18);(18;-1); (2;-9); (-9;2); (-2;9); (9;-2); (3;-6); (-6;3); (-3;6); (6;-3)}

2x^3-1=15 => 2x^3 = 15+1 = 16 => x^3=16:2=8 = 2^3 => x=2

Khi đó : y-25/16=z+9/25=x+16/9 = 2+16/9 = 2 => y = 57 ; z = 41 => x+y+z = 2+57+41 = 100