\(\frac{9x}{4}=\frac{16}{x}\)

=>\(x^2=\frac{16\cdot4}{9}=\frac{64}{9}\)

Mà x âm nên   \(x=-\frac{8}{3}\)

Ta có : \(\frac{9x}{4}=\frac{16}{x}\)

=> 9x² = 16 x 4

=> 9x² = 64

=> x² = 64 : 9

=> x² = \(\frac{64}{9}\)

=> x = \(\frac{8}{3}\)

Giá trị của x là -8/3

ta có: \(\frac{9x}{4}\)=\(\frac{16}{x}\)

suy ra 9xx=16.4

9x²=64

x²=64:9

x²=\(\frac{3}{8}\)

x=công trừ 3/8

mà theo đề bài thì x là số âm

vậy x=\(\frac{-3}{8}\)thì mới thỏa mãn 9x/4=16/x

\(\frac{16}{45}\)

\(\frac{16}{45}\)

\(\frac{16}{45}\)

\(\Rightarrow x=\frac{16}{45}\)

0,(12) =4/33

1,(6) = 5/3

11.0,(4) =11.4/9 =44/9

\(\frac{\frac{4}{33}}{\frac{5}{3}}=\frac{x}{\frac{44}{9}}\Leftrightarrow x=\frac{4.3}{33.5}.\frac{44}{9}=\frac{16}{45}\)

\(\frac{x+1}{x-1}=\frac{7}{3}\)

=> \(3.\left(x+1\right)=7.\left(x-1\right)\)

=> \(3x+3=7x-7\)

=> \(3x+10=7x\)

=> \(4x=10\)

=> \(x=\frac{10}{4}=\frac{5}{2}\)

Vậy \(x=\frac{5}{2}\)

\(\frac{x+1}{x-1}=\frac{7}{3}\)

\(\Rightarrow3\left(x+1\right)=7\left(x-1\right)\)

\(\Leftrightarrow3x+3=7x-7\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\frac{5}{2}\)

~~~!!!

\(\frac{x+1}{x-1}=\frac{7}{3}\)

\(\Leftrightarrow3x+3=7x-7\)

\(\Leftrightarrow3x=7x-7-3\)

\(\Leftrightarrow-4x=-10\)

\(\Rightarrow x=\frac{5}{2}\)