Ta có: \(\frac{9x}{4}=\frac{16}{x}\Rightarrow9.x^2=64\)

ủa 64 không chia hết cho 9 bạn ạ!

Đề sai rồi hay sao ớ

ta có \(\dfrac{9x}{4}=\dfrac{16}{x}\)=> 9x²=64 <=> (3x)²=64

<=> 3x=8 hoặc 3x=-8 <=> x=8/3 hoặc x=-8/3

mà theo đề là tìm x là số âm nên x=8/3 loại

Vậy x=-8/3

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

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

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

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

0,(12) =4/33

1,(6) = 5/3

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

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

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

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

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

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

\(\Leftrightarrow9x\cdot x=16\cdot4\)

\(\Leftrightarrow9x^2=64\)

\(\Leftrightarrow x^2=\frac{64}{9}\)

\(\Leftrightarrow x\in\left\{\frac{-8}{3};\frac{8}{3}\right\}\)

Mà x là số âm

\(\Rightarrow x=\frac{-8}{3}\)

Vậy.........

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

\(9x^2=16\times4\)

\(\Rightarrow9x^2=64\)

\(\Rightarrow x^2=64\div9\)

\(\Rightarrow x^2=\frac{64}{9}\)

\(\Rightarrow x^2\in\left\{\frac{-8}{3};\frac{8}{3}\right\}\)

\(\Rightarrow x=\frac{-8}{3}\)

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

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

=> \(3x+3=7x-7\)

=> \(3x+10=7x\)

=> \(4x=10\)

=> \(x=\frac{10}{4}=\frac{5}{2}\)

Vậy \(x=\frac{5}{2}\)

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

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

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

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

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

~~~!!!

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

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

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

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

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