giải nhanh giúp tui 

\(\dfrac{x}{3}=\dfrac{-10}{6}\)

\(\Leftrightarrow6x=-10.3\)

\(\Leftrightarrow6x=-30\)

\(\Leftrightarrow x=-30:6\)

\(\Leftrightarrow x=-5\)

Vậy \(x=-5\)

Ta có : \(\frac{x+4}{20}=\frac{5}{x+4}\)

=> (x + 4).(x + 4) = 20.5

=> (x + 4)² = 100

=> (x + 4)² = 10²

=> \(\orbr{\begin{cases}x+4=10\\x+4=-10\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-14\end{cases}}\)

Vậy \(x\in\left\{6;-14\right\}\)

\(\begin{array}{l}a)\dfrac{x}{6} = \dfrac{{ - 3}}{4}\\x = \dfrac{{( - 3).6}}{4}\\x = \dfrac{{ - 9}}{2}\end{array}\)

Vậy \(x = \dfrac{{ - 9}}{2}\)

\(\begin{array}{l}b)\dfrac{5}{x} = \dfrac{{15}}{{ - 20}}\\x = \dfrac{{5.( - 20)}}{{15}}\\x = \dfrac{{ - 20}}{3}\end{array}\)

Vậy \(x = \dfrac{{ - 20}}{3}\)

b: z=17y

y=21x

=>z/17=21x

=>z=357x

\(\text{a)}\frac{37-x}{x+13}=\frac{3}{7}\)

\(\Leftrightarrow3\left(x+13\right)=7\left(37-x\right)\)

\(\Leftrightarrow3x+39=259-7x\)

\(\Leftrightarrow3x+7x=259-39\)

\(\Leftrightarrow10x=220\)

\(\Leftrightarrow x=22\)

\(\text{b)}\frac{x+4}{20}=\frac{5}{x+4}\)

\(\Leftrightarrow\left(x+4\right)^2=100\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=10\\x+4=-10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=6\\x=-14\end{cases}}\)

tìm x trong tỉ lệ thức sau x/64=-3/0.8;20/17=x/81

\(\dfrac{x}{5}=\dfrac{20}{x}\)

⇔\(x.x=20.5\)

⇔\(x^2=100\)

⇔\(x^2=\left(+-10\right)^2\)

⇔\(\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

Đề thiếu rồi bạn ơi

$-\dfrac{3}{x-2} = \dfrac45$

$4(x-2) = (-3).5$

$4x - 8 = -15$

$x = -\dfrac74$

\(\dfrac{-3}{x-2}=\dfrac{4}{5}\\ =>\left(x-2\right)\cdot4=-3\cdot5\\ =>\left(x-2\right)\cdot4=-15\\ =>x-2=-15:4\\ =>x-2=-3,75\\ =>x=-3,75+2\\ =>x=-1,75\)