c) (0.01):2,5=(0,75x).(0,75)

         0,004 =(0,75x).(0,75)

         => x=0.0071

Đề này bn xem lại coi có đúng k

c) \(x:\left(-2,14\right)=\left(-3,12\right):1,2\)

=> \(x:\left(-2,14\right)=-2,6\)

=> \(x=\left(-2,6\right).\left(-2,14\right)\)

=> \(x=5,564\)

Vậy \(x=5,564.\)

d) \(2\frac{2}{3}:x=2\frac{1}{2}:\left(-0,06\right)\)

=> \(\frac{8}{3}:x=\frac{25}{12}:\left(-\frac{3}{50}\right)\)

=> \(\frac{8}{3}:x=-\frac{625}{18}\)

=> \(x=\frac{8}{3}:\left(-\frac{625}{18}\right)\)

=> \(x=-\frac{48}{625}\)

Vậy \(x=-\frac{48}{625}.\)

Chúc bạn học tốt!

a) Lm r nkoa!

b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)

\(=\left(0,25x\right):3=\frac{20}{3}\)

\(\Rightarrow0,25x=\frac{20}{3}\cdot3=20\)

\(\Rightarrow x=20:0,25=80\)

\(\Rightarrow x=80\)

c) \(0,01:2,5=\left(0,75x\right):0,75\)

\(=\frac{1}{250}=\left(0,75x\right):0,75\)

\(\Rightarrow0,75x=\frac{1}{250}\cdot0,75=\frac{3}{1000}\)

\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)

\(\Rightarrow x=\frac{1}{250}\)

d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)

\(=\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)

\(\Rightarrow0,1x=\frac{2}{3}:\frac{5}{3}=\frac{2}{5}\)

\(\Rightarrow x=\frac{2}{5}:0,1=4\)

\(\Rightarrow4\)

\(a,3,8:2x=\frac{1}{4}:2\frac{2}{3}\)

\(\Leftrightarrow3,8:2x=\frac{3}{32}\)

\(\Leftrightarrow2x=3,8:\frac{3}{32}\)

\(\Leftrightarrow2x=\frac{608}{15}\)

\(\Leftrightarrow x=\frac{304}{15}\)

\(b,0,25x:3=\frac{5}{6}:0,125\)

\(\Leftrightarrow0,25x:3=\frac{20}{3}\)

\(\Leftrightarrow0,25x=20\)

\(\Leftrightarrow x=80\)

\(c,0,01:2,5=0,75x:0,75\)

\(\Leftrightarrow0,75x:0,75=\frac{1}{250}\)

\(\Leftrightarrow x=\frac{1}{250}\)

Mik lm hơi tắt chỗ nào chw hiểu thì hỏi mik nhé

a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)

\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)

hay \(x\in\left\{30;-30\right\}\)

Vậy: \(x\in\left\{30;-30\right\}\)

b) Ta có: \(\left|x\right|+0.573=2\)

\(\Leftrightarrow\left|x\right|=1.427\)

hay \(x\in\left\{1.427;-1.427\right\}\)

Vậy: \(x\in\left\{1.427;-1.427\right\}\)

c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)

d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)

\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)

\(\Leftrightarrow x=\dfrac{1}{250}\)

Vậy: \(x=\dfrac{1}{250}\)