a) \(\dfrac{x}{4}=\dfrac{16}{128}\Rightarrow x=\dfrac{4.16}{128}=\dfrac{1}{2}=0,5\)

b)\(1\dfrac{5}{6}=\dfrac{-x}{5}\\ \Leftrightarrow\dfrac{11}{6}=\dfrac{-x}{5}\Rightarrow-x=\dfrac{5.11}{6}=\dfrac{55}{6}\Rightarrow x=-\dfrac{55}{6}\)

c) \(\dfrac{4,25}{8}=\dfrac{-3,5}{x}\Rightarrow x=\dfrac{8.\left(-3,5\right)}{4,25}=-\dfrac{112}{17}\)

a,X/4=16/28

=»x/4=4/7

=4/7.4

=7

b,1/5/6=-x/5

-x/5=11/6

X=11/6.5

X=55/6

Câu a :

\(\dfrac{x}{4}=\dfrac{16}{128}\) \(\Leftrightarrow x.128=4.16\) \(\Rightarrow x=\dfrac{1}{2}\)

Câu b :

\(1\dfrac{5}{6}=-\dfrac{x}{5}\) \(\Leftrightarrow11.5=-x.6\) \(\) \(\Rightarrow-x=-\dfrac{55}{6}\)

Câu c :

\(\dfrac{4,25}{8}=\dfrac{-3,5}{x}\) \(\Leftrightarrow4,25.x=8.-3,5\) \(\Rightarrow x=-\dfrac{112}{17}\)

a,Ta có : x.128=16.4

x.128=64

Suy ra : x=1/2

b,\(1\dfrac{5}{6}=\dfrac{11}{6}\)

Ta có ; 11.5=-x.6

-x.6=55

Suy ra : x=-55/6

c,Ta có : 4,25.x=-3,5.8

4,25.x=-28

Suy ra : x=-112/17

\(a.\)

\(\dfrac{x}{4}=\dfrac{16}{128}\)

\(\Rightarrow x=\dfrac{16.4}{128}=\dfrac{1}{2}\)

\(b.\)

\(1\dfrac{5}{6}=-\dfrac{x}{5}\)

\(\Rightarrow\dfrac{11}{6}=-\dfrac{x}{5}\)

\(\Rightarrow x=\dfrac{11.\left(-5\right)}{6}=-\dfrac{55}{6}\)

\(c.\)

\(4,25:8=-3,5:x\)

\(\Rightarrow\dfrac{4,25}{8}=\dfrac{-3,5}{x}\)

\(\Rightarrow x=\dfrac{-3,5.8}{4,25}=-\dfrac{112}{17}\)

\(a)\dfrac{x}{4}=\dfrac{16}{128}\)

\(\Rightarrow\dfrac{16}{128}.4=\dfrac{1}{2}\)

\(b)1\dfrac{5}{6}=\dfrac{-x}{5}\)

\(\Rightarrow x=\dfrac{11}{6}.\left(-5\right)=\dfrac{-55}{6}\)

\(c)4,25:8=-3,5:x\)

\(\Rightarrow\dfrac{4,25}{8}=\dfrac{-3,5}{x}\)

\(\Rightarrow x=\dfrac{-3,5.8}{4,25}=\dfrac{-112}{17}\)

Chúc bạn học tốt!

a) \(x-\dfrac{5}{6}=\dfrac{1}{2}\)

           \(x=\dfrac{1}{2}+\dfrac{5}{6}\)

           \(x=\dfrac{4}{3}\)

vậy x = ....

b) \(3\dfrac{1}{3}x+16\dfrac{3}{4}=-13,25\)

    \(\dfrac{10}{3}\)\(x+\dfrac{67}{4}=-13,25\)

     \(\dfrac{10}{3}x=\left(-13,25\right)-\dfrac{67}{4}\)

      \(\dfrac{10}{3}x=-30\)

          \(x=\left(-30\right):\dfrac{10}{3}\)

         \(x=-9\)

vậy x =...

sai mog bn thông cảm!!!

a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{4-9}=\dfrac{-16}{-5}=\dfrac{16}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.\dfrac{16}{5}\\y^2=9.\dfrac{16}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm\left(2.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{8\sqrt[]{5}}{5}\\y=\pm\left(3.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{12\sqrt[]{5}}{5}\end{matrix}\right.\)

\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow z=\dfrac{5}{4}y=\dfrac{5}{4}.\left(\pm\dfrac{12\sqrt[]{5}}{5}\right)=\pm3\sqrt[]{5}\)

b) \(\left|2x+3\right|=x+2\)

\(\Rightarrow\left[{}\begin{matrix}2x+3=x+2\\2x+3=-x-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-\dfrac{5}{3}\end{matrix}\right.\)

Đính chính

Dòng cuối \(3x=-\dfrac{5}{3}\rightarrow x=-\dfrac{5}{3}\)

a)\(\dfrac{x}{4}=\dfrac{16}{128}\)

\(\Leftrightarrow\dfrac{x}{4}=\dfrac{1}{8}\)

\(\Rightarrow x=\dfrac{1.4}{8}=\dfrac{1}{2}\)

b)\(1\dfrac{5}{6}=\dfrac{-x}{5}\)

\(\Leftrightarrow\dfrac{11}{6}=\dfrac{-x}{5}\)

\(\Rightarrow-x=\dfrac{11.5}{6}=\dfrac{-55}{6}\)

c)\(4,25:8=-3,5:x\)

\(\Leftrightarrow\dfrac{17}{32}=\dfrac{-3,5}{x}\)

\(\Rightarrow x=\dfrac{\left(-3,5\right).32}{17}=-\dfrac{112}{17}\)

Tìm x

a)\(\dfrac{x}{4}\)=\(\dfrac{16}{128}\)

\(\Rightarrow\)x.128=16.4

x.128=64

x=64:128

x=\(\dfrac{1}{2}\)

Vậy x=\(\dfrac{1}{2}\)

b)\(1\dfrac{5}{6}\)=\(\dfrac{-x}{5}\)

\(\dfrac{11}{6}\)=\(\dfrac{-x}{5}\)

-x:5=\(\dfrac{11}{6}\)

-x=\(\dfrac{11}{6}\).5

-x=\(\dfrac{55}{6}\)

\(\Rightarrow\)x=\(\dfrac{-55}{6}\)

Vậy x=\(\dfrac{-55}{6}\)

c)4,25:8 = -3,5:x

-3,5:x=\(\dfrac{17}{4}\)

x=-3,5:\(\dfrac{17}{4}\)

x=\(\dfrac{14}{17}\)

Vậy x=\(\dfrac{14}{17}\)

a, Ta có : \(\dfrac{x}{4}\)= \(\dfrac{16}{128}\)

\(\Rightarrow\)4 . 16 = x . 128

\(\Rightarrow\) x = (4 . 16 ) : 128

\(\Rightarrow\) x = \(\dfrac{1}{2}\)

b, Ta có : 1\(\dfrac{5}{6}\) = \(\dfrac{-x}{5}\)

\(\Leftrightarrow\dfrac{11}{6}\) = \(\dfrac{-x}{5}\)

\(\Rightarrow\)11 . 5 = 6 . ( \(-x\) )

\(\Rightarrow\)\(-x\) = (11. 5) : 6

\(\Rightarrow\) \(-x\) = \(\dfrac{-55}{6}\)

\(\Rightarrow\) \(x\) = \(\dfrac{55}{6}\)

c,Ta có : 4 , 25 : 8 = \(-3,5\) : x

\(\Rightarrow\) \(\dfrac{17}{32}\) = \(\dfrac{-3,5}{x}\)

\(\Rightarrow\) 17 . x = \(-3,5\) . 32

\(\Rightarrow\) x = ( \(-3,5\) . 32 ): 17

\(\Rightarrow\) x = \(\dfrac{-112}{17}\)

a) Vì \(\dfrac{x}{y} = \dfrac{5}{3} \Rightarrow \dfrac{x}{5} = \dfrac{y}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\begin{array}{l}\dfrac{x}{5} = \dfrac{y}{3} = \dfrac{{x + y}}{{5 + 3}} = \dfrac{{16}}{8} = 2\\ \Rightarrow x = 2.5 = 10\\y = 2.3 = 6\end{array}\)

Vậy x=10, y=6

b) Vì \(\dfrac{x}{y} = \dfrac{9}{4} \Rightarrow \dfrac{x}{9} = \dfrac{y}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

 \(\begin{array}{l}\dfrac{x}{9} = \dfrac{y}{4} = \dfrac{{x - y}}{{9 - 4}} = \dfrac{{ - 15}}{5} =  - 3\\ \Rightarrow x = ( - 3).9 =  - 27\\y = ( - 3).4 =  - 12\end{array}\)

Vậy x = -27, y = -12.

c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)

\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)

\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)

\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)

\(x=-\dfrac{15}{2}\)

d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)

\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)

A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)

\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)

\(x\)\(=\dfrac{11}{9}\)

B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)

\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)

\(x=\)\(\dfrac{-2}{3}\)

a)

\(\frac{4}{9} + x = \frac{5}{3}\)

=> \(x = \frac{5}{3}-\frac{4}{9}\)

=> \(x = \) \(\frac{11}{9}\)

Vậy \(x = \dfrac{11}{9}\)

b) 

\(\dfrac{3}{4} .x = \dfrac{-1}{2}\)

=> \(x = \dfrac{-1}{2} : \dfrac{3}{4}\)

=> \(x = \dfrac{-2}{3}\)

Vậy \(x = \dfrac{-2}{3}\)

c)

\( \dfrac{3}{7}+ \dfrac{5}{7}:x = \dfrac{1}{3}\)

=> \(\dfrac{5}{7}:x = \dfrac{1}{3}-\) \( \dfrac{3}{7}\)

=> \(\dfrac{5}{7}:x = \dfrac{-2}{21}\)

=> \(x = \dfrac{5}{7}:\dfrac{-2}{21}\)

=> \(x = \dfrac{-15}{2}\)

Vậy \(x = \dfrac{-15}{2}\)

d) 

\(3\dfrac{1}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \(\dfrac{13}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \( |2x - \dfrac{5}{12} | =\dfrac{13}{4} : \dfrac{39}{16}\)

=> \(|2x-\dfrac{5}{12} |= \dfrac{4}{3}\)

=> \(\left[\begin{matrix} 2x - \dfrac{5}{12} = \dfrac{4}{3}\\ 2x - \dfrac{5}{12} = \dfrac{4}{3}\end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{-4}{3}+\dfrac{5}{12}\\ 2x = \dfrac{-4}{3}+\dfrac{5}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{7}{4}\\ 2x = \dfrac{-11}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} x = \dfrac{7}{8}\\ x = \dfrac{-11}{24} \end{matrix}\right.\)

Vậy \(x \in \) { \(\dfrac{7}{8} ; \dfrac{-11}{24}\) }