a, \(70-5.\left(x-3\right)=45\\ =>5.\left(x-3\right)=70-45=25\\ =>x-3=25:5=5\\ =>x=5+3=8\)

b, \(10+2.x=4^5:4^3\\ =>2.x=\left(4^5:4^3\right)-10=16-10=6\\ =>x=6:2=3\)

a, 70 - 5 . ( \(x\) - 3) = 45

5 . ( \(x\)-3) = 70 - 45

5 . ( \(x\) - 3) = 25

\(x\) - 3 = 25: 5

\(x\) - 3 = 5

\(x\) = 5 + 3

\(x\) = 8

Vậy \(x\) = 8

b, 10 + 2 .\(x\) = \(4^5:4^3\)

10 + 2 . \(x\) = 1024 : 64

10 + 2 .\(x\) = 16

2 .\(x\) = 16 - 10

2 . \(x\) = 6

\(x\) = 6 : 2

\(x\) = 3

Vậy \(x\) = 3

Chúc bạn học tốt !!!

a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)

\(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\)

hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\)

Vậy: \(x=\frac{9}{10}\)

b) Ta có: \(5\frac{4}{7}:x=13\)

\(\Leftrightarrow\frac{39}{7}:x=13\)

\(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\)

Vậy: \(x=\frac{3}{7}\)

c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=84\)

\(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\)

Vậy: x=30

d) Ta có: \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

\(\Leftrightarrow\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{-1}{15}\)

hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\)

Vậy: x=-5

e) Ta có: \(8\frac{2}{3}:x-10=-8\)

\(\Leftrightarrow\frac{26}{3}:x=2\)

hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\)

Vậy: \(x=\frac{13}{3}\)

g) Ta có: \(x+30\%=-1.3\)

\(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\)

hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\)

Vậy: \(x=\frac{-8}{5}\)

i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\)

\(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\)

\(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\)

\(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\)

Vậy: x=-9

k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\)

hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\)

Vậy: x=30

m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\)

\(\Leftrightarrow\left|2x-1\right|=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)

thank you nha!

cái câu a) kiểu j đấy

Bài 1

a, 2³ + ( x - 3² ) = 1

x - 3² = 1 - 2³ = -7

x = -7 + 3²

x = 2

b, 5 . (x+7) -10 = 40

5 . (x+7) = 50

x+7 = 50 :5 =10

x = 10 - 7

x = 3

b) \(\frac{3}{x-5}=\frac{-4}{x+2}\)

=> 3.(x + 2) = (x - 5). (-4)

= 3x + 6 = -4x + 20

= 3x - 4x = 26

=> -x = 26

=> x = -26

c) \(\frac{x}{-2}=\frac{-8}{x}\)

=> x.x = -2 . (-8)

 => x² = 16

=> x = 4

t i c k nha!!! 645646677778879078452352543546456457567564546567

còn câu a

\(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)

\(\frac{2}{3}x\)            = \(\frac{1}{10}+\frac{1}{2}\)

\(\frac{2}{3}x\)            =          \(\frac{3}{5}\)

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

     \(x\)             =         \(\frac{9}{10}\)

Câu b:

(2 4/5x - 50) : 2/3 = 51

2 4/5x - 50 = 51 x 2/3

14/5x - 50 = 34

14/5x = 34 + 50

14/5x = 84

x = 84 : 14/5

x = 30

Vậy x = 30

4a) \(\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}=\frac{1}{10}\)

\(\Leftrightarrow x=\frac{1}{10}:\frac{-2}{3}=\frac{1}{10}.\frac{3}{-2}=\frac{3}{-20}\)

Vậy x=\(\frac{3}{-20}\)

b) \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)

\(\Leftrightarrow\left(\frac{2}{3}-\frac{3}{2}\right)x=\frac{5}{12}\)

\(\Leftrightarrow\frac{-5}{6}x=\frac{5}{12}\)

\(\Leftrightarrow x=\frac{5}{12}:\frac{-5}{6}=\frac{5}{12}.\frac{6}{-5}=\frac{1}{-2}\)

Vậy x=\(\frac{1}{-2}\)

g)Sửa đề: \(\left|4x-1\right|=\left(-3\right)^2\)

\(\Leftrightarrow\left|4x-1\right|=9\)

\(\Rightarrow\left[{}\begin{matrix}4x-1=9\\4x-1=\left(-9\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{5}{2};-2\right\}\)

i) \(\left(x-1^3\right)=125\)

\(\Leftrightarrow x-1=125\)

\(\Leftrightarrow x=125+1=126\)

Vậy x=126

k) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)

xin lỗi câu h tui xin chữa lại là:\(|x+70\%|=2\frac{1}{5}\)

\(\left(2\frac{4}{5}x-50\right)\div\frac{2}{3}=51\)

\(\frac{14}{5}x-50=51\times\frac{2}{3}\)

\(\frac{14}{5}x-50=34\)

\(\frac{14}{5}x=84\)

\(x=84\times\frac{5}{14}\)

\(x=30\)

\(a,\left(2\frac{4}{5}.x-50\right):\frac{2}{3}=51\)

    \(\left(\frac{14}{5}.x-50\right):\frac{2}{3}=51\)

   \(\frac{14}{5}.x-50=51\times\frac{2}{3}\)

   \(\frac{14}{5}.x-50=34\)

   \(\frac{14}{5}.x=34+50\)

   \(\frac{14}{5}.x=84\)

    \(x=84:\frac{14}{5}\)

   \(x=\frac{405}{14}\)