a: 2x(x-1/7)=0

=>x(x-1/7)=0

=>x=0 hoặc x=1/7

b: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

nên \(x=\dfrac{-1}{4}:\dfrac{7}{20}=\dfrac{-20}{4\cdot7}=\dfrac{-5}{7}\)

c: \(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(3.2:3.2+\dfrac{45}{10}\cdot\dfrac{31}{45}\right):\left(-21.5\right)\)

\(\Leftrightarrow2-7< x< \dfrac{\left(1+3.1\right)}{-21.5}\)

\(\Leftrightarrow-5< x< \dfrac{-41}{215}\)

mà x là số nguyên

nên \(x\in\left\{-4;-3;-2;-1\right\}\)

Bn k có máy tính ạ/

nóa pải ghi cách lm bn

a: =>1/6x=-49/60

=>x=-49/60:1/6=-49/60*6=-49/10

b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2

=>x=17/15 hoặc x=-13/15

c: =>1,25-4/5x=-5

=>4/5x=1,25+5=6,25

=>x=125/16

d: =>2^x*17=544

=>2^x=32

=>x=5

i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5

=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2

=>x=14,4 hoặc x=9,6

j: =>(2x-1)(2x+1)=0

=>x=1/2 hoặc x=-1/2

a. \(\dfrac{-39}{7}:x=26\)

x = \(\dfrac{-39}{7}:26\)

x = \(\dfrac{-3}{14}\)

b. \(x:\dfrac{13}{5}=\dfrac{7}{4}\)

x = \(\dfrac{7}{4}.\dfrac{13}{5}\)

x = \(\dfrac{91}{20}\)

c. x = \(\dfrac{-3}{5}-\dfrac{1}{2}\)

x = \(\dfrac{-11}{10}\)

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

x = \(\dfrac{9}{4}+\dfrac{3}{4}\)

x = 3

e. \(\dfrac{7}{8}:x=\dfrac{14}{3}\)

x = \(\dfrac{7}{8}:\dfrac{14}{3}\)

x = \(\dfrac{3}{16}\)

f. \(x:\dfrac{8}{3}=\dfrac{13}{3}\)

x = \(\dfrac{13}{3}.\dfrac{8}{3}\)

x = \(\dfrac{104}{9}\)

g. x = \(\dfrac{4}{10}-\dfrac{2}{5}\)

x = 0

chúc bạn học tốt

Giải:

a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)

\(\Leftrightarrow x=\dfrac{-63}{10}\)

Vậy ...

b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)

\(\Leftrightarrow x=\dfrac{-4}{11}\)

Vậy ...

Các câu sau làm tương tự câu b)

2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)

a) \(\dfrac{x}{48}=-\dfrac{4}{7}\Rightarrow x=-\dfrac{192}{7}\)

b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\Rightarrow x+\dfrac{4}{5}=1\)

\(\Rightarrow x=\dfrac{1}{5}\)

c) \(2\left|x-1\right|^2=72\Rightarrow\left|x-1\right|^2=36\)

\(\Rightarrow\left|x-1\right|=6\)

TH1: x - 1 = -6 => x = -5

TH2: x - 1 = 6 => x = 7

e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\Rightarrow x=2\)

f) | x - 2 | = 1 + 4 = 5

TH1: x - 2 = -5 => x = -3

TH2: x - 2 = 5 => x = 7

a) \(\dfrac{x}{48}=\dfrac{-4}{7}\)

⇒ x.7=48.(-4)

7x = -192

x=\(\dfrac{-192}{7}\) Vậy x=\(\dfrac{-192}{7}\)

b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\)

\(\left(x+\dfrac{4}{5}\right)=\dfrac{3}{5}+\dfrac{2}{5}\)

\(x+\dfrac{4}{5}=1\)

\(x=1-\dfrac{4}{5}\)

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

c) chưa từng gặp dạng với giá trị tuyệt đối sory

d) \(\dfrac{1}{6}x-\dfrac{2}{3}=2\)

\(\dfrac{1}{6}x=2+\dfrac{2}{3}\)

\(\dfrac{1}{6}x=\dfrac{8}{3}\)

\(x=\dfrac{8}{3}:\dfrac{1}{6}\)

\(x=16\)

e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\)

=> x.5 = 4.2,5

5x=10

x=10:5

x=2

f) |x-2|-4=1

|x-2|=1+4

|x-2|=5

=>\(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=5+2\\x=-5+2\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

đôi khi cũng có sai sót , hãy xem lại thật kĩ

Cậu không làm được hay cần gấp con nào nhỉ ?

Bài 1:

a: \(\Leftrightarrow\dfrac{x+2}{2}=x-5\)

=>2x-10=x+2

=>x=12

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

=>x+2=10 hoặc x+2=-10

=>x=-12 hoặc x=8

c: \(\Leftrightarrow\left(2x-5\right)^3=27\)

=>2x-5=3

=>2x=8

=>x=4

Bài 1: a) 4 . ( \(\dfrac{1}{3}\) - x ) + \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) . x

=> \(\dfrac{4}{3}\) - 4.x + \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) . x

=> \(\dfrac{11}{6}\) = \(\dfrac{29}{6}\) . x

=> x = \(\dfrac{11}{29}\) .

b) \(\dfrac{5}{2}\) - 3 . ( \(\dfrac{1}{3}\) - x ) = \(\dfrac{1}{4}\) - 7.x

=> \(\dfrac{1}{4}\) - \(\dfrac{3}{2}\) = 7.x + 3.x

=> \(\dfrac{-5}{4}\) = 10.x

=> x = \(\dfrac{-1}{8}\).