\(||12x-\dfrac{1}{2}|-2|\) : \(\dfrac{3}{4}\)
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8 tháng 8 2018

mọi người trả lời nhanh đi ạ, e tối đi học rùi, thanks trc

a: =>||12x-1/2|-2|=-2/3x3/4=-6/12=-1/2(loại)

b: =>2/3-1/3x-1/2+2/3x=2x+2/3

=>-5/3x=1/2

=>x=-1/2:5/3=-1/2x3/5=-3/10

c: =>|3/2x+1/4|=2+3/4=11/4

=>3/2x+1/4=11/4 hoặc 3/2x+1/4=-11/4

=>3/2x=5/2 hoặc 3/2x=-3

=>x=3/5 hoặc x=-3:3/2=-2

18 tháng 6 2018

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)

20 tháng 8 2017

Bn k có máy tính ạ/

20 tháng 8 2017

nóa pải ghi cách lm bn

5 tháng 8 2018

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 banhquahahaleuleuok

2 tháng 8 2018

a, 1/3-3/4+3/5+1/4-2/9-1/36+1/15

=(1/3+3/5+1/15)-(3/4-1/4+2/9+1/36)

=1 - 3/4

=1/4

b, 3-1/4+2/3-5-1/3+6/5-6+7/4-3/2

=(3-5-6)-(1/4-7/4)+(2/3-1/3)+(6/5-3/2)

=-8 +3/2 +1/3 -3/10

=-97/15

Bài 1:

a: \(=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{3}+\dfrac{4}{3}=\dfrac{4}{3}-1+\dfrac{1}{2}=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\)

b: \(=\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{1}{9}-1-\dfrac{2}{5}+\dfrac{5}{4}=2-1+\dfrac{1}{9}=\dfrac{10}{9}\)

c: \(=\left(\dfrac{-3}{2}\cdot\dfrac{4}{3}\right)\cdot\dfrac{-9}{2}-\dfrac{1}{2}=9-\dfrac{1}{2}=8.5\)

12 tháng 1 2018

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: \(\Leftrightarrow\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{7}{5}\)

=>10<=x<=-13/7

hay \(x\in\varnothing\)

b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =\dfrac{-2}{3}\cdot\dfrac{-11}{12}\)

=>-13/9<=x<=22/36

hay \(x\in\left\{-1;0\right\}\)

a: =>x/3=-5/2

hay x=-15/2

b: \(\Leftrightarrow\dfrac{7}{3}:x=\dfrac{1}{5}-\dfrac{4}{9}=\dfrac{9-20}{45}=\dfrac{-11}{45}\)

\(\Leftrightarrow x=\dfrac{7}{3}:\dfrac{-11}{45}=\dfrac{7}{3}\cdot\dfrac{-45}{11}=\dfrac{-105}{11}\)

c: \(\Leftrightarrow x=\dfrac{-7}{2}\cdot2=-7\)

d: =>x/27=-1/3+2/9=2/9-3/9=-1/9=-3/27

=>x=-3

26 tháng 7 2018

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

26 tháng 7 2018

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

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

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