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24 tháng 7 2018

a, Ta có: \(|x-1|\ge0\forall x;|x-4|\ge0\forall x\)

\(\Rightarrow|x-1|+|x-4|\ge0\forall x\)

\(\Rightarrow3x\ge0\)

\(\Rightarrow x-1+x-4=3x\)

\(\Rightarrow\left(x+x\right)-\left(1+4\right)=3x\)

\(\Rightarrow2x-5=3x\)

\(\Rightarrow3x-2x=5\)

\(\Rightarrow x=5\)

Vậy x=5

b, Ta có: \(|x+1|\ge0\forall x;|x+4|\ge0\forall x\)

\(\Rightarrow|x+1|+|x+4|\ge0\)

\(\Rightarrow3x\ge0\)

\(\Rightarrow\left(x+1\right)+\left(x+4\right)=3x\)

\(\Rightarrow x+1+x+4=3x\)

\(\Rightarrow\left(x+x\right)+\left(1+4\right)=3x\)

\(\Rightarrow2x+5=3x\)

\(\Rightarrow5=3x-2x\)

\(\Rightarrow5=x\)

Vậy x=5

24 tháng 7 2018

a) Lập bảng xét dấu, ta được kết quả sau:

Nếu \(x\le1\Rightarrow\left|x-1\right|+\left|x-4\right|=-\left(x-1\right)-\left(x-4\right)=3x\)

                                                       \(=-x+1-x+4=3x\Rightarrow-5x=-5\Rightarrow x=1\)  (nhận)

Nếu \(1\le x< 4\Rightarrow\left|x-1\right|+\left|x-4\right|=x-1-x+4=3x\) 

                                                         \(-3x=-3\Rightarrow x=1\) (nhận)

Nếu \(x\ge4\Rightarrow\left|x-1\right|+\left|x-4\right|=x-1+x-4=3x\)

                                                 \(\Rightarrow-x=5\Rightarrow x=-5\) (loại)

Vậy x = 1

3 tháng 2 2022

a) \(\dfrac{-x}{4}=\dfrac{-9}{x}\)

\(\Rightarrow-x^2=-36\)

\(\Rightarrow x^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Vậy: \(x\in\left\{6;-6\right\}\)

b) \(\dfrac{5}{9}+\dfrac{x}{-1}=-\dfrac{1}{3}\)

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

\(\Rightarrow5-9x=-3\)

\(\Rightarrow-9x=-8\)

\(\Rightarrow x=\dfrac{8}{9}\)

Vậy: \(x=\dfrac{8}{9}\)

c) \(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)

\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{3}{2}.\dfrac{16}{5}\)

\(\Rightarrow x=\dfrac{24}{5}\)

Vậy: \(x=\dfrac{24}{5}\)

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

\(\Rightarrow\left(3x-1\right)^2=25\)

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

\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)

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

Vậy: \(x\in\left\{2;-\dfrac{4}{3}\right\}\)

13 tháng 10 2021

\(a,\Rightarrow\left(35x+3\right)\cdot19=152\\ \Rightarrow35x+3=8\\ \Rightarrow x=\dfrac{1}{7}\\ b,\Rightarrow3\left(x+7\right)=42\\ \Rightarrow x+7=14\Rightarrow x=7\\ c,\Rightarrow3\left(x+1\right)=48\\ \Rightarrow x+1=16\Rightarrow x=15\\ d,\Rightarrow120-5x+100\cdot2:5=4\cdot15\\ \Rightarrow120-5x+40=60\\ \Rightarrow5x=100\Rightarrow x=20\\ e,\Rightarrow4x-10=30\\ \Rightarrow4x=40\\ \Rightarrow x=10\\ g,\Rightarrow10x+10=70\\ \Rightarrow10x=60\\ \Rightarrow x=6\)

20 tháng 12 2021

b: =>x-3=12

hay x=15

20 tháng 12 2021

a. 4.(x+41) = 7

x + 41 = 7 : 4 = 1,75

x = 1,75 - 41 = -39,25

b. 4.(x-3) = 72 - 110 = 49 - 1 = 48

x - 3 = 48 : 4 = 12

x = 12 + 3 = 15

a) \(4\left(x+41\right)=400\)
\(\Rightarrow x+41=400:4\)
\(\Rightarrow x+41=100\)
\(\Rightarrow x=100-41\)
\(\Rightarrow x=59\)

20 tháng 12 2021

a/ \(4.\left(x+41\right)=400\)

\(x+41=\dfrac{400}{4}=100\)

\(\Rightarrow x=59\)

b/ \(4.\left(x-3\right)=7^2-1^{10}\)

\(4.\left(x-3\right)=49-1\)

\(4.\left(x-3\right)=48\)

\(x-3=\dfrac{48}{3}=16\)

\(\Rightarrow x=19\)

#AEZn8

b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)

=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)

=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)

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

=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)

=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)

=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)

d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)

=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)

=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)

=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)

c) Ta có: \(\left|x-\dfrac{2}{3}\right|+2.25=\dfrac{3}{4}\)

\(\Leftrightarrow\left|x-\dfrac{2}{3}\right|=\dfrac{3}{4}-\dfrac{9}{4}=\dfrac{-3}{2}\)(vô lý)

Vậy: \(x\in\varnothing\)

 

a) Ta có: \(x+\dfrac{-3}{7}=\dfrac{4}{7}\)

\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{4}{7}\)

hay x=1

Vậy: x=1

31 tháng 5 2023

\(a,50\%x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)

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

\(\Leftrightarrow\dfrac{3}{2}x=1\)

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

\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)

\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)

\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{25}{4}\)

31 tháng 5 2023

Bài lớp \(6\) chưa sử dụng dấu \(\Leftrightarrow\) chị nhé ! Vẫn phải sử dụng dấu \(\Rightarrow\) Khi nào bài lớp \(8\) trở lên thì cj hãy dùng \(\Leftrightarrow\) ạ

21 tháng 4 2022

a) \(\left(12-12\dfrac{1}{3}\right):x+\dfrac{1}{6}=-\dfrac{2}{3}\)

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

\(-\dfrac{1}{3}x=-\dfrac{5}{6}\)

\(x=-\dfrac{5}{6}:\left(-\dfrac{1}{3}\right)\)

\(x=\dfrac{5}{2}\)

b) \(\dfrac{4}{x}=\dfrac{x}{16}\)

\(x^2=4.16\)

\(x^2=64\)

\(\Rightarrow x=8;x=-8\)

21 tháng 4 2022

`a)=>(12-37/3):x+1/6=-2/3`

`=>(12-37/3):x=-5/6`

`=>(-1/3):x=-5/6`

`=>x=(-1/3):(-5/6)`

`=>x=6/15=2/5`

`b)4/x=x/16`

`=>x^2=4*16`

`=>x^2=64`

`=>x^2=(+-8)^2`

9 tháng 3 2022

x - 2/3 = 3/5 - 1/4

x - 2/3 = 7/20

x = 7/20 + 2/3 

x = 61/60

- 2/3 . x + 1/5 = 1/10

-2/3 . x = 1/10 - 1/5

- 2/3 . x = -1/10

x = -1/10 : -2/3

x = 3/20