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8 tháng 8 2018

(x + 1) + (x + 4) + (x + 7) + ... + (x + 28) = 155

=> (x + x + x + ... + x) + (1 + 4 + 7 + ... + 28) = 155

=> 10x + 145 = 155

=> 10x = 10

=> x = 1

vậy_

\(\left(x-2\right)\div37=\left(1008+1\right)\times\frac{2}{37}\)

\(\left(x-2\right)\div37=1009\times\frac{2}{37}\)

\(\left(x-2\right)\div37=\frac{2018}{37}\)

\(x-2=\frac{2018}{37}\times37\)

\(x-2=2018\)

\(x=2018+2\)

\(x=2020\)

7 tháng 5 2022

+) \(\dfrac{1}{3}x=-\dfrac{4}{3}-\dfrac{1}{2}=-\dfrac{11}{6}\)

\(x=-\dfrac{11}{6}:\dfrac{1}{3}=-\dfrac{11}{2}\)

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

\(x=-\dfrac{1}{6}:\dfrac{4}{3}=-\dfrac{1}{8}\)

+) \(2\left(x-1\right)=\dfrac{5}{2}+\dfrac{2}{3}=\dfrac{19}{6}\)

\(x-1=\dfrac{19}{12}\)

\(x=\dfrac{31}{12}\)

7 tháng 5 2022

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

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

\(\dfrac{1}{3}x=-\dfrac{11}{6}\)

\(x=\left(-\dfrac{11}{6}\right):\dfrac{1}{3}\)

\(x=-\dfrac{11}{2}\)

 

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

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

\(\dfrac{4}{3}x=-\dfrac{1}{6}\)

\(x=\left(-\dfrac{1}{6}\right):\dfrac{4}{3}\)

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

 

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

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

\(2\left(x-1\right)=\dfrac{19}{6}\)

\(\left(x-1\right)=\dfrac{19}{6}:2\)

\(x-1=\dfrac{19}{12}\)

\(x=\dfrac{19}{12}+1\)

\(x=\dfrac{31}{12}\)

 

a) \(x+\left(-7\right)=-20\)

\(\Rightarrow x=-20+7\)

\(\Rightarrow x=-13\)

Vậy \(x=-13\)

b) \(8-x=-12\)

\(\Rightarrow x=8-\left(-12\right)\)

\(\Rightarrow x=20\)

Vậy \(x=20\)

c) \(|x|-7=-6\)

\(\Rightarrow|x|=-6+7\)

\(\Rightarrow|x|=1\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy \(x\in\left\{1;-1\right\}\)

d) \(5^2.2^2-7.|x|=65\)

\(\Rightarrow\left(5.2\right)^2-7.|x|=65\)

\(\Rightarrow10^2-7.|x|=65\)

\(\Rightarrow100-7.|x|=65\)

\(\Rightarrow7.|x|=35\)

\(\Rightarrow|x|=5\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

Vậy \(x\in\left\{5;-5\right\}\)

e) \(37-3.|x|=2^3-4\)

\(\Rightarrow37-3.|x|=8-4\)

\(\Rightarrow37-3.|x|=4\)

\(\Rightarrow3.|x|=33\)

\(\Rightarrow|x|=11\)

\(\Rightarrow\orbr{\begin{cases}x=11\\x=-11\end{cases}}\)

Vậy \(x\in\left\{11;-11\right\}\)

f) \(|x|+|-5|=|-37|\)

\(\Rightarrow|x|+5=37\)

\(\Rightarrow|x|=32\)

\(\Rightarrow\orbr{\begin{cases}x=32\\x=-32\end{cases}}\)

Vậy \(x\in\left\{32;-32\right\}\)

g)\(5.|x+9|=40\)

\(\Rightarrow|x+9|=8\)

\(\Rightarrow\orbr{\begin{cases}x+9=8\\x+9=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-17\end{cases}}\)

Vậy \(x\in\left\{-1;-17\right\}\)

h) \(-\frac{5}{6}+\frac{8}{3}+\frac{-29}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)

\(\Rightarrow\frac{-5}{6}+\frac{16}{6}+\frac{-29}{6}\le x\le\frac{-1}{2}+\frac{4}{2}+\frac{5}{2}\)

\(\Rightarrow-3\le x\le4\)

Vậy \(-3\le x\le4\)

16 tháng 2 2021

câu a

x+(-7)=-20

x=-20-(-7)

x=-13

a)\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\frac{1}{120}.120+x:\frac{1}{3}=-4\)

\(\Rightarrow1+x:\frac{1}{3}=-4\)

\(\Rightarrow x:\frac{1}{3}=-4-1=-5\)

\(\Rightarrow x=-5.\frac{1}{3}=\frac{-5}{3}\)

b)\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)

\(\Rightarrow\frac{8}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)

\(\Rightarrow\frac{8}{5}+\frac{2}{5}.x=\frac{16}{5}\)

\(\Rightarrow\frac{2}{5}.x=\frac{16}{5}-\frac{8}{5}=\frac{8}{5}\)

\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=\frac{8}{5}.\frac{5}{2}=\frac{8}{2}=4\)

\(\Rightarrow x=4\)

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

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

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...

 

3 tháng 9 2017

a) Đặt A = 1 + 4 + 7 + ..... + 100

Số số hạng của A là : 

           (100 - 1) : 3 + 1 = 34 

Tổng A có giá trị là  :

           (100 + 1) x 34 : 2 = 1717

Thay A vào đề bài ta có : x + 1717 = 1996

=> x = 1996 - 1717

=> x = 279

b) 786 + (x + 2 + 4 + ......... + 48 + 50) = 2007

Đặt B = 2 + 4 + ......... + 50

Số số hạng của B là : 

                (50 - 2) : 2 + 1 = 25 (số)

Tổng B có giá trị là : 

               (50 + 2) x 25 : 2 = 650 

Thay B vào đề bài ta có : 786 + x + 650 = 2007

=> x + 1436 = 2007

=> x = 571 

Vậy x = 571