1, (\(x\)+1)+(\(x\)+4)+(\(x\)+7)+...+(\(x\)+28)=155
2, (\(x\)-2):37=(1008+1) x \(\frac{2}{37}\)
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\(\left(x-2\right)\div37=\left(1008+1\right)\times\frac{2}{37}\)
\(\left(x-2\right)\div37=1009\times\frac{2}{37}\)
\(\left(x-2\right)\div37=\frac{2018}{37}\)
\(x-2=\frac{2018}{37}\times37\)
\(x-2=2018\)
\(x=2018+2\)
\(x=2020\)
+) \(\dfrac{1}{3}x=-\dfrac{4}{3}-\dfrac{1}{2}=-\dfrac{11}{6}\)
\(x=-\dfrac{11}{6}:\dfrac{1}{3}=-\dfrac{11}{2}\)
+) \(\dfrac{4}{3}x=-\dfrac{2}{3}+\dfrac{1}{2}=-\dfrac{1}{6}\)
\(x=-\dfrac{1}{6}:\dfrac{4}{3}=-\dfrac{1}{8}\)
+) \(2\left(x-1\right)=\dfrac{5}{2}+\dfrac{2}{3}=\dfrac{19}{6}\)
\(x-1=\dfrac{19}{12}\)
\(x=\dfrac{31}{12}\)
\(\dfrac{1}{3}x+\dfrac{1}{2}=-\dfrac{4}{3}\)
\(\dfrac{1}{3}x=\left(-\dfrac{4}{3}\right)-\dfrac{1}{2}\)
\(\dfrac{1}{3}x=-\dfrac{11}{6}\)
\(x=\left(-\dfrac{11}{6}\right):\dfrac{1}{3}\)
\(x=-\dfrac{11}{2}\)
\(-\dfrac{2}{3}-\dfrac{4}{3}x=-\dfrac{1}{2}\)
\(\dfrac{4}{3}x=\left(-\dfrac{2}{3}\right)-\dfrac{-1}{2}\)
\(\dfrac{4}{3}x=-\dfrac{1}{6}\)
\(x=\left(-\dfrac{1}{6}\right):\dfrac{4}{3}\)
\(x=-\dfrac{1}{8}\)
\(\dfrac{5}{2}-2\left(x-1\right)=-\dfrac{2}{3}\)
\(2\left(x-1\right)=\dfrac{5}{2}-\left(-\dfrac{2}{3}\right)\)
\(2\left(x-1\right)=\dfrac{19}{6}\)
\(\left(x-1\right)=\dfrac{19}{6}:2\)
\(x-1=\dfrac{19}{12}\)
\(x=\dfrac{19}{12}+1\)
\(x=\dfrac{31}{12}\)
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 \(-3\le x\le4\)
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 ...
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
(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_