(3x-7)-(-5+x)=10
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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 ...
\(a,2-x=17-\left(-5\right)\)
\(\Rightarrow2-x=17+5\)
\(\Rightarrow2-x=22\)
\(\Rightarrow x=2-22\)
\(\Rightarrow x=-20\)
Vay...
\(b,11-\left(15+11\right)=x-\left(25-9\right)\)
\(\Rightarrow11-26=x-16\)
\(\Rightarrow-15=x-16\)
\(\Rightarrow-15+16=x\)
\(\Rightarrow x=1\)
d) \(PT\Leftrightarrow x\left(2x-7\right)-4\left(x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{7}{2};4\right\}\)
e) \(PT\Leftrightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
Vậy: \(S=\left\{7;1\right\}\)
f) \(PT\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{1;3\right\}\)
1,-12 + x = 5x - 20
-12 + 20 = 5x - x
8 = 4x
x = 2
2, 7x - 4 = 20 + 3x
7x - 3x = 20 + 4
4x = 24
x = 6
3 , 5x - 7 = -21 - 2x
5x + 2x = -21 + 7
7x = -14
x = -14 : 7 = -2
4 , x + 15 = 20 - 4x
x + 4x = 20 - 15
5x = 5
x = 5 : 5 = 1
5, 17 - x = 7 - 6x
-x + 6x = 7 - 17
5x = -10
x = -10 : 5 = -2
x-(17-8) = 5+(10-3x)
x-9=5+10-3x
x+3x=5+10+9
x+3x=24
4x=24
x=24/4
x=6
vậy x=6
\(\dfrac{3}{5}\)\(x\) - \(\dfrac{11}{5}\) = \(\dfrac{-3}{14}\) : \(\dfrac{5}{7}\)
\(\dfrac{3}{5}\)\(x\) - \(\dfrac{11}{5}\) = - \(\dfrac{3}{10}\)
\(\dfrac{3}{5}\)\(x\) = - \(\dfrac{3}{10}\) + \(\dfrac{11}{5}\)
\(\dfrac{3}{5}\)\(x\) = \(\dfrac{19}{10}\)
\(x\) = \(\dfrac{19}{10}\) : \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{19}{6}\)
\(\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{14}:\dfrac{5}{7}\)
\(\Rightarrow\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{14}\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{10}\)
\(\Rightarrow\dfrac{3}{5}x=-\dfrac{3}{10}+\dfrac{11}{5}\)
\(\Rightarrow\dfrac{3}{5}x=\dfrac{19}{10}\)
\(\Rightarrow x=\dfrac{19}{10}:\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{19}{6}\)
=> 3x-7+5-x=10
=> 2x-2 = 10
=> 2x = 10+2 = 12
=> x = 12 : 2 = 6
Vậy x = 6
Tk mk nha
(3x-7)-(-5+x)=10
3x-7+5-x=10
3x-x-7+5=10
2x-2=10
2x=10+2
2x=12x
x=12:2
x+6
Vật x=6