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\(x-\dfrac{1}{4}=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}+\dfrac{1}{4}\)
\(x=\dfrac{-42}{20}+\dfrac{5}{20}=\dfrac{-37}{20}\)
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\(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)^2=8.2\)
=>(x+1)2=16
=>x+1=\(\sqrt{16}\)
=>x+1=4
=>x=3
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\(x+\left(x+1\right)+\left(x+2\right)+...+19+20=20\)
\(\Leftrightarrow x+\left(x+1\right)+\left(x+2\right)+...+19=0\)
\(\Leftrightarrow\frac{x\left(19+x\right)}{2}=0\)
\(\Leftrightarrow x\left(x+19\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+19=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-19\end{cases}}\)
Mà \(x\ne0\)nên \(x=-19\)
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\(\Leftrightarrow x-\left[3-x+3+x-2\right]=0\)
=>x=4
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Ta có: x(1+30%)=-1,31
⇒ \(\dfrac{13}{10}\)x =-1,31
⇒ x =-1,31 : \(\dfrac{13}{10}\)
⇒ x = -\(\dfrac{131}{130}\)
x + 30% x = -1,31
\(x+\dfrac{30}{100}x=-1,31\)
x+0,3x=-1,31
x.(0,3+1)=-1,31
x.1,3=-1,31
\(x=\dfrac{-1,31}{1,3}\)
\(x=-\dfrac{131}{130}\)
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\(x^{15}=x\)
\(\Rightarrow x^{15}-x=0\)
\(\Rightarrow x.\left(x^{14}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^{14}=0+1=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x}{5}=\dfrac{5}{6}-\dfrac{19}{30}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{25}{30}-\dfrac{19}{30}=\dfrac{6}{30}=\dfrac{1}{5}\)
\(\Leftrightarrow x=1\)
Ta có : \(\dfrac{x}{5}=\dfrac{5}{6}+\left(-\dfrac{19}{30}\right)=\dfrac{25}{30}-\dfrac{19}{30}=\dfrac{1}{5}\)
\(\Leftrightarrow5x=5\)
\(\Leftrightarrow x=1\)
Vậy ...
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\(x\left(y-1\right)-y=2\)
\(\Rightarrow x\left(y-1\right)-\left(y-1\right)=3\)
\(\Rightarrow\left(x-1\right)\left(y-1\right)=3\)
Ta xét bảng sau
x-1 | 1 | 3 | -1 | -3 |
y-1 | 3 | 1 | -3 | -1 |
x | 2 | 4 | 0 | -2 |
y | 4 | 2 | -2 | 0 |
Vậy \(x\in\left\{2;4;0;-2\right\}\)
x = 10,5