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\(#V\)
\(⇔ x ( 1 + 0 , 3 ) = 1 , 3\)
\(⇔ x .1 , 3 = 1 , 3\)
\(⇔ x = 1 , 3 : 1 , 3\)
\(⇔ x = 1\)
x+30%.x=-1,3
=>x+310.x=-1,3
=>x.(310+1)=-1,3
=>x.1,3=-1,3
=>x=-1,3:1,3
=>x=-1
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\(1-\left(x-1\right):3=\dfrac{2}{3}\)
\(\Rightarrow1-\left(x-1\right)=\dfrac{2}{3}.3\)
\(\Rightarrow1-\left(x-1\right)=2\)
\(\Rightarrow x-1=1-2\)
\(\Rightarrow x-1=\left(-1\right)\)
\(\Rightarrow x=\left(-1\right)+1\)
\(\Rightarrow x=0\)
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\(\dfrac{1}{2}\left(x-2\right)+\dfrac{1}{3}\left(2-x\right)=x\\ \Leftrightarrow\dfrac{1}{2}\left(x-2\right)-\dfrac{1}{3}\left(x-2\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{3-2}{6}\right)=x\\ \Leftrightarrow\left(x-2\right).\dfrac{1}{6}=x\\ \Leftrightarrow\dfrac{1}{6}x-\dfrac{1}{3}-x=0\\ \Leftrightarrow\left(\dfrac{1}{6}-1\right)x=\dfrac{1}{3}\\ \Leftrightarrow\left(\dfrac{1-6}{6}\right)x=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{-5}{6}x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}:\left(-\dfrac{5}{6}\right)\\ \Leftrightarrow x=-\dfrac{2}{5}\)
Vậy \(x=-\dfrac{2}{5}\)
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\(\Rightarrow x\left(2y+1\right)-3\left(2y+1\right)=7\)
\(\Leftrightarrow\left(x-3\right)\left(2y+1\right)=7=1.7=7.1=-1.-7=-7.-1\)
x-3 | -7 | -1 | 1 | 7 |
2y+1 | -1 | -7 | 7 | 1 |
x | -4 | 2 | 4 | 10 |
y | -1 | -4 | 3 | 0 |
vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(\Rightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(\Rightarrow A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(\Rightarrow A=\left(2+2^2\right)\left(1+2^2+...+2^{98}\right)\)
\(\Rightarrow A=6\left(1+2^2+...+2^{98}\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
|x+3|+|y-1|=0
\(\Rightarrow\hept{\begin{cases}\left|x+3\right|\ge0\\\left|y-1\right|\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x+3=0\\y-1=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
Vậy x=-3 ; y=1
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\frac{1}{x}=\frac{y}{-5}\)
\(\Leftrightarrow x\cdot y=1\cdot\left(-5\right)=-5\)
Mà x,y thuộc Z
\(\Rightarrow x\inƯ\left(-5\right)=\left\{-5;-1;1;5\right\}\)
Lập bảng
x | -5 | -1 | 1 | 5 |
y | -1 | -5 | 5 | 1 |
KL | c | c | c | c |
Vậy (x;y)=(-5;-1);(-1;-5);(1;5);(5;1)
\(x\left(x-2\right)-1\left(x-2\right)=2\)
\(x^2-2x-x+2-2=0\)
\(x^2-3x=0\)
\(x\left(x-3\right)=0\)
\(Th1:x=0\)
\(Th2:x-3=0=>x=3\)
Vậy\(x\in\left\{0;3\right\}\)
Toán 8 mà
(x-1).(x-2)=2
x2-2x-x+2=2
x2-3x=0
x(x-3)=0
TH1: x=0
TH2: x-3=0 =>x=3