1 x 0= ?
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a) Ta có: \(\left|x-3\right|+\left|y-2x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x=2\cdot3=6\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)x^2-2x-1=0
<=> (x^2-2x+1)-2=0
<=>(x-1)2 =2
=>x-1 = \(\pm\sqrt{2}\)
=> x= \(\pm\sqrt{2}\) +1
2) x^2-x-1=0
<=> (x^2-x+1/4) -5/4=0
<=>(x+1/2)2= 5/4
=> x+1/2 = \(\pm\sqrt{\dfrac{5}{4}}\)
=>x=\(\pm\sqrt{\dfrac{5}{4}}\) - 1/2
3)x^2+x-3=0
<=> (x^2 + x + 1/4) -13/4=0
<=>(x+1/2)2 = 13/4
=> x+1/2 = \(\sqrt{\dfrac{13}{4}}\)
=> x= \(\sqrt{\dfrac{13}{4}}\) -1/2
4) 4x^2-4x-1=0
<=> (4x^2-4x+1)-2=0
<=>(2x-1)2 -2=0
<=> (2x-1)2 - \(\left(\sqrt{2}\right)^2\) =0
<=> (2x-1 - \(\sqrt{2}\) ) . (2x-1 +\(\sqrt{2}\) )=0
=> 2x-1-\(\sqrt{2}\) =0 hoặc 2x-1+\(\sqrt{2}\) =0
=> 2x= 1+\(\sqrt{2}\) hoặc 2x= 1 - \(\sqrt{2}\)
=> x=\(\dfrac{1+\sqrt{2}}{2}\) hoặc x=\(\dfrac{1-\sqrt{2}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 4x(x+1)=8(x+1)
<=>4x(x+1)-8(x+1)=0
<=>(4x-8)(x+1)=0
<=>\(\left[\begin{array}{} 4x-8=0\\ x+1=0 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=2\\ x=-1 \end{array} \right.\)
Vậy...
b)x(x-1)-2(1-x)=0
<=>(x+2)(x-1)=0
<=>\(\left[\begin{array}{} x+2=0\\ x-1=0 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=-2\\ x=1 \end{array} \right.\)
Vậy...
c)5x(x-2)-(2-x)=0
<=>(5x+1)(x-2)=0
<=>\(\left[\begin{array}{} 5x+1=0\\ x-2 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=-1/5\\ x=2 \end{array} \right.\)
d)5x(x-200)-x+200=0
<=>(5x-1)(x-200)=0
<=>\(\left[\begin{array}{} 5x-1=0\\ x-200=0 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=1/5\\ x=200 \end{array} \right.\)
e)\(x^3+4x=0 \)
\(\Leftrightarrow x(x^2+4)=0 \)
\(\Leftrightarrow \left[\begin{array}{} x=0\\ x^2+4=0 (loại vì x^2+4>=0 với mọi x) \end{array} \right.\)
Vậy x=0
f)\((x+1)=(x+1)^2\)
\(\Leftrightarrow (x+1)-(x+1)^2=0\)
\(\Leftrightarrow (x+1)(1-x-1)=0\)
\(\Leftrightarrow (x+1)(-x)=0\)
\(\Leftrightarrow \left[\begin{array}{} x=-1\\ x=0 \end{array} \right.\)
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(3x\left(x-1\right)+x-1=0\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\Leftrightarrow\hept{\begin{cases}x-1=0\\3x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)
\(S=\left\{1;\frac{1}{3}\right\}\)
b)\(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\Leftrightarrow\hept{\begin{cases}2-x=0\\x+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=-3\end{cases}}}\)
\(S=\left\{2;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Vậy tập nghiệm \(S=\left\{-2;3\right\}\)
2) \(\left(2x+3\right)\left(-x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=7\end{matrix}\right.\)
Vậy...
3) \(\left(x-1\right)\left(x+5\right)\left(-3x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\\-3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy...
4) \(x\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Do x ∈ Z và 0 < x < 3
⇒ x ∈ {1; 2}
2) Do x ∈ Z và 0 < x ≤ 3
⇒ x ∈ {1; 2; 3}
3) Do x ∈ Z và -1 < x ≤ 4
⇒ x ∈ {0; 1; 2; 3; 4}
1x0=0
1 x 0 = 0
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