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\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5< 0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5>0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 5\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x>5\\x< -2\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 5\\ \Rightarrow x\in\left\{-1;0;1;2;3;4\right\}\\ b,\Rightarrow5< x^2< 14\\ \Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
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a) (Tớ đọc đề không hiểu, không nhớ :v)
b) a. `2(x-5) = 0`
`x-5 = 0:2`
`x-5 = 0`
`x=0+5`
`x=5`
b) `12(x-35) = 0`
`x-35=0:12`
`x-35 =0`
`x=0+35`
`x=35`
c) `(x-10).(x-13) = 0`
`=> x-10=0`
`x-13=0`
`=> x=0+10`
`x=0+13`
`=>x=10`
`x=13`.
a) Tích 2 số bằng không khi 1 trong 2 số bằng 0
b) \(2\left(x-5\right)=0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
\(12\left(x-35\right)=0\)
\(\Rightarrow x-35=0\)
\(\Rightarrow x=35\)
\(\left(x-10\right)\left(x-13\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-10=0\\x-13=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=13\end{matrix}\right.\)
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a)
\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+98\right)=0\)
\(x+x+2+x+4+...+x+98=0\)
\(50x+\left(98+2\right).\left[\left(98-2\right):2+1\right]:2=0\)
\(50x+100.49:2=0\)
\(50x+49.50=0\)
\(50x=0-49.50\)
\(50x=-2450\)
\(x=-2450:50\)
\(x=-49\)
b)
\(\left(x-5\right)+\left(x-4\right)+\left(x-3\right)+...+\left(x+11\right)+\left(x+12\right)=99\)
\(x+x+x+...+x-5-4-3-...+11+12=99\)
\(18x+6+7\text{+ 8 + 9 + 10 + 11 + 12 = 99}\)
\(18x+63=99\)
\(18x=99-63\)
\(18x=36\)
\(x=36:18\)
\(x=2\)
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a) \(6x-5=613\)
\(\Leftrightarrow6x=618\)
\(\Leftrightarrow x=103\)
b) \(74\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
\(a,\Rightarrow6x=613+5=618\\ \Rightarrow x=103\\ b,\Rightarrow x-3=0\Rightarrow x=3\)
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a)
\(\begin{array}{l}\left( {9x - {2^3}} \right):5 = 2\\9x - {2^3} = 2.5\\9x - 8 = 10\\9x = 18\\x = 2\end{array}\)
Vậy \(x = 2\)
b)
\(\begin{array}{l}\left[ {{3^4} - \left( {{8^2} + 14} \right):13} \right]x = {5^3} + {10^2}\\\left[ {81 - \left( {64 + 14} \right):13} \right]x = 125 + 100\\\left[ {81 - 78:13} \right]x = 125 + 100\\\left[ {81 - 6} \right]x = 225\\75x = 225\\x = 3\end{array}\)
Vậy \(x = 3\)
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a) (x+5)(x-4)=0
<=> x+5=0 hoặc x-4=0
<=> x=-5 hoặc x=4
b) (x-1)(x-3)=0
<=> x-1=0 hoặc x-3=0
<=> x=1 hoặc x=3
a) (x+5).(9x-4)=0
=> x+5=0 hoặc 9x-4=0
Nếu x+5=0: x=0-5=-5
Nếu 9x-4=0: 9x=0+4=4
x=4/9
b) (x-1).(x-3)=0
=> x-1=0 hoặc x-3=0
Nếu x-1=0: x=0+1=1
Nếu x-3=0: x=0+3=3
c) (3-x).(x-3)=0
=> 3-x=0 hoặc x-3=0
Nếu 3-x=0: x=3-0=0
Nếu x-3=0: x=0+3=3
d) x.(x+1)=0
=> x=0 hoặc x+1=0
Nếu x+1=0: x=0-1=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
a ) \(3x^2-9x=0\)
\(\Leftrightarrow3x\left(x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x=0\\x-3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=3\end{array}\right.\)
Vậy \(x\in\left\{0;3\right\}\)
b ) \(\left(x-3\right)\left(x-5\right)=0\)
\(\Rightarrow x-3\) và \(x-5\) trái dấu
Mà \(x-3>x-5\) nên \(x-3>0\) và \(x-5< 0\)
\(\begin{cases}x-3>0\Rightarrow x>3\\x-5< 0\Rightarrow x< 5\end{cases}\)
\(\Rightarrow3< x< 5;x\) nguyên \(\Rightarrow x=4\).
Vậy \(x=4\)