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\(1)|5-2x|=|x+4|\)
\(\Leftrightarrow\orbr{\begin{cases}5-2x=x+4\\5-2x=-x-4\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x-x=4-5\\-2x+x=-4-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}-3x=-1\\-x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=9\end{cases}}}\)
Vậy \(x=\frac{1}{3};x=9\)
\(2)|x-1|=|2x+5|\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2x+5\\x-1=-2x-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2x=5+1\\x+2x=-5+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}-x=4\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=-\frac{4}{3}\end{cases}}}\)
Vậy \(x=-4;x=-\frac{4}{3}\)
\(3)|x+1|+|x+2|+|x+3|=0\left(1\right)\)
Ta có: \(|x+1|\ge0\forall x;|x+2|\ge0\forall x;|x+3|\ge0\forall x\)
\(\Leftrightarrow|x+1|+|x+2|+|x+3|\ge0\forall x\)
\(\left(1\right)\Leftrightarrow|x+1|+|x+2|+|x+3|=0\)
\(\Leftrightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=0\)
\(\Leftrightarrow x+1+x+2+x+3=0\)
\(\Leftrightarrow\left(x+x+x\right)+\left(1+2+3\right)=0\)
\(\Leftrightarrow3x+6=0\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-6:3\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
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a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)
b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)
c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)
d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)
e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)
g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)
h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)
i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)
k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)
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a) Suy ra : \(x\in BCNN\left(14,21\right)\)
14 = 2.7
21 = 3.7
=> BCNN (14,21)=7.2.3=42
Vậy x = 42
b) 200-(2x+6)=4^3
=>200-2x-6=64
=>194-2x=64
=>2x=194-64=130
=>x=65
a) x ⋮ 14 và x ⋮ 21
Mà x ≠ 0 và x là số nhỏ nhất nên:
\(x\in BCNN\left(14,21\right)\)
Ta có:
\(14=2\cdot7\)
\(21=3\cdot7\)
\(\Rightarrow x=BCNN\left(14,21\right)=7\cdot2\cdot3=42\)
b) \(200-\left(2x+6\right)=4^3\)
\(\Rightarrow2x+6=200-4^3\)
\(\Rightarrow2x+6=200-64\)
\(\Rightarrow2x+6=136\)
\(\Rightarrow2x=136-6\)
\(\Rightarrow2x=130\)
\(\Rightarrow x=130:2\)
\(\Rightarrow x=65\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 150 : [17 - (2x - 3)] = 350
=> 17 - (2x - 3) = \(\frac{150}{350}=\frac{3}{7}\)
=> 2x - 3 = \(17-\frac{3}{7}=\frac{116}{7}\)
=> 2x = \(\frac{116}{7}+3\)
=> 2x = \(\frac{137}{7}\)
=> x = \(\frac{137}{7}:2=\frac{137}{14}\)
b) 4 . [105 - (x - 9)] - 486 = 0
=> 4. [105 - (x - 9)] = 486
=> 105 - (x - 9) = 243/2
=> x - 9 = \(105-\frac{243}{2}=-\frac{33}{2}\)
=> x = \(-\frac{33}{2}+9=-\frac{15}{2}\)
c) 1000 - 2{600 - 4.[198 - 7.(21 - 7)]}.x = 200
=> 1000 - 2{600 - 4.[198 - 7.14]}.x = 200
=> 1000 - 2{600 - 4.[198 - 98].x = 200
=> 1000 - 2{600 - 4.100}.x = 200
=> 1000 - 2{600 -400}.x = 200
=> 1000 - 2.200.x = 200
=> 2.200.x = 800
=> 400.x = 800
=> x = 2
Câu 2 :
a) 350 : [19 - (7x - 3)] = 350
=> 19 - (7x - 3) = 1
=> 7x - 3 = 18
=> 7x = 21
=> x = 3
b) 7.[109 - (x - 9)] - 420 = 0
=> 7.[109 - (x - 9)] = 420
=> 109 - (x - 9) = 60
=> x - 9 = 49
=> x = 49 + 9 = 58
c) 254 - 2{200 - 4.[90 - 7.(19 - 9)]}.x = 54
=> 254 - 2{200 - 4.[90 - 7.10]}.x = 54
=> 254 - 2{200 - 4[90 - 70]} . x= 54
=> 254 - 2{200 - 4.20} . x = 54
=> 254 - 2{200 - 80}.x = 54
=> 254 - 2.120 . x= 54
=> 2.120.x = 200
=> 240.x = 200
=> \(x=\frac{5}{6}\)
Ta có :\(\left(2x-4\right)^{200}+\left|y-3\right|=0\)
Lại có : \(\left(2x-4\right)^{200}\ge0;\left|y-3\right|\ge0=>\left(2x-4\right)^{200}+\left|y-3\right|\ge0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}2x-4=0\\y-3=0\end{cases}=>\hept{\begin{cases}x=2\\y=3\end{cases}}}\)
Ta có : \((2x-4)^{200}\ge0\forall x\)
\(\left|y-3\right|\ge0\forall y\)
\(\Leftrightarrow(2x-4)^{200}+\left|y-3\right|\ge0\forall x,y\)
Dấu " =" xảy ra khi : \(\hept{\begin{cases}2x-4=0\\\left|y-3\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-4=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)
Vậy:...