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a,Ư(25)={1;5;25}
\(\Rightarrow\)2x+1\(\in\left\{1;5;25\right\}\)
\(\Rightarrow x\in\left\{0;2;12\right\}\)
bTa có : \(x\inƯC\left(60;150;210\right)\)
Ư(60)={1;2;3;4;5;6;10;12;15;20;30;60}
Ư(150)={1;2;3;5;6;10;15;25;30;50;125;150}
Ư(210)={1;2;3;5;6;7;10;15;30;35;42;70;105;210}
\(\Rightarrow x\in\left\{1;2;3;5;6;10;15;30\right\}\)
mà x>25 nên x=30
a) (5x - 1) : 3 + 1 = 4
=> (5x - 1) : 3 = 3
=> (5x - 1) = 9
=> 5x - 1 = 9
=> 5x = 10
=> x = 2
b) 54 : (16 - x) - 1=5
=> 54:(16-x) = 6
=> 16-x = 9
=> x = 7
a, x : [(1800 + 600) : 30] = 560 : (315 - 35)
<=> x : (2400 : 30) = 560 : 280
<=> x : 80 = 2
<=> x = 160
b, x - 6 : 2 - (48 - 24) : 2 : 6 - 3 = 0
<=> x - 3 - 24 : 2 : 6 - 3 = 0
<=> x - 3 - 2 - 3 = 0
<=> x - 8 = 0
<=> x = 8
c, 390 - (x - 7) = 169 : 13
<=> 390 - (x - 7) = 13
<=> x - 7 = 377
<=> x = 384
d, (x - 140) : 7 = 33 - 23.3
<=> (x - 140) : 7 = 33 - 24
<=> (x - 140) : 7 = 3
<=> x - 140 = 21
<=> x = 161
@Đỗ Thị Huyền Trang
1.Tim x:
a)| x + 1 | = 5 -> Th1: x+1=5-> x= 5-1=4
Th2: x+1=-5-> x= (-5) -1=-6(Loại. vì x lớn hơn hoặc bằng 0)
Vậy x= 4
b)| x - 3 | = 7 -> TH1: x-3=7-> x=7+3=10(Loại. Vì x<3)
TH2: x-3=-7-> x=-7+3=-4
Vậy x= -4
c) x + | 2 - x | = 6
-> | 2 - x | =6 -x
-> TH1: 2-x = 6-x
-> -x+ x= 2-6
-> 0x =-4(LOẠI)
TH2: 2-x= -6+x
->(-x)-x= 2+6
-> -2.x=8
-> x=8: -2=-4
Vậy x=-4
Tick cho mik nha!!!
2. Tìm x
a) | x | = 7-> x=-7 hoặc x=7
b) | x | < 7.Vì| x | lớn hơn hoặc bằng 0
-> | x | =(0;1;2;3;4;5;6)
-> x= (-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6)
c) | x | > 7
-> | x | =(8;9;10;11;12;13.............)
-> x= (...............;-9;-8;8;9;10;.............)
bài 2) a) \(2\left(x+1\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\) vậy \(x=-1\)
b) \(x\left(x-2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) vậy \(x=0;x=2\)
c) \(\left(x-1\right)\left(x+7\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) vậy \(x=1;x=-7\)
d) \(\left(x+2\right)\left(x^2-9\right)=0\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x^2-9=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x^2=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\end{matrix}\right.\) vậy \(x=-2;x=3;x=-3\)
e) \(x^2\left(x-5\right)+2\left(x-5\right)=0\Leftrightarrow\left(x^2+2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2=0\\x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\) vậy \(x=5\)
bài 1) \(A=48+\left(-48-174\right)+\left|-74\right|=48-48-174+74=-100\)
\(B=\left(-123\right)+77+\left(-257\right)-23-43=-123+77-257-23-43=-369\)
\(C=\left(-57\right)+\left(-159\right)+47+169=-57-159+47+169=0\)
quá hợp lí 
a) \(\left(x^2-5\right)\left(x^2-25\right)< 0\)
Vì \(x^2-5>x^2-25\) nên \(\left\{{}\begin{matrix}x^2-5>0\\x^2-25< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2>5\\x^2< 25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{5}< x< -\sqrt{5}\left(vl\right)\\-5< x< 5\end{matrix}\right.\)
b) \(\left(x+5\right)\left(9+x^2\right)< 0\)
Vì \(9+x^2>0\) nên \(x+5< 0\Leftrightarrow x< -5\)
c) \(\left(x+3\right)\left(x^2+1\right)=0\)
Vì \(x^2+1>0\) nên \(x+3=0\Leftrightarrow x=-3\)
d) \(\left(x+5\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x+2\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-2\\x=2\end{matrix}\right.\)
\(a,\)
\(3-\left|x\right|=5\)
\(\Rightarrow\left|x\right|=3-5=-2\)
\(\Rightarrow x\) không có giá trị
\(b,\)
\(\left|x+3\right|=0\)
\(\Rightarrow x+3=0\)
\(\Rightarrow x=-3\)
\(c,\)
\(\left|x-3\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
\(d,\)
\(\left|x+5\right|=-3\)
\(\Rightarrow x\) không có giá trị
a, \(\dfrac{x}{13}=\dfrac{35}{91}\Leftrightarrow\dfrac{x}{13}=\dfrac{5}{13}\Leftrightarrow x=5\)
b, \(\dfrac{9+x}{13-x}=\dfrac{5}{6}\Leftrightarrow\dfrac{9+x}{13-x}=\dfrac{10}{12}\Leftrightarrow x=1\)
@đỗ hương giang
a.
\(\dfrac{x}{13}=\dfrac{35}{91}\)
\(\Rightarrow x=\dfrac{35.13}{91}=5\)
Vậy x = 5
b.
\(\dfrac{9+x}{13-x}=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{9+x}{13-x}=\dfrac{10}{12}\)
\(\Rightarrow\dfrac{x=10-9}{13-12=x}\Rightarrow x=\dfrac{1}{1}=1\)
Vậy x =1
c.
x + x : 5 . 7,5 + x : 2.9 = 315
\(\Rightarrow x+\dfrac{x}{5}7,5+\dfrac{x}{2}9=7x\)
\(\Rightarrow7x=3^3.5\)
\(\Rightarrow7x=135\)
\(\Rightarrow x=\dfrac{135}{7}\)
Vậy x = \(\dfrac{135}{7}\)
Chúc bạn học tốt!!!
a) x = 200.
b) x = 70.
c) x = 0.
d) x = 0.
a) x = 200
b) x = 70
c) x = 0
d) x = 0