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\(b,x=ƯCLN\left(45,30\right)=15\\ c,x=BCNN\left(6,8\right)=24\\ d,x\in\left\{10;25;50\right\}\)
a: \(x\in\left\{25;30;35\right\}\)
b: \(x\in\left\{24;32;40;48;56;64\right\}\)
c: \(x\in\left\{3;4;6\right\}\)
1)(x+1)thuộc ước của -2
ư(2)={1;2;-1;-2}
x+1 | 1 | 2 | -1 | -2 |
x | 0 | 1 | -2 | -3 |
vậy x =0;x=1;x=-2;x=-3
2)ta có : 2x+7=2(x+3)+1
2(x+3)chia hết cho x+3
=>để 2x+7chia hết cho x+3
<=>1chia hết cho x+3
=>x+3 thuộc ư(1)
u(1)={1;-1}
x+3 | 1 | -1 |
2 | -2 | -4 |
vậy x=-2;x=-4
a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)
mà -3<x<30
nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)
b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)
mà -16<=x<20
nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)
c: \(\Leftrightarrow x-1+4⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)
d: \(\Leftrightarrow2x+4-5⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-1;-3;3;-7\right\}\)
(\(x-1,25\))(\(x-8\)) = 0
\(\left[{}\begin{matrix}x-1,25=0\\x-8=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1,25\\x=8\end{matrix}\right.\)
Vậy \(x\) \(\in\) { 1,25; 8}
Ta có : x - 1,25 = 0 hoặc x - 8 = 0
x = 0 + 1,25 x = 0 + 8
x = 1,25 x = 8
Vậy : x = 1,25 hoặc x = 8
a) 2+3𝑥=−15−19
3x= -15 - 19 -2
3x = -36
x= -12
b) 2𝑥−5=−17+12
2x = -17 + 12 + 5
2x = 0
x = 0
c) 10−𝑥−5=−5−7−11
-x = -5 - 7 - 11 - 10 + 5
-x = -28
x = 28
d) |𝑥|−3=0
|x|= 3
x = \(\pm\)3
e) (7−|𝑥|).(2𝑥−4)=0
th1 : ( 7 - | x| ) = 0
|x|= 7
x=\(\pm\)7
th2: ( 2x-4) = 0
2x = 4
x= 2
f) −10−(𝑥−5)+(3−𝑥)=−8
-10 - x + 5 + 3 - x = -8
-10 + 5 + 3 + 8 = 2x
2x= 6
x = 3
g) 10+3(𝑥−1)=10+6𝑥
10 + 3x - 3 = 10 + 6x
3x - 6x = 10 - 10 + 3
-3x = 3
x= -1
h) (𝑥+1)(𝑥−2)=0
th1: x+1= 0
x = -1
x-2=0
x=2
hok tốt!!!
1.
\(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)
\(\frac{x}{3}=-\frac{5}{6}+\frac{1}{4}\)
\(\frac{x}{3}=-\frac{7}{12}\)
\(\Rightarrow\)\(x\times12=-7\times3\)
\(\Rightarrow\)\(x\times12=-21\)
\(\Rightarrow\)\(x=-\frac{7}{4}\)
Vậy \(x=-\frac{7}{4}\)
2.
\(\frac{x+3}{15}=\frac{1}{3}\)
\(\Rightarrow\)\(\left(x+3\right)\times3=1\times15\)
\(\Rightarrow\)\(\left(x+3\right)\times3=15\)
\(\Rightarrow\)\(x+3=5\)
\(\Rightarrow\)\(x=2\)
Vậy \(x=2\)
3.
\(\frac{x-12}{4}=\frac{1}{2}\)
\(\Rightarrow\)\(\left(x-12\right)\times2=4\times1\)
\(\Rightarrow\)\(x-12=2\)
\(\Rightarrow\)\(x=14\)
Vậy \(x=14\)
x= {0 ; 5 ; 10 ; 15 ; 20 ; 25 }