a) 

B(14) = 0; 14; 28; 42; 56; 70; 84; …..

Vì 20 < x < 80 => x ∈ { 28; 42; 56; 70．}

b)

Vì 70 chia hết cho x ѵà 80 chia hết cho x => x ∈ ƯC(70; 80)

Phân tích:

70 = 2 ．5 ．7

80 = 24 ．5

ƯCLN (70; 80) = 2.5=10

ƯC ( 70; 80) = Ư(10) ={1;2;5;10}

Mà x > 8 => x = 10

c)

Vì 126 chia hết cho x ѵà 210 chia hết cho x => x ∈ ƯC(126; 210)

Phân tích

126 = 2 ．3² ．7

210 = 2 ．3 ．5 ．7

ƯCLN(126; 210) = 2 ．3 ．7 = 42

ƯC(126; 210) = { 1; 2; 3; 6; 7; 14; 21; 42 }

Vì 15 < x < 30 => x = 21

a) \(B\left(14\right)=\left\{0;14;28;42;56;70;84;.......\right\}\)

Vì \(20< x< 8\Rightarrow x\in\left\{28;42;56;70\right\}\)

b) Vì 70 chia hết cho x và 80 chia hết cho x nên \(\Rightarrow x\inƯC\left(70;80\right)\)

Phân tích :

\(70=2.5.7\)

\(80=2^4.5\)

Mà \(x>8\Rightarrow x=10\)

c) Vì 126 chia hết cho x  và 210 chia hết cho x nên \(\Rightarrow x\inƯC\left(126;210\right)\)

Phân tích :

\(126=2.3^2.7\)

\(210=2.3.5.7\)

\(ƯC\left(126;210\right)=\left\{1;2;3;6;7;14;21;42\right\}\)

\(ƯCLN\left(126;210\right)=2.3.7=42\)

Theo đề : \(x\in\left(>15< 30\right)\Rightarrow x=21\)

`a)`

`(x - 14) - 20 = 0`

`=> x - 14 = 0 + 20`

`=> x - 14 = 20`

`=> x = 20 + 14`

`=> x = 34`

Vậy, `x = 34.`

Đáp án cần chọn là: B

Đáp án là B

Ta có:Ư(60) = {1; 2; 3; 4; 5; 6; 10; 12; 15; 20; 30; 60}

x ∈ Ư(60) ⇒ x ∈ {1; 2; 3; 4; 5; 6; 10; 12; 15; 20; 30; 60}

Vì x > 20 ⇒ x ∈ {30; 60}

Đáp án: B

Tìm x thuộc ước của 60 và x > 20

Ư(60) = {1;2;3;4;5;6;10;12;15;20;30;60}

Mà x > 20 nên x ∈ {30; 60}

\(a>\)\(\left(x+2\right)\) thuộc \(Ư\left(20\right)\)

\(\left(x+1\right)\inƯ\left(20\right)=\left\{1;2;4;5;10;20\right\}\)

\(+>x+1=1\)

\(\Rightarrow x=0\)

\(+>x+1=2\)

\(\Rightarrow x=1\)

\(+>x+1=4\)

\(\Rightarrow x=3\)

\(+>x+1=5\)

\(\Rightarrow x=4\)

\(+>x+1=10\)

\(\Rightarrow x=9\)

\(+>x+1=20\)

\(\Rightarrow x=19\)

Vậy \(x\in\left\{0;1;3;4;9;19\right\}\)

\(b>\left(x-2\right)\) là ước của 6

\(\left(x-2\right)\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)

\(+>x-2=1\)

\(\Rightarrow x=3\)

\(+>x-2=2\)

\(\Rightarrow x=4\)

\(+>x-2=3\)

\(\Rightarrow x=5\)

\(+>x-2=6\)

\(\Rightarrow x=8\)

Vậy \(x\in\left\{3;4;5;8\right\}\)

\(c>\left(2x+3\right)\) là \(Ư\left(10\right)\)

\(\left(2x+3\right)\inƯ\left(10\right)=\left\{1;2;5;10\right\}\)

\(+>2x+3=1\)

\(\Rightarrow x=-1\)

\(+>2x+3=2\)

\(\Rightarrow x=-\dfrac{1}{2}\)

\(+>2x+3=5\)

\(\Rightarrow x=1\)

\(+>2x+3=10\)

\(\Rightarrow x=\dfrac{7}{2}\)

Vậy \(x\in\left\{-1;-\dfrac{1}{2};1;\dfrac{7}{2}\right\}\)

a,x\(\in\){12;24;36;48;56}

b,x\(\in\){1;2;4}

c,x\(\in\){1;2;3;4;6;12}

a: =>3^x=3^4*3=3^5

=>x=5

b: =>\(2^{x+1}=2^5\)

=>x+1=5

=>x=4

c: \(\Leftrightarrow3^{x+2-3}=3\)

=>x-1=1

=>x=2

d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)

=>x=4 hoặc x=-4

e: (2x-1)^4=81

=>2x-1=3 hoặc 2x-1=-3

=>2x=4 hoặc 2x=-2

=>x=-1 hoặc x=2

f: (2x-6)^4=0

=>2x-6=0

=>x-3=0

=>x=3

a) \(3^x=81\cdot3\)

\(\Rightarrow3^x=3^4\cdot3\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

b) \(2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)

c) \(3^{x+2}:27=3\)

\(\Rightarrow3^{x+2}:3^3=3\)

\(\Rightarrow3^{x+2-3}=3\)

\(\Rightarrow3^{x-1}=3\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x=2\)

d) \(2x^2=32\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

e) \(\left(2x-1\right)^4=81\)

\(\Rightarrow\left(2x-1\right)^4=3^4\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

f)  \(\left(2x-6\right)^4=0\)

\(\Rightarrow2x-6=0\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=6:2\)

\(\Rightarrow x=3\)

\(a,3^x=81\cdot3\\ \Leftrightarrow3^x=3^4\cdot3\\ \Leftrightarrow3^x=3^5\\ \Leftrightarrow x=5\\ d,2^{x+1}=32\\ \Leftrightarrow x+1=5\\ \Leftrightarrow x=4\\ c,3^{x+2}:27=3\\ \Leftrightarrow3^{x+2}:3^3=3\\ \Leftrightarrow3^{x-1}=3\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\\ d,2x^2=32\\ \Leftrightarrow x^2=16\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ e,\left(2x-1\right)^4=81\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ f,\left(2x-6\right)^4=0\\ \Leftrightarrow2x-6=0\\ \Leftrightarrow x=3\)

x thuộc { 0 ; 12 ; 24 ; 36 ; 48 ; 52 }

b) x thuộc { 1 ; 2 ; 4 }

c) x thuộc { 1 ; 2 ; 4 ; 6 ; 12 }

a;x thuoc -60;-48;-36;-24;-12;0;12;24;36;48;60

b;x thuoc:-4;-2;-1;1;2;4

c;x thuoc:-12;-6;-4;-3;-2;-1;1;2;3;4;6;12