Lời giải:
Ta có:

Ư$(8)=\left\{\pm 1; \pm 2; \pm 4; \pm 8\right\}$

Ư$(10)=\left\{\pm 1; \pm 2; \pm 5; \pm 10\right\}$

Do đó: ƯC$(10,8)=\left\{\pm 1; \pm 2\right\}$

$\Rightarrow x\in \left\{\pm 1; \pm 2\right\}$

a) \(5\times\left(3+7\times x\right)=400\)

\(3+7\times x=80\)

\(7\times x=77\)

\(x=11\)

b) \(x\times37+x\times63=1200\)

\(x\times\left(37+63\right)=1200\)

\(x\times100=1200\)

\(x=12\)

c) \(x\times6+12:3=40\)

\(x\times6+4=40\)

\(x\times6=36\)

\(x=6\)

d) \(4+6\times\left(x+1\right)=70\)

\(6\times\left(x+1\right)=66\)

\(x+1=11\)

\(x=10\)

e) \(163:x+34:x=10\)

\(\left(163+34\right):x=10\)

\(197:x=10\)

\(x=19,7\)

X={1;2;3;4;6;12}

x ∈ ƯC { 36 ; 24 ) và x ≤ 20 

Ta có : 

36 = 2² . 3²

24 = 2³ . 3

ƯCLN ( 36 ; 24 ) = 2² . 3 = 4 . 3 = 12

ƯC ( 24 ; 36 ) = Ư ( 12 ) 

Mà Ư ( 12 ) = { 1 ; 2 ; 3 ; 4 ; 6 ;12 }

ƯC ( 24 ; 36 ) = { 1 ; 2 ; 3 ; 4 ; 6 ; 12 }

Mà x ≤ 20 nên x ∈ { 1 ; 2 ; 3 ; 4 ; 6 ; 12 }

Vậy x ∈ { 1 ; 2 ; 3 ; 4  ; 6 ; 12 }

\(a,2x+34=56\\ \Rightarrow2x=56-34\\ \Rightarrow x=22:2\\ \Rightarrow x=11\\ b,87-\left(x-654\right):3=21\\ \Rightarrow\left(x-654\right):3=87-21\\ \Rightarrow x-654=66:3\\ \Rightarrow x=22+654\\ \Rightarrow x=676\\ c,7^{65}:7^x=7^{43}.7^{21}\\ \Rightarrow7^{65-x}=7^{43+21}\\ \Rightarrow65-x=64\\ \Rightarrow x=65-64\\ \Rightarrow x=1\)

a) \(\Leftrightarrow2x+5=3^6\\ \Leftrightarrow2x+5=729\\ \Leftrightarrow x=362\)

b) \(\Leftrightarrow x+55=60\\ \Leftrightarrow x=5\)

c) \(x=\left\{12;24;36;48\right\}\)

c) x ⋮ 12 và x < 60

x ∈ B₍₁₂₎ = { 0 ; 12 ; 24 ; 36 ; 48 ; 60 ;...}

mà x ⋮ 12 và x < 60

nên x ∈ B₍₁₂₎= { 0 ; 12 ; 24 ; 36 ; 48 }

a: =>(2x-1)^3=4^12:4^10=4^2=8

=>2x-1=2

=>2x=3

=>x=3/2(loại)

b: 6x+5 chia hết cho 3x-1

=>6x-2+7 chia hết cho 3x-1

=>7 chia hết cho 3x-1

mà x là số tự nhiên

nên 3n-1=-1

=>n=0

\(\sqrt{4\left(x+1\right)}=\sqrt{8}\)

⇒4(x+1)=8

⇒x+1=2

⇒x=1

a. \(\sqrt{4\left(x+1\right)}=\sqrt{8}\)                    ĐKXĐ: \(x\ge-1\)

<=> \(\left(\sqrt{4\left(x+1\right)}\right)^2=\left(\sqrt{8}\right)^2\)

<=> 4(x + 1) = 8

<=> 4x + 4 = 8

<=> 4x = -4

<=> x = -1 (TM)

Vậy nghiệm của PT là S = \(\left\{-1\right\}\)

\(\left(-\dfrac{3}{4}x+1\right)\div\dfrac{2}{3}=1\)

\(-\dfrac{3}{4}x+1=1\times\dfrac{2}{3}\)

\(-\dfrac{3}{4}x+1=\dfrac{2}{3}\)

\(-\dfrac{3}{4}x=\dfrac{2}{3}-1\)

\(-\dfrac{3}{4}x=-\dfrac{1}{3}\)

\(x=-\dfrac{1}{3}\div\left(-\dfrac{3}{4}\right)\)

\(x=\dfrac{4}{9}\)

x+3=6

x=6-3

x=3

a: 78x(x-97)-x+97=0

=>(x-97)(78x-1)=0

=>\(\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)

b: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

=>\(x\left(x-2\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

c: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

=>\(x^2+4x+4-x^2+4=0\)

=>4x+8=0

=>x+2=0

=>x=-2

\(a,78x\left(x-97\right)-x+97=0\)

\(\Leftrightarrow78x\left(x-97\right)-\left(x-97\right)=0\)

\(\Leftrightarrow\left(x-97\right)\left(78x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-97=0\\78x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)

\(b,\dfrac{2}{3}x\left(x^2-4\right)=0\)

\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=0\\x-2=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

\(c,\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[\left(x+2\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\cdot4=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)