\(2\cdot3^x=10\cdot3^{12}+7\cdot27^4\)

\(\Rightarrow2\cdot3^x=10\cdot3^{12}+7\cdot3^{12}\)

\(\Rightarrow2\cdot3^x=3^{12}\cdot\left(10+7\right)\)

\(\Rightarrow2\cdot3^x=3^{12}\cdot17\)

Xem lại đề

\(2\times3^x=10\times3^{12}+7\times27^4\)

\(2\times3^x=10\times531441+7\times531441\)

\(2\times3^x=531441\times\left(10+7\right)\)

\(2\times3^x=531441\times17\)

\(2\times3^x=9034497\)

\(3^x=4517248,5\)

\(3^x\sim3^{13,948}\)

\(x\sim13,948\)

Đặt \(ƯCLN\left(5m+1,4m+1\right)=d\) (với \(d\inℕ^∗\))

\(\Rightarrow\left\{{}\begin{matrix}5m+1⋮d\\4m+1⋮d\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}4\left(5m+1\right)⋮d\\5\left(4m+1\right)⋮d\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}20m+4⋮d\\20m+5⋮d\end{matrix}\right.\) 

\(\Rightarrow\left(20m+5\right)-\left(20m+4\right)⋮d\) 

\(\Rightarrow1⋮d\)

\(\Rightarrow d=1\)

 Vậy \(ƯCLN\left(5m+1,4m+1\right)=1\), suy ra \(5m+1\) và \(4m+1\) là 2 số nguyên tố cùng nhau.

   Gọi ƯCLN(5m+1,4m+1) là d \(\left(d\ne0\right)\) 

=> \(5m+1⋮d;4m+1⋮d\) 

=> \(4\left(5m+1\right)⋮d;5\left(4m+1\right)⋮d\) 

=> \(20m+4⋮d;20m+5⋮d\) 

=> \(\left(20m+5\right)-\left(20m+4\right)⋮d\) 

=> \(1⋮d\) 

=> \(d=1\) 

Vậy 5m +1 và 4m +1 là hai số nguyên tố cùng nhau

\(A=5+5^2+...+5^{30}\)

\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)\)

\(A=\left(5+25\right)+5\cdot\left(5+25\right)+...+5^{28}\cdot\left(5+25\right)\)

\(A=30+5\cdot30+...+5^{28}\cdot30\)

\(A=30\cdot\left(1+5+...+5^{28}\right)\)

Vậy A chia hết cho 30

\(A=5+5^2+....+5^{30}\)

\(A=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{28}+5^{29}+5^{30}\right)\)

\(A=5\cdot\left(1+5+25\right)+5^4\cdot\left(1+5+25\right)+...+5^{28}\cdot\left(1+5+25\right)\)

\(A=5\cdot31+5^4\cdot31+...+5^{28}\cdot31\)

\(A=31\cdot\left(5+5^4+...+5^{28}\right)\)

Vậy A chia hết cho 31

X là ước của 32

 \(Ư\left(32\right)=\left\{1;2;4;8;16;32\right\}\)

Mà: \(10\le X\le25\)

\(\Rightarrow X\in\left\{16\right\}\)

Ước của 32 = {1;2;4;8;16;32}

\(\Rightarrow x=16\)

X chia hết cho 25 vậy X là \(B\left(25\right)\)

\(\Rightarrow X\in B\left(25\right)=\left\{0;25;50;75;100;125;150;...\right\}\)

Mà: X < 100 

Nên:

\(\Rightarrow X\in\left\{0;25;50;75\right\}\)

X chia hết cho 25 vậy X là $B\left(25\right)$

$\Rightarrow X\in B\left(25\right)=\left\{0;25;50;75;100;125;150;...\right\}$

Mà: X < 100  Nên:⇒X∈{0;25;50;75}

Vậy X ∈ {0;25;50;75}

⇒�∈{0;25;50;75}⇒X∈{0;25;50;75}

Ta có 38 chia hết cho X nên X là \(Ư\left(38\right)\)

\(X\inƯ\left(38\right)=\left\{1;2;19;38\right\}\) 

Mà: X > 20

Nên:

\(X\in\left\{38\right\}\)

=>x=38.

3 = 3

4 = 2²

10 = 2.5

BCNN(3; 4; 10) = 2².3.5 = 60

Các bội nhỏ hơn 200 của 3; 4; 10 là:

BC(3; 4; 10) = {0; 60; 120; 180}

\(BCNN\left(3;4;10\right)=30\)

Mà Bội số của 3;4;10 nhỏ hơn 200

\(\Rightarrow BC\left(3;4;10\right)=\left\{0;30;60;90;120;150;180\right\}\)

\(\left(x+32\right)-17=52:4\)

\(\left(x+32\right)-17=13\)

\(\left(x+32\right)=13+17\)

\(x+32=30\)

\(x=30-32\)

\(x=-2\)

Vậy x=\(-2\)

(χ+32)-17 =52:4

(x+32)-17=13

(x+32)=13+17

x+32=30

x=30-32

x=−2

Vậy x=$-2$

315+[125-x]=435

125-x=435-315

125-x=120

x=125-120

x=5 

\(315+\left(125-x\right)=435\)

\(125-x=435-315\)

\(125-x=120\)

\(x=125-120\)

\(x=5\)

= 12+400-1500

=412-1500

= -1088

14/30+8/50--15/100=7/15+4/25+3/20=140/300+48/300+45/300

=233/300.