a,

15^12=(3*5)^12=3^12*5^12

81^3*125^5=(3^4)^3*(5^3)^5=3^12*5^15

Vì 12<15 suy ra 5^12<5^15

Suy ra 3^12*5^12<3^12*5^15

\(a.81^3.125^5=\left(3^4\right)^3.\left(5^3\right)^5=3^{12}.5^{15}=3^{12}.5^{12}.5^3=\left(3.5\right)^{12}.5^3=15^{12}.5^3>15^{12}\)

\(b.4^{20}.81^{12}=\left(2^2\right)^{20}.\left(9^2\right)^{12}=2^{40}.9^{24}=2^{20}.2^{20}.9^{20}.9^4=\left(2.9\right)^{20}.2^{20}.9^4=18^{20}.2^{20}.9^4>18^{20}\)

\(c.73^{75}=\left(73^3\right)^{25}=389017^{25}\)

\(107^{50}=107^{2.50}=\left(107^2\right)^{25}=11449^{25}\)

Vì \(389017^{25}>11449^{25}\Rightarrow73^{75}>107^{50}\)

x - 12/81 = 50/75

x             = 50/75 + 12/81

x             = 22/27

\(\dfrac{2}{3}\)= x - \(\dfrac{4}{27}\)

\(\dfrac{18}{27}\) = x - \(\dfrac{4}{27}\)

x = \(\dfrac{18}{27}\) + \(\dfrac{4}{27}\)

x = \(\dfrac{22}{27}\)

Ta có: \(\dfrac{50}{75}=\dfrac{x-12}{81}\)

\(\Leftrightarrow x-12=54\)

hay x=66

Bài 1:

a. $=(-25)(-4)(-35)=100(-35)=-3500$

b. $=16-10=6$

c. $=180-(-16)-(-36)=180+16+36=232$

d. $=250-200:[1(-3)^2+(-8)]$

$=250-200:(9-8)=250-200=50$

2.

$60+2(12-x)=-48$

$2(12-x)=60-(-48)=60+48=108$

$12-x=108:2=54$

$x=12-54=-42$
 

Bài 3:

$A=(7+7^2)+(7^3+7^4)+....+(7^{17}+7^{18})$

$=7(1+7)+7^3(7+1)+...+7^{17}(1+7)$

$=(1+7)(7+7^3+....+7^{17})=8(7+7^3+....+7^{17})\vdots 8$

b.

$A=(7+7^2+7^3)+(7^4+7^5+7^6)+...+(7^{16}+7^{17}+7^{18})$

$=7(1+7+7^2)+7^4(1+7+7^2)+....+7^{16}(1+7+7^2)$

$=(1+7+7^2)(7+7^4+....+7^{16})$

$=57(7+7^4+...+7^{16})\vdots 57$

c.

$A=(7+7^2+7^3+7^4)+(7^5+7^6+7^7+7^8)+(7^9+7^{10}+7^{11}+7^{12})+(7^{13}+7^{14}+7^{15}+7^{16})+(7^{17}+7^{18})$
$=7(1+7+7^2+7^3)+7^5(1+7+7^2+7^3)+7^9(1+7+7^2+7^3(+7^{13}(1+7+7^2+7^3)+7^{17}(1+7)$

$=(1+7+7^2+7^3)(7+7^5+7^9+7^{13})+8.7^{17}$

$=400(7+7^5+7^9+7^{13})+8.7^{17}$

Ta thấy $400(7+7^5+7^9+7^{13})\vdots 50$ nhưng $8.7^{17}\not\vdots 50$ nên $A\not\vdots 50$

Bài 1:

a. $=(-25)(-4)(-35)=100(-35)=-3500$

b. $=16-10=6$

c. $180-(-16)-(-36)=180+16+36=196+36=232$

d. $=250-200:[2000.(-3).2-6]$

$=250-200:[2000.(-6)+(-6)]$

$=250-200:[(-6)(2000+1)]=250-200[(-6).2001]$

$=250+200.6.2001=250+2401200=2401450$

Bài 2:

$60+2(12-x)=-48$

$2(12-x)=-48-60=-108$
$12-x=-108:2=-54$

$x=12-(-54)=66$

Bài 3:

Số số hạng của A: $(718-71):1+1=648$

$A=(718+71)\times 648:2=789\times 324$

a. Có: $A=789\times 324=789\times 81\times 4$

$\Rightarrow A\not\vdots 8$ (bạn xem lại đề)

b. $A=789\times 81\times 4\not\vdots 57$ (bạn xem lại đề)

c. $A=789\times 81\times 4\not\vdots 50$ 

Vậy có vẻ đề câu này sai rồi. Bạn xem lại.

97,5 co phai violympic ?

Chúng mình hãy làm quen nhé!

97.5

Bài 2:

a: \(17-x=3\)

=>\(x=17-3\)

=>x=14(nhận)

b: \(2\cdot\left(x-1\right):3=6\)

=>\(2\left(x-1\right)=6\cdot3=18\)

=>x-1=18/2=9

=>x=9+1=10(nhận)

c: \(x+\left(-2\right)=\left(-11\right)+7\)

=>\(x-2=-4\)

=>\(x=-4+2=-2\left(loại\right)\)

d: \(\left(x-1\right)^2-5=20\)

=>\(\left(x-1\right)^2=25\)

=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)

Câu 3:

a: Đặt *=a

\(\overline{57a3}⋮9\)

=>\(5+7+a+3⋮9\)

=>\(a+15⋮3\)

mà 0<=a<=9

nên a=3

=>*=a

b: \(A=123\cdot7+8+9\)

123*7 là số lẻ

9 là số lẻ

=>123*7+9 chia hết cho 2

mà 8 chia hết cho 2

nên \(A=123\cdot7+9+8⋮2\)

\(123\cdot7⋮3;9⋮3;8⋮̸3\)

=>\(A=123\cdot7+9+8⋮̸3\)

c: \(B=3\cdot5\cdot7+10^{50}\)

\(=5\cdot3\cdot7+5\cdot5^{49}\cdot2^{49}\)

\(=5\left(3\cdot7+5^{49}\cdot2^{49}\right)⋮5\)

=>B là hợp số

a+ax2+ax3+...+ax49+ax50=12750

ax(1+2+3+...+49+50)=12750

ax1275=12750

a=12750:1275

a=10

Ta có : a + a × 2 + a × 3 + a × 4 + … + a × 49 + a × 50 = 12750

<=> a.(1 + 2 + 3 + ...... + 50) = 12750

=> a.1275 = 12750

=> a = 12750 : 1275

=> a = 10

a x ( 2 + 3 + 4 + ....+ 49 + 50) = 12750

a x 1275                                = 12750

a                                           = 12750 : 1275

a                                           =  10

Vậy a = 10