Đặt \(A=5+5^3+...+5^{99}\)

=>\(25A=5^3+5^5+...+5^{101}\)

=>\(25A-A=5^5+5^5+...+5^{101}-5-5^3-...-5^{99}\)

=>\(24A=5^{101}-5\)

=>\(A=\dfrac{5^{101}-5}{24}\)

 \(\dfrac{1988\text{x}1996+1996\text{x}1995}{1995\text{x}1998-1995\text{x}1996}\)

\(=\dfrac{1996\text{x}\left(1988+1995\right)}{1995\text{x}\left(1998-1996\right)}\)

\(=\dfrac{1996\text{x}3983}{1995\text{x}2}=\dfrac{998\text{x}3983}{1995}=\dfrac{3975034}{1995}\)

\(\left(846-2\times X\right):12=51\)

=>\(846-2\times X=51\cdot12=612\)

=>2x=846-612=234

=>x=234:2=117

Gọi số cần tìm là \(\overline{ab312c}\)  ( a,b,c là các chữ số, a ≠ 0)

Ta có:

+) Số cần tìm chia hết cho 4 và 5 nên nó sẽ có số tận cùng là 0

\(\Rightarrow c=0\)

+) Số cần tìm chia hết cho 6 

\(\Rightarrow a+b+3+1+2+0⋮3\)

\(\Rightarrow a+b+6⋮3\)

Vì \(6⋮3\) nên \(a+b⋮3\)

Mà số cần tìm có giá trị nhỏ nhất có thể nên:

\(a=1;b=2\)

Vậy số cần tìm là \(123120\)

a) \(\dfrac{2}{3}+\dfrac{-2}{18}=\dfrac{12}{18}+\dfrac{-2}{18}=\dfrac{10}{18}=\dfrac{5}{9}\)

b) \(\left(-12,5\right)+3,4+12,5+\left(-3,4\right)=\left(12,5-12,5\right)+\left(3,4-3,4\right)=0\)

c) 

\(\dfrac{-6}{11}:\dfrac{7}{4}+\dfrac{-6}{11}:\dfrac{7}{3}+1\dfrac{1}{11}\\ =\dfrac{-6}{11}\cdot\dfrac{4}{7}+\dfrac{-6}{11}\cdot\dfrac{3}{7}+\dfrac{12}{11}\\ =-\dfrac{6}{11}\cdot\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{12}{11}\\ =\dfrac{-6}{11}+\dfrac{12}{11}\\ =\dfrac{6}{11}\)

a) \(\dfrac{2}{3}+\dfrac{-2}{18}=\dfrac{2}{3}-\dfrac{1}{9}\\ =\dfrac{6}{9}-\dfrac{1}{9}=\dfrac{5}{9}\)

b) \(\left(-12,5\right)+3,4+12,5+\left(-3,4\right)\\ =\left(12,5-12,5\right)+\left(3,4-3,4\right)\\ =0\)

c) \(\dfrac{-6}{11}:\dfrac{7}{4}+\dfrac{-6}{11}:\dfrac{7}{3}+1\dfrac{1}{11}\\ =-\dfrac{6}{11}\times\dfrac{4}{7}+\dfrac{-6}{11}\times\dfrac{3}{7}+\dfrac{12}{11}\\ =-\dfrac{6}{11}\times\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{12}{11}\\ =-\dfrac{6}{11}\times1+\dfrac{12}{11}\\ =-\dfrac{6}{11}+\dfrac{12}{11}=\dfrac{6}{11}\)

cho tớ hỏi giữa 9 và 11 là kí hiệu j vậy ak

lưu ý:9 11 là 9/11 nhé

\(B=\left[\left(-0,5\right):0,07-0,2:0,07\right]\cdot1,5-2024^0\\ =\left[\left(-0,5\right):\dfrac{7}{100}-0,2:\dfrac{7}{100}\right]\cdot1,5-1\\ =\left[\left(-0,5\right)\cdot\dfrac{100}{7}-0,2\cdot\dfrac{100}{7}\right]\cdot1,5-1\\ =\dfrac{100}{7}\cdot\left(-0,5-0,2\right)\cdot1,5-1\\ =\dfrac{100}{7}\cdot-0,7\cdot1,5-1\\ =\dfrac{100}{7}\cdot\dfrac{-7}{10}\cdot1,5-1\\ =-10\cdot1,5-1\\ =-15-1\\ =-16\)

cứu với

a) \(\left(-32\right)^9=-32^9=-\left(2^5\right)^9=-2^{45}\)

\(\left(-16\right)^{13}=-16^{13}=-\left(2^4\right)^{13}=-2^{52}\)

Vì \(2^{45}< 2^{52}=>-2^{45}>-2^{52}\)

b) \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)

\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)

Vì \(243>125=>243^{10}>125^{10}>\left(-3\right)^{50}>\left(-5\right)^{30}\) 

c) \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)

\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)

Vì: \(\dfrac{1}{8}>\dfrac{1}{9}=>\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}=>\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)

d) 

\(\left(\dfrac{1}{5}\right)^{199}>\left(\dfrac{1}{5}\right)^{200}=\left[\left(\dfrac{1}{5}\right)^2\right]^{100}\\ =\left(\dfrac{1}{25}\right)^{100}>\left(\dfrac{1}{27}\right)^{100}=\left[\left(\dfrac{1}{3}\right)^3\right]^{100}=\left(\dfrac{1}{3}\right)^{300}\) 

\(32< 2^n< 128\\ =>2^5< 2^n< 2^7\\ =>5< n< 7\)

Vì n là số nguyên nên n=6

b) \(2.16\ge2^n>4\\ =>2.2^4\ge2^n>2^2\\ =>2^5\ge2^n>2^2\\ =>5\ge n>2\)

Vì n là số nguyên nên \(n\in\left\{3;4;5\right\}\)

\(8^{12}=\left(8^3\right)^4=512^4\\ 12^8=\left(12^2\right)^4=144^4\\ \)

Nhận thấy: \(512^4>144^4\Rightarrow8^{12}>12^8\)

\(8^{12}=\left(2^3\right)^{12}=2^{36}\)

\(12^8=\left(2^2\cdot3\right)^8=\left(2^2\right)^8\cdot3^8\\ =2^{16}\cdot3^8< 2^{16}\cdot4^8=2^{16}\cdot\left(2^2\right)^8=2^{16}\cdot2^{16}=2^{32}< 2^{36}\) 

=> \(12^8< 8^{12}\)