Đặt: \(A=\frac{3^6.21^{12}}{175^9.7^3}=\frac{3^{18}.7^{12}}{7^{12}.25^9}=\frac{3^{18}}{5^{18}}=\left(\frac{3}{5}\right)^{18}\)

\(B=\frac{3^{10}.6^7.4}{10^9.5^8}=\frac{3^{10}.2^7.3^7.2^2}{2^9.5^9.5^8}=\frac{3^{17}.2^9}{2^9.5^{17}}=\left(\frac{3}{5}\right)^{17}\)

Vì: \(\left(\frac{3}{5}\right)^{18}< \left(\frac{3}{5}\right)^{17}\Rightarrow A< B\)

cảm ơn bạn nhiều nha!

Lời giải:

$A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$

$\Rightarrow 2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}$

$\Rightarrow A=2A-A=1-\frac{1}{32}< 1-\frac{1}{2004}$

Hay $A< \frac{2003}{2004}$

Hay $A< B$

6 mũ 37>16 mũ 12

Hoặc 6 mũ 37<16 mũ 12

Hoặc 6 mũ 37=16 mũ 12 

6 mũ 37 > 16 MŨ 12

\(\frac{12}{9}x\frac{15}{10}=\frac{12x15}{9x10}=\frac{3x4x3x5}{3x3x2x3}=\frac{4}{2}=2\)

Ta có

 \(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{17.18}>A=\frac{1}{2.3}+\frac{1}{5.4}+...+\frac{1}{18.19}\)

\(C< =>\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{18-17}{17.18}\)\(>A\)

\(C< =>\frac{1}{2}-\frac{1}{18}\)\(>A\)

\(C< =>\frac{4}{9}\)\(>A\left(1\right)\)

Lại có  \(C=\frac{4}{9}< \frac{9}{19}=B\left(2\right)\)

Từ (1),(2) => B>A

x ⋮ 12 và x ⋮ 18 

⇒ x ∈ BC(12, 18) 

Ta có:

\(B\left(12\right)=\left\{0;12;24;36;48;60;72;84;96;108;120;...\right\}\)

\(B\left(18\right)=\left\{0;18;36;54;72;90;...\right\}\)

\(\Rightarrow x\in\left\{0;36;72;108;144;180;216;252;...\right\}\)

Mà: x < 250

\(\Rightarrow x\in\left\{0;36;72;108;144;180;216\right\}\)

x ⋮ 12; x ⋮ 18 nên x ∈ BC(12; 18)

Ta có:

12 = 2².3

18 = 2.3²

⇒ BCNN(12; 18) = 2².3² = 36

⇒ x ∈ BC(12; 18) = {0; 36; 72; 108; 144; 180; 216; 252; ...}

Mà x < 250

⇒ x ∈ {0; 36; 72; 108; 144; 180; 216}