A = \(\dfrac{3^{10}.11+3^{10}.5}{3^9.2^4}\) 

\(A=\dfrac{3^{10}.\left(11+5\right)}{3^9.16}\) 

\(A=\dfrac{3^{10}.16}{3^9.16}\)

 \(A=\dfrac{3^{10}}{3^9}=3\) 

Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh rằng ta có các tỉ lệ thức sau (giả thiết các tỉ lệ thức là có nghĩa ) : 

a) \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)     

b) \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\) 

Chọn D : did you do 

Đáp án :  B. Kí sinh 

a) Ta có : 

\(\dfrac{13}{33}=\dfrac{13.11}{33.11}=\dfrac{143}{363}\) 

\(\Rightarrow\dfrac{13}{33}=\dfrac{143}{363}\) ( điều phải chứng minh ) 

Vậy ...... 

b) Ta có : 

\(\dfrac{1212}{2727}=\dfrac{1212:101}{2727:101}=\dfrac{12}{27}\) 

\(\Rightarrow\dfrac{1212}{2727}=\dfrac{12}{27}\) ( điều phải chứng minh ) 

Vậy ...... 

Chúc bạn học tốt 

\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^n\) 

\(\Rightarrow\dfrac{4^5.4}{3^5.3}.\dfrac{6^5.6}{2^5.2}=2^n\) 

\(\Rightarrow\dfrac{4^5.4.6^5.6}{3^5.3.2^5.2}=2^n\) 

\(\Rightarrow\dfrac{\left(2.2\right)^5.2.2.\left(3.2\right)^5.3.2}{3^5.3.2^5.2}=2^n\) 

\(\Rightarrow\dfrac{2^5.2^5.2.2.3^5.2^5.3.2}{3^5.3.2^5.2}=2^n\) 

Rút gọn vế trái ta có :

\(2^5.2.2.^5=2^n\)

\(\Rightarrow2^{12}=2^n\) 

\(\Rightarrow n=12\) ( Thỏa mãn điều kiện \(n\in N\) ) 

Vậy n =12 

a)  80 - (4.5² - 3.2³ )

= 80 - ( 4 . 25 - 3. 8 ) 

= 80 - ( 100 - 24 )

= 80 - 76 = 4 

b) 5⁶ : 5⁴ + 2³ . 2² - 1²⁰¹⁷ 

= 5² + 2⁵ - 1 

= 25 + 32 - 1 

= 57 -1 = 56 

Ta có  \(A=\dfrac{2}{1.3}-\dfrac{2}{2.4}+\dfrac{2}{3.5}-\dfrac{2}{4.6}+\dfrac{2}{5.7}-\dfrac{2}{6.8}+\dfrac{2}{7.9}-\dfrac{2}{8.10}+\dfrac{2}{9.11}-\dfrac{2}{10.12}\) 

\(\Rightarrow A=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)-\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+\dfrac{2}{10.12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\) 

\(\Rightarrow A=1-\dfrac{1}{11}-\dfrac{1}{2}+\dfrac{1}{12}\) 

\(\Rightarrow A=\dfrac{9}{22}+\dfrac{1}{12}\) 

\(\Rightarrow A=\dfrac{65}{132}\) 

Mà \(\dfrac{65}{132}< 1\) \(\Rightarrow A< 1\) 

Vậy \(A< 1\)