Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{31\cdot\left(31^{12}-1\right)}{31\left(31^{13}+1\right)}=\dfrac{31^{13}+1-32}{31\left(31^{13}+1\right)}=\dfrac{1}{31}-\dfrac{32}{31^{14}+31}\)
\(B=\dfrac{31\left(31^{13}-1\right)}{31\left(31^{14}+1\right)}=\dfrac{1}{31}-\dfrac{32}{31^{15}+31}\)
Dễ thấy \(31^{14}+31< 31^{15}+31\Rightarrow\dfrac{32}{31^{14}+31}>\dfrac{32}{31^{15}+31}\\ \Rightarrow\dfrac{1}{31}-\dfrac{32}{31^{14}+31}< \dfrac{1}{31}-\dfrac{32}{31^{15}+31}\)
Vậy A < B
h1) Ta có : ^A + ^B + ^C = 1800
=> 750 + 660 + x = 1800
<=> x = 1800 - 750 - 660 = 390
h2) Ta có : ^E + ^D + ^F = 1800
=> 630 + 370 + x = 1800
<=> x = 1800 - 1000 = 800
h3) Ta có : ^N + ^M + ^P = 1800
=> 1360 + x + x = 1800 <=> 2x = 440 <=> x = 220
=> ^N = ^P = x = 220
h4) Ta có : ^A + ^B + ^C = 1800
=> 1000 + 550 + x = 1800
<=> x = 1800 - 1000 - 550 = 250
\(I=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2009.2010}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2010}\\ =1-\dfrac{1}{2010}\\ =\dfrac{2009}{2010}\)
\(K=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{2008.2010}\\ =2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2008.2010}\right)\\ =2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\\ =2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\\ =2.\dfrac{502}{1005}\\ =\dfrac{1004}{1005}\)
\(F=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+...+\dfrac{3}{30.33}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}.\dfrac{10}{33}\\ =\dfrac{10}{99}\)
\(I=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}=\dfrac{2010-1}{2010}=\dfrac{2008}{2010}\)
\(K=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}=\dfrac{502}{1005}\)
\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{99.100}=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{100}\right)=\dfrac{97}{900}\)
`@` `\text {Ans}`
`\downarrow`
`j)`
\(x^{17}\div x^{12}=x^{17-12}=x^5\)
`k)`
\(x^8\div x^5=x^{8-5}=x^3\)
`r)`
\(a^5\div a^5=a^{5-5}=a^0=1\)
`l)`
\(x^4\div x=x^{4-1}=x^3\)
`m)`
\(x^7\div x^6=x^{7-6}=x\)
`n)`
\(x^9\div x^9=x^{9-9}=x^0=1\)
`o)`
\(a^{12}\div a^5=a^{12-5}=a^7\)
`p)`
\(a^8\div a^6=a^{8-6}=a^2\)
`q)`
\(a^{10}\div a^7=a^{10-7}=a^3\)
`r(2),`
\(1024\div4=2^{10}\div2^2=2^8\)
`t)`
\(512\div2^3=2^9\div2^3=2^6\)
TRẢ LỜI:
Đáp án: C
Trong hiện tượng giao thoa sóng trên mặt nước, khoảng cách giữa hai cực đại liên tiếp nằm trên đường nối tâm hai sóng có độ dài là một nửa bước sóng.
a: \(3\dfrac{3}{7}:1\dfrac{5}{7}\)
\(=\dfrac{24}{7}:\dfrac{12}{7}\)
\(=\dfrac{24}{7}\cdot\dfrac{7}{12}=\dfrac{24}{12}=2\)
b: \(\dfrac{2}{3}+\dfrac{-3}{5}=\dfrac{2}{3}-\dfrac{3}{5}\)
\(=\dfrac{10-9}{15}\)
\(=\dfrac{1}{15}\)
c: \(\dfrac{2}{9}-\left(\dfrac{1}{20}+\dfrac{2}{9}\right)\)
\(=\dfrac{2}{9}-\dfrac{1}{20}-\dfrac{2}{9}\)
\(=-\dfrac{1}{20}\)
d: \(\dfrac{11}{23}\cdot\dfrac{12}{17}+\dfrac{11}{23}\cdot\dfrac{5}{17}+\dfrac{12}{23}\)
\(=\dfrac{11}{23}\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\dfrac{12}{23}\)
\(=\dfrac{11}{23}+\dfrac{12}{23}=\dfrac{23}{23}=1\)