Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(8\in A\)
b) \(\left\{8\right\}\subset A\)
c) \(\left\{8;10\right\}=A\)
Chúc bạn làm bài tốt
Cho tập hợp A = { 8; 10 }. Điền kí hiệu ∈,⊂ hoặc = vào chộ chấm:
a) 8 \(\in\) A
b) { 10 } \(\subset\) A
c) {8 ; 10 } \(=\) A
b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)
\(B=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)
mà \(10^7-8< 10^8-7\)
nên A>B
c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)
mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)
nên A<B
a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
a)\(\dfrac{1}{4}=\dfrac{2}{8}=\dfrac{4}{16}=\dfrac{8}{32}=......\)
b)\(\dfrac{-3}{4}=\dfrac{-6}{8}=\dfrac{-9}{12}=\dfrac{-12}{16}.......\)
\(1=\dfrac{2}{2}=\dfrac{-4}{-4}\)\(=\dfrac{6}{6}=\dfrac{-8}{-8}=\dfrac{10}{10}\)
Ta có: \(\frac{n}{n+1}< 1\)
\(\Rightarrow\frac{n}{n+1}< \frac{n+2}{n+1+2}\)
\(\Rightarrow\frac{n}{n+1}< \frac{n+2}{n+3}\)
\(\Rightarrow A< B\)
b. mình ko biết làm
c. mình cũng ko biết làm
d.Ta có :\(\frac{10^{1993}+1}{10^{1992}+1}>1\)
\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}\)
\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1992}.10+10.1}{10^{1991}.10+10.1}\)
\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)
\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1992}+1}{10^{1991}+1}\)
\(\Rightarrow A>B\)
Chúc bạn học tốt nhé
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
