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\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\)\(\left(a,b,m\in N\cdot\right)\)
Ta có:
\(B=\dfrac{10^9+1}{10^{10}+1}< 10\left(10^9< 10^{10}\right)\)
\(\Leftrightarrow B=\dfrac{10^9+1}{10^{10}+1}< \dfrac{10^9+1+9}{10^{10}+1+9}=\dfrac{10^9+10}{10^{10}+10}=\dfrac{10\left(10^8+1\right)}{10\left(10^9+1\right)}=\dfrac{10^8+1}{10^9+1}=A\)
\(\Leftrightarrow A>B\)
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
mình chỉ làm được 2 cách thôi một cách mình chưa nghĩ ra
Ta có:
\(10A=10.\dfrac{10^7+1}{10^8+1}=\dfrac{10.10^7+1}{10^8+1}=\dfrac{10^8+1}{10^8+1}=1\)
\(10B=\dfrac{10.10^8+1}{10^9+1}=\dfrac{10^9+1}{10^9+1}=1\)
\(\Rightarrow10A=10B\)
\(\Rightarrow A=B\)
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
\(A=\dfrac{8^9+13}{8^9+7}=\dfrac{8^9+7+6}{8^9+7}=1+\dfrac{6}{8^9+7}\)
\(B=\dfrac{8^{10}-1+5}{8^{10}-1}=1+\dfrac{5}{8^{10}-1}\)
Vì \(1+\dfrac{6}{8^9+7}>1+\dfrac{5}{8^{10}-1}\) \(\Rightarrow A>B\)
\(A=\dfrac{8^9+12}{8^9+7}=\dfrac{8^9+7+5}{8^9+7}=\dfrac{8^9+7}{8^9+7}+\dfrac{5}{8^9+7}=1+\dfrac{5}{8^9+7}\left(1\right)\)
\(B=\dfrac{8^{10}+4}{8^{10}-1}=\dfrac{8^{10}-1+5}{8^{10}-1}=\dfrac{8^{10}-1}{8^{10}-1}+\dfrac{5}{8^{10}-1}=1+\dfrac{5}{8^{10}-1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A>B\)
mk giải cho câu A rồi tự suy mấy câu khác nhé!
ta có : A = 10^8 + 2/10^8 - 1
=> A = 10^8 - 1 + 3/10^8 - 1
=> A = 1+ 3/10^8 - 1
B = 10^8/10^8 - 3
=> B = 10^8 - 3 + 3/10^8 - 3
=> B = 1+ 3/10^8 - 3
vì 3/10^8 - 1 < 3/10^8 - 3
=> 1 + 3/10^8 - 1 < 1 + 3/10^8 - 3
=> A < B
vậy A < B
cách này cô dạy mk đó
Bạn kham khảo nhé!
\(A=\dfrac{10^9+1}{10^{10}+1}>B=\dfrac{10^8+1}{10^9+1}\)