Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{10^{2015}-1}{10^{2016}^{ }-1}=\frac{10^{2015}}{10^{2016}}=\frac{1}{1},B=\frac{10^{2014}-1}{10^{2015}-1}=\frac{10^{2014}}{10^{2015}}=\frac{1}{1}A=B\Rightarrow\)
10A=(10^2014+1).10/10^2015+1=10^2015+10/10^2015+1=10^2015+1+9/10^2015+1=1+(9/10^2015+1) 10B=(10^2015+1).10/10^2016+1=10^2016+10/10^2016+1=10^2016+1+9/10^2016+1=1+(9/10^2016+1) Vì 9/10^2015+1>9/10^2016+1 nên 10A>10B .Từ đó suy ra A>B
xét A ta có
\(10A=\frac{10.\left(10^{2014}+1\right)}{10^{2015}+1}=\frac{10^{2015}+10}{10^{2015}+1}=\frac{\left(10^{2015}+1\right)+9}{10^{2015}+1}\)suy ra \(10A=1+\frac{9}{10^{2015}+1}\)
xét B ta có
\(10B=\frac{10.\left(10^{2015}+1\right)}{10^{2016}+1}=\frac{10^{2016}+10}{10^{2016}+1}=\frac{\left(10^{2016}+1\right)+9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)
Vì 10A>10B suy ra A >B
10A = 10 2015 + 1 10. 10 2014 + 1
= 10 2015 + 1 10 2015 + 10
= 10 2015 + 1 10 2015 + 1 + 9
suy ra 10A = 1 + 10 2015 + 1 9
b, 2000A = \(\frac{2000\left(2000^{2015}+1\right)}{2000^{2016}+1}\)
= \(\frac{2000^{2016}+2000}{2000^{2016}+1}\)
= \(\frac{\left(2000^{2016}+1\right)+1999}{2000^{2016}+1}\)
= \(\frac{2000^{2016}+1}{2000^{2016}+1}\) + \(\frac{1999}{2000^{2016}+1}\)
= 1 + \(\frac{1999}{2000^{2016}+1}\)
2000B = \(\frac{2000\left(2000^{2014}+1\right)}{2000^{2015}+1}\)
= \(\frac{2000^{2015}+2000}{2000^{2015}+1}\)
= \(\frac{\left(2000^{2015}+1\right)+1999}{2000^{2015}+1}\)
= \(\frac{2000^{2015}+1}{2000^{2015}+1}\) + \(\frac{1999}{2000^{2015}+1}\)
= 1 + \(\frac{1999}{2000^{2015}+1}\)
So sanh
câu b tiếp
So sánh 2000A với 2000B
Vì \(\frac{1999}{2000^{2016}+1}\) < \(\frac{1999}{2000^{2015}+1}\)
→ 2000A< 2000B
→ A<B
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(B=\frac{10^{2014}+1}{10^{2015}+1}< \frac{10^{2014}+1+9}{10^{2015}+1+9}=\frac{10^{2014}+10}{10^{2015}+10}=\frac{10\left(10^{2013}+1\right)}{10\left(10^{2014}+1\right)}=\frac{10^{2013}+1}{10^{2014}+1}=A\)
\(\Rightarrow\)\(B< A\) hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
áp dụng tính chất
nếu a/b>1thì a/b<(a+n)/(b+n)
=)))))))))))))))))
Ta có: \(10A=10.\left(\frac{10^{2014}+1}{10^{2015}+1}\right)=\frac{10^{2015}+10}{10^{2015}+1}=\frac{10^{2015}+1+9}{10^{2015}+1}=1+\frac{9}{10^{2015}+1}\)
\(10B=10.\left(\frac{10^{2015}+1}{10^{2016}+1}\right)=\frac{10^{2016}+10}{10^{2016}+1}=\frac{10^{2016}+1+9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)
Vì 1 = 1; 9 = 9 ta so sánh mẫu:
Ta có: 102015 < 102016 => 102015+1 < 102016+1
=> \(1+\frac{9}{10^{2015}+1}>1+\frac{9}{10^{2016}+1}\)
=> 10A > 10B
=> A > B.
2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)
Vậy \(2^{332}< 3^{223}\)
1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)
\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)
Nên suy ra \(10A>10B\Rightarrow A>B\)
Ta có công thức :
\(\frac{a}{b}>\frac{a+c}{b+c}\)\(\left(\frac{a}{b}>1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(B=\frac{10^{2016}+1}{10^{2015}+1}>\frac{10^{2016}+1+9}{10^{2015}+1+9}=\frac{10^{2016}+10}{10^{2015}+10}=\frac{10\left(10^{2015}+1\right)}{10\left(10^{2014}+1\right)}=\frac{10^{2015}+1}{10^{2014}+1}=A\)
\(\Rightarrow\)\(B>A\) hay \(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~