So sánh:\(A=\frac{10^{2015}-1}{10^{2016}-1}vàB=\frac{10^{2014}+1}{10^{2015}+1}\)
so sanh A = \(\frac{10^{2014}+2016}{10^{2015}+2016}vàB=\frac{10^{2015}+2016}{10^{2016}+2016}\)
so sánh A=\(\frac{10^{2015-1}}{10^{2016-1}}\)và B=\(\frac{10^{2014+1}}{10^{2015+1}}\)
cho A=\(\frac{10^{2015}-1}{10^{2016}-1}\)và B=\(\frac{10^{2014}+1}{10^{2015}+1}\). So sánh A và B
\(A=\frac{10^{2015}-1}{10^{2016}^{ }-1}=\frac{10^{2015}}{10^{2016}}=\frac{1}{1},B=\frac{10^{2014}-1}{10^{2015}-1}=\frac{10^{2014}}{10^{2015}}=\frac{1}{1}A=B\Rightarrow\)
SO SÁNH A =\(\frac{10^{2014}+1}{10^{2015}+1}\) VÀ B = \(\frac{10^{2015}+1}{10^{2016}+1}\)
So sánh A = \(\frac{10^{2014}+1}{10^{2015}+1}\) và B = \(\frac{10^{2015}+1}{10^{2016}+1}\)
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
Cho A=\(\frac{10^{2015}+1}{10^{2014}+1}\) ; B=\(\frac{10^{2016}+1}{10^{2015}+1}\)
Hãy so sánh A và 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 ~
So sánh:
A=\(\frac{10^{2014}+1}{10^{2015}+1}\) và B=\(\frac{10^{2015}+1}{10^{2016}+1}\)
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
10A=10.(102014+1102015+1 )=102015+10102015+1 =102015+1+9102015+1 =1+9102015+1
10B=10.(102015+1102016+1 )=102016+10102016+1 =102016+1+9102016+1 =1+9102016+1
Vì 1 = 1; 9 = 9 ta so sánh mẫu:
Ta có: 102015 < 102016 => 102015+1 < 102016+1
=> 1+9102015+1 >1+9102016+1
=> 10A > 10B
=> A > B.
Đúng 20 Sai 0 Lê Thị Như Quỳnh đã chọn câu trả lời này.
so sanh A=\(\frac{10^{2015}-1}{10^{2016}-1}\) B=\(\frac{10^{2014}+1}{10^{2015}+1}\)
So sánh A và B
a) A = \(\frac{10^{11}-1}{10^{12}-1}\) Và \(B=\frac{10^{10}+1}{10^{11}+1}\)
b) \(A=\frac{2000^{2015}+1}{2000^{2016}+1}\) Và \(B=\frac{2000^{2014}+1}{2000^{2015}+1}\)
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