Cho A= 10^2017+3/10^2016+3 và B=10^2016+1/10^2015+1
so sánh A và B
Những câu hỏi liên quan
So sánh A và B
A= 10^2015/(10^2016-1)
B= 10^2016/(10^2017-1)
Cho A = 10^2015+1/10^2016+1 và B = 10^2016+1/10^2017+1
So sánh A và B
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A=\(\frac{10^{2015}+1}{10^{2016}+1}\)=>10A=\(\frac{10.\left(10^{2015}+1\right)}{10^{2016}+1}\)= \(\frac{10^{2016}+10}{10^{2016}+1}\)=\(\frac{\left(10^{2016}+1\right)+9}{10^{2016}+1}\)=\(\frac{10^{2016}+1}{10^{2016}+1}+\frac{9}{10^{2016}+1}\)=1+\(\frac{9}{10^{2016}+1}\)
B=\(\frac{10^{2016}+1}{10^{2017}+1}\)=>10B=\(\frac{10.\left(10^{2016}+1\right)}{10^{2017+1}}=\frac{10^{2017}+10}{10^{2017}+1}\)= \(\frac{\left(10^{2017}+1\right)+9}{10^{2017}+1}\)=\(\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}\)= 1+\(\frac{9}{10^{2017}+1}\)
Vì \(10^{2016}+1< 10^{17}+1\)=>\(\frac{9}{10^{2016}+1}\)>\(\frac{9}{10^{2017}+1}\)nên \(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)=>10A>10B
Vậy A>B
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SO SÁNH A VÀ B BIẾT:
A = \(\frac{10^{2016}+1}{10^{2015}+1}\)VÀ B = \(\frac{10^{2017}+1}{10^{2016}+1}\)
Áp dung công thức \(a>b\Leftrightarrow\frac{a}{b}>\frac{a+m}{b+m}\)
\(B=\frac{10^{2017}+1}{10^{2016}+1}>\frac{10^{2017}+1+9}{10^{2016}+1+9}=\frac{10^{2017}+10}{10^{2016}+10}=\frac{10\left(10^{2016}+1\right)}{10\left(10^{2015}+1\right)}=\frac{10^{2016}+1}{10^{2015}+1}=A\)
\(\Leftrightarrow B>A\)
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So sánh
A= 102015+1 \ 102016 +1 và B= 102016 +1 / 102017+1
So sánh:
a) A = 102016 - 2 / 102017 - 2 và B = 202015 + 1 / 102016 + 1
b) A = 20162017 - 3 / 20162018 - 3 và B = 20162016 + 3 / 20162017 + 3
c) A = 20172016 - 2015 / 20172017 - 2015 và B = 20172015 + 1 / 20172016 + 1
So sánh
A = 10^2016 + 1 / 10^2015 + 1
B = 10^2017 + 1 / 10^2016 + 1
Ta có :
\(A=\frac{10^{2016}+1}{10^{2015}+1}=\frac{\left(10^{2016}+1\right).10}{\left(10^{2015}+1\right).10}=\frac{10^{2017}+10}{10^{2016}+10}=\frac{10^{2017}+10}{10^{2016}+10}\)
Vì \(10^{2017}=10^{2017}\)và \(10>1\)nên \(10^{2017}+10>10^{2017}+1\)( 1 )
Vì \(10^{2016}=10^{2016}\)và \(10>1\)nên \(10^{2016}+10>10^{2016}+1\)( 2 )
Từ ( 1 ) và ( 2 ) , suy ra : \(\frac{10^{2017}+10}{10^{2016}+10}>\frac{10^{2017}+1}{10^{2016}+1}\)
Vậy \(A>B\)
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\(B=\frac{10^{2016}+1}{10^{2017}+1}=\frac{10^{2016}+1+9}{10^{2017}+1+9}=\frac{10^{2016}+10}{10^{2017}+10}=\frac{10.\left(10^{2015}+1\right)}{10.\left(10^{2016}+1\right)}=\frac{10^{2015}+1}{10^{2016}+1}\)
lm tương tự vs B ta có
\(A=\frac{10^{2015}+1}{10^{2014}+1}\)
suy ra A>B
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Ta có: A=\(\frac{10^{2016}+1}{10^{2015}+1}\)
=>\(\frac{1}{A}=\frac{10^{2015}+1}{10^{2016}+1}=\frac{10\left(10^{2015}+1\right)}{10\left(10^{2016}+1\right)}=\frac{10^{2016}+10}{10\left(10^{2016}+1\right)}=\frac{10^{2016}+1+9}{10\left(10^{2016}+1\right)}\)
\(=\frac{1}{10}+\frac{9}{10^{2017}+10}\)
\(B=\frac{10^{2017}+1}{10^{2016}+1}\)
=>\(\frac{1}{B}=\frac{10^{2016}+1}{10^{2017}+1}=\frac{10\left(10^{2016}+1\right)}{10\left(10^{2017}+1\right)}=\frac{10^{2017}+10}{10\left(10^{2017}+1\right)}\)
\(=\frac{10^{2017}+1+9}{10\left(10^{2017}+1\right)}=\frac{1}{10}+\frac{9}{10^{2018}+10}\)
Vì\(10^{2017}< 10^{2018}=>10^{2017}+10< 10^{2018}+10\)
\(=>\frac{9}{10^{2017}+10}>\frac{9}{10^{2018}+10}=>\frac{1}{10}+\frac{9}{10^{2017}+10}>\frac{1}{10}+\frac{9}{10^{2017}+10}\)
\(=>\frac{1}{A}>\frac{1}{B}=>A< B\)
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Hãy so sánh 2 biểu thức A và B
A=102015+1/102016+1
B=102016+1/102017+1
So sánh: A = 102015 +1 phần 102016 +1 và B = 102016 +1 phần 102017 +1
\(A=\frac{10^{2015}+1}{10^{2016}+1}\Rightarrow10A=\frac{10.\left(10^{2015}+1\right)}{10^{2016}+1}=\frac{10^{2016}+10}{10^{2016}+1}\)
\(A=\frac{10^{2016}+1+9}{10^{2016}+1}=\frac{10^{2016}+1}{10^{2016}+1}+\frac{9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)
\(B=\frac{10^{2016}+1}{10^{2017}+1}\Rightarrow10B=\frac{10.\left(10^{2016}+1\right)}{10^{2017}+1}=\frac{10^{2017}+10}{10^{2017}+1}\)
\(B=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
Vì 102016+1 < 102017+1
=>\(\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\)
=>\(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)
=>10A > 10B
=>A > B
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\(B=\frac{10^{2016}+1}{10^{2017}+1}<\frac{10^{2016}+1+9}{10^{2017}+1+9}\)
\(=\frac{10^{2016}+10}{10^{2017}+10}\)
\(=\frac{10.\left(10^{2015}+1\right)}{10.\left(10^{2016}+1\right)}\)
\(=\frac{10^{2015}+1}{10^{2016}+1}=A\)
\(\Rightarrow\) B<A
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\(10A=\frac{10\left(10^{2015}+1\right)}{10^{2016}+1}=\frac{10^{2016}+10}{10^{2016}+1}=\frac{10^{2016}+1+9}{10^{2016}+1}=\frac{10^{2016}+1}{10^{2016}+1}+\frac{9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)
\(10B=\frac{10\left(10^{2016}+1\right)}{10^{2017}+1}=\frac{10^{2017}+10}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
vì 102016+1<102017+1
=>\(\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\)
=>A>B
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Xem thêm câu trả lời
Cho A= 10^2014+1/10^2015+1 và B= 10^2015+1/10^2016+1. So sánh A và B