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\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Nhận thấy: \(\frac{9}{10^{17}+1}< \frac{9}{10^{16}+1}\)=> 10B < 10A
=> A > B
A = ( 10^15+1 ) / ( 10^16+1 ) => 10A = ( 10^16+10 ) / ( 10^16+1 ) = 1 + ( 9/10^15+1 )
B = ( 10^16+1 ) / ( 10^17+1 ) => 10B = ( 10^17+10 ) / ( 10^17+1 ) = 1 + ( 9/10^16+1 )
Vì 10^15+1 < 10^16+1 nên 9/10^15+1 > 9/10^16+1 => 1 + ( 9/10^15+1 ) > 1 + ( 9/10^16+1 )
Vậy A > B
Ta có :
\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{\left(10^{16}+1\right)+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{\left(10^{17}+1\right)+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Vì \(10^{16}+1< 10^{17}+1\) nên \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\) \(\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\)
=> 10A > 10B Do đó A > B
Vậy A > B
\(A=\frac{10^{15}+1}{10^{16}+1}\)
\(B=\frac{10^{16}+1}{10^{17}+1}\)
Ta có:
\(A=\frac{10^{15}+1}{10^{16}+1}=\frac{\left(10^{15}+1\right).10}{\left(10^{16}+1\right).10}=\frac{10^{16}+10}{10^{17}+10}=\frac{10^{16}+1+9}{10^{17}+1+9}\)
Vì \(B=\frac{10^{16}+1}{10^{17}+1}< 1\)
\(\Rightarrow B=\frac{10^{16}+1}{10^{17}+1}< \frac{10^{16}+1+9}{10^{17}+1+9}=A\)
Vậy B < A
\(A=\frac{10^{15}+1}{10^{16}+1}\)
\(\Rightarrow10A=\frac{10^{16}+10}{10^{16}+1}=\frac{\left(10^{16}+1\right)+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(A=\frac{10^{16}+1}{10^{17}+1}\)
\(\Rightarrow10B=\frac{10^{17}+10}{10^{17}+1}=\frac{\left(10^{17}+1\right)+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Vì \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\left(Do10^{16}+1< 10^{17}+1\right)\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
c1: Ta có \(\frac{10^{15}+1+9}{10^{16}+1+9}\)=\(\frac{10^{15}+10}{10^{16}+10}\)=\(\frac{10\left(10^{14}+1\right)}{10\left(10^{15}+1\right)}\)=\(\frac{10^{14}+1}{10^{15}+1}\)
Vì \(\frac{10^{15}+1}{10^{16}+1}\)<1 nên \(\frac{10^{15}+1+9}{10^{16}+1+9}\)>\(\frac{10^{15}+1}{10^{16}+1}\)Vậy A<B
c2 Ta có 10A=\(\frac{10^{16+10}}{10^{16}+1}\)= \(\frac{10^{16}+1+9}{10^{16+1}}\)=\(\frac{10^{16}+1}{10^{16}+1}\)+ \(\frac{9}{10^{16}+1}\)=1+\(\frac{9}{10^{16}+1}\)
Tương tự câu B cung nhân vs 10 đc 1 +\(\frac{9}{10^{17}+1}\) vì 1+\(\frac{9}{10^{17}+1}\)< 1 + \(\frac{9}{10^{16}+1}\)\(\Rightarrow\)10B<10A\(\Rightarrow\)B<A
Vậy A>B
chuc ban hoc tot
Ta có:
10A=1016+10/1016+1=1+(9/1016+1)
10B=1017+10/1017+1=1+(9/1017+1)
Vì 9/1016+1 > 9/1017+1 nên 10A>10B,do đó A>B
Ta có:
\(A=\frac{10^{15}+1}{10^{16}+1}\)
\(10A=\frac{10^{16}+10}{10^{16}+1}\)
\(B=\frac{10^{16}+1}{10^{17}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}\)
Ta so sánh \(10A\) và \(10B\)
Có:
\(10A:\) Mẫu - tử = 9
\(10B:\) Mẫu - tử = 9
Lại có:
\(\frac{10^{16}+10}{10^{16}+1}\) \(-1\)\(=\frac{9}{10^{16}+1}\)
\(\frac{10^{17}+10}{10^{17}+1}-1=\frac{9}{10^{17}+1}\)
Vì \(\frac{9}{10^{16}+1}\)\(>\frac{9}{10^{17}+1}\)nên \(10A>10B\)
\(\Rightarrow\)\(A>B\)
Vậy \(A>B\)
Theo bải ra ta có:
A=\(\frac{10^{15}+1}{10^{16}+1}\)=> 10A =.\(\frac{10.\left(10^{15}+1\right)}{10^{16}+1}\)= \(\frac{10.10^{15}+1.10}{10^{16}+1}\)
= \(\frac{10.10^{15}+10}{10^{16}+1}\)=\(\frac{10^{16}+1+9}{10^{16}+1}\)= \(1+\frac{9}{10^{16}+1}\)
B= \(\frac{10^{16}+1}{10^{17}+1}\)=> 10B = \(\frac{10.\left(10^{16}+1\right)}{10^{17}+1}\)=\(\frac{10.10^{16}+1.10}{10^{17}+1}\)
= \(\frac{10.10^{16}+10}{10^{17}+1}\)= \(\frac{10^{17}+1+9}{10^{17}+1}\)= \(1+\frac{9}{10^{17}+1}\)
Vì 1=1 mà \(\frac{9}{10^{16}+1}\)> \(\frac{9}{10^{17}+1}\)nên => 10A > 10B => A>B
Vậy A>B.
Ta có:
\(A=\frac{10^{15}+1}{10^{16}+1}=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(B=\frac{10^{16}+1}{10^{17}+1}=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Vì \(\frac{1}{10^{16}+1}>\frac{1}{10^{17}+1}\)
\(\Rightarrow\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\)
\(\Rightarrow1+\frac{9}{10^{16}+1}>1+\frac{9}{10^{17}+1}\)
\(\Rightarrow A>B\)
Cảm ơn bạn nhiều nha!