\(Sosánh\)
\(\frac{2017^{10}+1}{2017^{10}-1}\) và \(\frac{2017^{10-1}}{2017^{10-3}}\)
So sánh
A = \(\frac{10^{2017}-2}{10^{2017}+1}\) và B = \(\frac{10^{2017}}{10^{2017}+3}\)
Có \(A=\frac{10^{2017}+1-3}{10^{2017}+1}=1-\frac{3}{10^{2017}+1}\)
\(B=\frac{10^{2017}+3-3}{10^{2017}+3}=1-\frac{3}{10^{2017}+3}\)
Có 102017+1<102017+3
=> \(\frac{3}{10^{2017}+1}>\frac{3}{10^{2017}+3}\)
=>A<B
So sánh:
\(A=\frac{10^{2016}+1}{10^{2017}+1}\) và \(B=\frac{10^{2017}+1}{10^{2018}+1}\)
Ta có : \(A=\frac{10^{2016}+1}{10^{2017}+1}\)
Suy ra \(10A=\frac{10^{2017}+10}{10^{2017}+1}\)
Suy ra \(10A=1+\frac{9}{10^{2017}+1}\)
Ta lại có : \(B=\frac{10^{2017}+1}{10^{2018}+1}\)
Suy ra : \(10B=\frac{10^{2018}+10}{10^{2018}+1}\)
Suy ra : \(10B=1+\frac{9}{10^{2018}+1}\)
Vì \(\frac{9}{10^{2017}+1}>\frac{9}{10^{2018}+1}\)
Nên \(1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)
Suy ra \(10A>10B\)
Suy ra \(A>B\)
\(B< \frac{10^{2017}+1+9}{10^{2018}+1+9}=\frac{10^{2017}+10}{10^{2018}+10}=\frac{10\left(10^{2016}+1\right)}{10\left(10^{2017}+1\right)}=\frac{10^{2016}+1}{10^{2017}+1}=A\)
vậy A > B
So sánh A và B, biết:
A =\(\frac{10^{2016}+1}{10^{2017}+1}\)và B =\(\frac{10^{2017}+1}{10^{2018}+1}\)
Nhân cả hai tử của \(A\)và \(B\)với 2 , ta được :
\(10A=10.\left(\frac{10^{2016}+1}{10^{2017}+1}\right)=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{2^{2017}+1}\)
\(10B=10\left(\frac{10^{2017}+1}{10^{2018}+1}\right)=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}}=1+\frac{9}{10^{2018}+1}\)
Vì \(1=1;9=9\)
\(\Rightarrow\)Ta so sánh mẫu , ta có:
\(10^{2017}< 10^{2018}\)
\(\Rightarrow10^{2017}+1< 10^{2018}+1\)
\(\Rightarrow1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)
\(\Rightarrow10A>10B\)
Hay \(A>B\)
so sánh
a, A=\(\frac{10^{17}-1}{10^{16}-1}vaB=\frac{10^{16}+2}{10^{15}+2}\)
b,\(C=\frac{2017^{15}+1}{2017^{16}+1}vaO=\frac{2017^{16}-1}{2017^{17}-1}\)
c,\(E=\frac{99^{15}-1}{99^{16}-1}vaF=\frac{99^{16}+2}{99^{17}+2}\)
So sánh A=\(\frac{10^{2017}+1}{10^{ }^{2016}+1}\)B=\(\frac{10^{2018}+1}{10^{2017}^{ }+1}\)
Anh hiền àaaaaaaaaaaaaaaaaaaaaaaaaa
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 :
\(B=\frac{10^{2018}+1}{10^{2017}+1}>\frac{10^{2018}+1+9}{10^{2018}+1+9}=\frac{10^{2018}+10}{10^{2018}+10}=\frac{10\left(10^{2017}+1\right)}{10\left(10^{2016}+1\right)}=\frac{10^{2017}+1}{10^{2016}+1}=A\)
\(\Rightarrow\)\(B>A\) hay \(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
So sánh A=\(\frac{10^{2017}+1}{10^{2018}+1}\), B=\(\frac{10^{2016}+1}{10^{2017}+1}\)
So sánh A và B, biết rằng:
A = \(\frac{10^{2017}+1}{10^{2018}+1}\)
Và B = \(\frac{10^{2016}+1}{10^{2017}+1}\)
Ta có :
A = \(\frac{10^{2017}+1}{10^{2018}+1}\)< 1 => A < \(\frac{10^{2017}+1+9}{10^{2018}+1+9}\)= \(\frac{10^{2017}+10}{10^{2018}+10}\)= \(\frac{10^{2016}+1}{10^{2017}+1}\)= B
Vậy A < B
A<B. lời giải thích khó viết lắm nên bạn tự tìm cách làm nhé
So sánh A và B biết :
A = \(\frac{10^{2016}+1}{10^{2017}+1}\)
B = \(\frac{10^{2017}+1}{10^{2018}+1}\)
Ta có: \(\hept{\begin{cases}A=\frac{10^{2016}+1}{10^{2017}+1}\\B=\frac{10^{2017}+1}{10^{2018}+1}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}10A=\frac{10^{2017}+10}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\\10B=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\end{cases}}\)
Vì \(\frac{9}{10^{1017}+1}>\frac{9}{10^{2018}+1}\)
nên \(10A>10B\Rightarrow A>B\)
\(A=\frac{10^{2016}+1}{10^{2017}+1}\Rightarrow10A=\frac{10\cdot(10^{2016}+1)}{10^{2017}+1}=\frac{10^{2017}+10}{10^{2017}+1}\)
\(A=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
Vì \(10^{2016}+1< 10^{2017}+1\)
\(\Rightarrow\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\)
\(\Rightarrow\)\(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)
....
A= \(\frac{10^{2016}+1}{10^{2016}+1}=\frac{10^{2016}+1}{10\cdot10^{2016}+1}=\frac{1}{10}\cdot\frac{10^{2016}+1}{10^{2016}+1}=\frac{1}{10}\)(1)
B=\(\frac{10^{2017}+1}{10^{2018}+1}=\frac{10^{2017}+1}{10\cdot10^{2017}+1}=\frac{1}{10}\cdot\frac{10^{2017}+1}{10^{2017}+1}=\frac{1}{10}\)(2)
Từ (1) và (2) \(\Rightarrow\)A=B
Bài 3 : Không quy đồng , hãy so sánh :
\(P=\frac{10^{2017}-1}{10^{2018}-1}\) và \(Q=\frac{10^{2016}+1}{10^{2017}+1}\)
\(P=\frac{10^{2017}-1}{10^{2018}-1}\) \(>\) \(Q=\frac{10^{2016}+1}{10^{2017}+1}\)