So sánh hai số A và B biết rằng: A=\(\frac{10^{17}+5}{10^{17}-8}\)và B=\(\frac{10^{17}}{10^{17}-13}\)
So sánh 2 số A và B biết:
\(A=\frac{10^{17}+5}{10^{17}-8}\)
\(B=\frac{10^{17}}{10^{17}-3}\)
\(A=\frac{10^{17}+5}{10^{17}-8}=\frac{10^{17}-8+13}{10^{17}-8}=\frac{10^{17}-8}{10^{17}-8}+\frac{13}{10^{17}-8}=1+\frac{13}{10^{17}-8}\)
\(B=\frac{10^{17}}{10^{17}-3}=\frac{10^{17}-3+13}{10^{17}-3}=\frac{10^{17}-3}{10^{17}-3}+\frac{13}{10^{17}-3}=1+\frac{13}{10^{17}-3}\)
Nhận xét: \(10^{17}-8\frac{13}{10^{17}-3}\Rightarrow1+\frac{13}{10^{17}-8}>1+\frac{13}{10^{17}-3}\Rightarrow A>B\)
\(A=\frac{10^{17}+5}{10^{17}-8}=\frac{10^{17}-8+13}{10^{17}-8}=\frac{10^{17}-8}{10^{17}-8}+\frac{13}{10^{17}-8}=2+\frac{3}{10^{17}-8}\)
\(B=\frac{10^{17}}{10^{17}-3}=\frac{10^{17}-3+3}{10^{17}-3}=\frac{10^{17}-3}{10^{17}-3}+\frac{3}{10^{17}-3}=1+\frac{3}{10^{17}-3}\)
Do \(2+\frac{3}{10^{17}-8}>1+\frac{3}{10^{17}-3}\)n\(A>B\)
So sánh A và B biết:
a) A =\(\frac{10^7+5}{10^7-8}\) B =\(\frac{10^8+6}{10^8-7}\)
B) A =\(\frac{15^{16}-13}{15^{16}+7}\) B = \(\frac{16^{17}-12}{16^{17}+8}\)
So sánh 2 biểu thức sau:
a)A = 10^8+2/10^8-1 và B = 10^8/10^8-3
b)C=17^203+1/17^204+1 và D = 17^202+1/17^203+1
\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=1+\frac{3}{10^8-1}\)
\(B=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=1+\frac{3}{10^8-3}\)
Nhận thầy 108 - 1 > 108 - 3
=> \(\frac{3}{10^8-1}< \frac{3}{10^8-3}\)
=> \(1+\frac{3}{10^8-1}< \frac{3}{10^8-3}+1\)
=> A < B
b) 17C = \(\frac{17\left(17^{203}+1\right)}{17^{204}+1}=\frac{17^{204}+1+16}{17^{204}+1}=1+\frac{16}{17^{204}+1}\)
17D = \(\frac{17\left(17^{202}+1\right)}{17^{203}+1}=\frac{17^{203}+1+16}{17^{203}+1}=1+\frac{16}{17^{203}+1}\)
Nhận thầy 17203 + 1 < 17204 + 1
=> \(\frac{16}{17^{203}+1}>\frac{16}{17^{204}+1}\)
=> \(\frac{16}{17^{203}+1}+1>\frac{16}{17^{204}+1}+1\Rightarrow17C>17D\Rightarrow C>D\)
So sánh A và B biết:
\(A=\frac{10^{15}+1}{10^{16}+1}\)và \(B=\frac{10^{16}+1}{10^{17}+1}\)
\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=1+\frac{9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
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
So sánh A và B biết:\(A=\frac{10^{15}+1}{10^{16}+1}vàB=\frac{10^{16}+1}{10^{17}+1}\)
so sánh A và B biết:
A=\(\frac{10^{15}+1}{10^{16}+1}\)
B=\(\frac{10^{16}+1}{10^{17}+1}\)
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=1/10
B=1/10
Vậy hai phân số A và B bằng nhau
so sánh hai phân số:
a. 3/5 và 1/5
b. 9/10 và 11/10
c. 13/17 và 15/17
d. 25/19 và 22/19
giúp mik ik, lm đúng mik tick cho nha
\(\frac{3}{5}>\frac{1}{5}\)
\(\frac{9}{10}< \frac{11}{10}\)
\(\frac{13}{17}< \frac{15}{17}\)
\(\frac{25}{19}\)>\(\frac{22}{19}\)
3/5 > 1/5
9/10 < 11/10
13/17 < 15/17
25/19 > 22/19
T i c k cho mình nhé bạn!
ko quy đồng hãy so sánh \(A=\frac{-5}{10^{2009}}+\frac{-17}{10^{2010}}vàB=\frac{-17}{10^{2009}}+\frac{-5}{10^{2010}}\)
cu lay phep tinh nay tru phep tinh kia hk ra thi nt hoi mink
So sánh hai phân số \(A=\frac{10^{15}+1}{10^{16^{ }}+1}\) và \(B=\frac{10^{16+1}}{10^{17^{ }}+1}\)
\(\frac{10^{15}+1}{10^{16}+1}=\frac{10^{16}+10}{10^{17}+10}\)
Vì B<1 suy ra B<\(\frac{10^{16}+1+9}{10^{17}+1+9}=\frac{10^{16}+10}{10^{17}+10}=A\)
Vậy B<A
Ta có: \(10A=\frac{10^{16}+10}{10^{16}+1}=1+\frac{9}{10^{16}+1}\) ; \(10B=\frac{10^{17}+10}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Mà \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\) nên \(10A>10B\) => \(A>B\)