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a)\(\dfrac{2}{3}giờ< \dfrac{3}{4}giờ\)
b)\(\dfrac{7}{10}m< \dfrac{3}{4}m\)
c)\(\dfrac{7}{8}kg< \dfrac{9}{10}kg\)
d)\(\dfrac{5}{6}\)km/h >\(\dfrac{7}{9}\)km/h
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b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)
\(B=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)
mà \(10^7-8< 10^8-7\)
nên A>B
c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)
mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)
nên A<B
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a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
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d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
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mình ghi nhầm nên các bạn cứ hết hai phân số là một câu nhé ví dụ như \(\dfrac{-5}{8}\):\(\dfrac{15}{4}\)
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a) \(\dfrac{4}{5}\) giờ dài hơn
b) \(\dfrac{3}{5}\) mét ngắn hơn
c) \(\dfrac{7}{8}\) kg lớn hơn
a, \(\dfrac{1}{2}\)= \(\dfrac{5}{10}\) giờ \(\dfrac{4}{5}\)= \(\dfrac{8}{10}\)giờ
=> \(\dfrac{1}{5}\) < \(\dfrac{4}{5}\)
Vậy \(\dfrac{4}{5}\)giờ dài hơn \(\dfrac{1}{2}\) giờ
b, \(\dfrac{2}{3}\)m = \(\dfrac{10}{15}\) m \(\dfrac{3}{5}\) m = \(\dfrac{9}{15}\) m
=>\(\dfrac{10}{15}\) > \(\dfrac{9}{15}\)
Vậy \(\dfrac{2}{3}\) mét dài hơn \(\dfrac{3}{5}\) mét
c, \(\dfrac{6}{7}\) kg = \(\dfrac{48}{56}\)kg \(\dfrac{7}{8}\)kg = \(\dfrac{49}{56}\)kg
=> \(\dfrac{48}{56}\)kg < \(\dfrac{49}{56}\)kg
Vậy \(\dfrac{6}{7}\)kg nhỏ hơn \(\dfrac{7}{8}\)
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Gọi mẫu của các phân số cần tìm là x
Theo đề, ta có:
\(\dfrac{-1}{7}< \dfrac{-3}{x}< \dfrac{-1}{8}\)
\(\Leftrightarrow\dfrac{1}{7}>\dfrac{3}{x}>\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{3}{21}>\dfrac{3}{x}>\dfrac{3}{24}\)
nên \(x\in\left\{22;23\right\}\)
Vậy: Tổng là \(\dfrac{-3}{22}+\dfrac{-3}{23}=\dfrac{-69-66}{22\cdot23}=\dfrac{-135}{506}\)
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Vì 18/91 < 18/90 =1/5
23/114>23115=1/5
vậy 18/91<1/5<23/114
suy ra 18/91<23/114
vì 21/52=210/520
Mà 210/520=1-310/520
213/523=1-310/523
310/520>310/523
vậy 210/520<213/523
suy ra 21/52<213/523
ta có : `7/(-10)= -7/10 ; 8/(-10)=-8/10`
mà `-7>(-8)`
`-> 7/(-10) >8/(-10)`
Ta có : \(\dfrac{7}{-10}\text{=}\dfrac{-7}{10}\) và \(\dfrac{8}{-10}\text{=}\dfrac{-8}{10}\)
do hai phân số đã có mẫu bằng nhau nên ta xét tử của chúng :
Do \(-7>-8\)
\(\Rightarrow\dfrac{7}{-10}>\dfrac{8}{-10}\)