So sánh hai số hữu tỉ sau:
-1và3/10 và -(-13/-10
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ta có:-13/38<-13/39=-1/3(1)
29/-88>-29/87=-1/3(2)
từ (1) và (2) \(\Rightarrow\)-13/38<29/-88
\(1\frac{3}{7}\times1\frac{3}{10}\times1\frac{3}{13}\times...\times1\frac{3}{67}\)
\(=\frac{10}{7}\times\frac{13}{10}\times\frac{16}{13}\times...\times\frac{70}{67}\)
\(=\frac{10\times13\times16\times...\times70}{7\times10\times13\times...\times67}\)
\(=\frac{70}{7}=10\)
\(-\frac{13}{15}+-\frac{2}{15}=-1;-\frac{14}{16}+-\frac{2}{16}\)
Vì \(-\frac{2}{15}< -\frac{2}{16}\Rightarrow\frac{-13}{15}< -\frac{14}{16}\)
2.Gọi 3 p/số đó là x;y;z
\(-\frac{5}{8}< x< y< z< -\frac{3}{5}\)
\(-\frac{100}{160}< x< y< z< -\frac{96}{160}\)
\(\Rightarrow x=-\frac{99}{160};y=-\frac{98}{160}=-\frac{49}{80};z=-\frac{97}{160}\)
Vd 3:
a) 9/10 > 5/42 b) -4/27 < 10/-73
Vd 4:
5/-6: -7/12; 5/8; 3/4
Vd 5:
x<y
Vd 6:
-16/27= -16/27> -16/29
1)
-2/15 < 0
-10/-11 > 0
nên x < y
2)
ví dụ 1 < 2 => -1 > -2
ta có 16/29 < 16/27 < 19/27
suy ra -16/29 > -16/27 > -19/27
\(\frac{11}{13}\)và \(\frac{22}{27}\)
Ta có:
\(\frac{11}{13}=\frac{297}{351}\)
\(\frac{22}{27}=\frac{242}{351}\)
Mà: \(\frac{297}{351}>\frac{242}{351}\)
Vậy \(\frac{11}{13}>\frac{22}{27}\)
\(\frac{-5}{11}\)và \(\frac{-9}{23}\)
Ta có:
\(\frac{-5}{11}=\frac{-115}{253}\)
\(\frac{-9}{23}=\frac{-99}{253}\)
Mà: \(\frac{-115}{253}< \frac{-99}{253}\)
Vậy \(\frac{-5}{11}< \frac{-9}{23}\)
\(\dfrac{97}{100}\) và \(\dfrac{98}{99}\)
\(\dfrac{97}{100}=\dfrac{97\times99}{100\times99}=\dfrac{9603}{9900}\)
\(\dfrac{98}{99}=\dfrac{98\times100}{99\times100}=\dfrac{9800}{9900}\)
Vì: \(9603< 9800\) nên => \(\dfrac{97}{100}< \dfrac{98}{99}\)
\(\dfrac{13}{17}\) và \(\dfrac{131}{171}\)
\(\dfrac{13}{17}=\dfrac{13\times171}{17\times171}=\dfrac{2223}{2907}\)
\(\dfrac{131}{171}=\dfrac{131\times17}{171\times17}=\dfrac{2227}{2907}\)
Vì: \(2227>2223\) nên: => \(\dfrac{13}{17}< \dfrac{131}{171}\)
\(\dfrac{51}{61}\) và \(\dfrac{515}{616}\)
\(\dfrac{51}{61}=\dfrac{51\times616}{61\times616}=\dfrac{31416}{37576}\)
\(\dfrac{515}{616}=\dfrac{515\times61}{616\times61}=\dfrac{31415}{37576}\)
Vì: \(31416>31415\) Nên => \(\dfrac{51}{61}>\dfrac{515}{616}\)
a/
$\frac{97}{100}< \frac{98}{100}< \frac{98}{99}$
c/
$\frac{131}{171}=1-\frac{40}{171}> 1-\frac{40}{170}=1-\frac{4}{17}=\frac{13}{17}$
d/
$\frac{51}{61}=1-\frac{10}{61}=1-\frac{100}{610}$
$\frac{515}{616}=1-\frac{101}{616}$
Xét hiệu:
$\frac{100}{610}-\frac{101}{616}=\frac{100.616-101.610}{610.616}$
$=\frac{100(610+6)-101.610}{610.616}$
$=\frac{600-610}{610.616}<0$
$\Rightarrow \frac{100}{610}< \frac{101}{616}$
$\Rightarrow 1-\frac{100}{610}> 1-\frac{101}{616}$
$\Rightarrow \frac{51}{61}> \frac{515}{616}$
\(-1\dfrac{3}{10}=-\left(1+\dfrac{3}{10}\right)=-\left(\dfrac{10}{10}+\dfrac{3}{10}\right)=-\dfrac{13}{10}\\ -\left(\dfrac{-13}{-10}\right)=-\dfrac{13}{10}\)
Do đó nên : \(-1\dfrac{3}{10}=-\left(\dfrac{-13}{-10}\right)\)
= -( -13/-10 )