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a) \(\dfrac{8,5-8,2}{16}=\dfrac{0,3}{16}=\dfrac{0,3\cdot10}{16\cdot10}=\dfrac{3}{160}\)
b) \(\dfrac{17\cdot5-17}{3-20}=\dfrac{17\cdot\left(5-1\right)}{-17}=\dfrac{1\cdot4}{-1}=-4\)
a)
\(\dfrac{8\cdot5-8\cdot2}{16}=\dfrac{8\left(5-2\right)}{16}=\dfrac{3}{2}\)
b)
\(\dfrac{17\cdot5-17}{3-20}=\dfrac{17\left(5-1\right)}{-17}=\dfrac{4}{-1}=-4\)
\(\dfrac{-63}{72}=\dfrac{-7}{8}\)
\(\dfrac{20}{-140}=\dfrac{-1}{7}\)
\(\dfrac{3.10}{5.24}=\dfrac{30}{120}=\dfrac{1}{4}\)
\(\dfrac{3.7.11}{22.9}=\dfrac{231}{198}=\dfrac{7}{6}\)
\(\dfrac{8.5-8.2}{16}=\dfrac{24}{16}=\dfrac{3}{2}\)
\(\dfrac{11.4-11}{2-13}=\dfrac{33}{-11}=-3\)
a. \(\dfrac{8,5-8,2}{16}=\dfrac{0,3}{16}=\dfrac{3}{160}\)
b. \(\dfrac{2\cdot14}{7\cdot8}=\dfrac{1\cdot2}{1\cdot4}=\dfrac{2}{4}=\dfrac{1}{2}\)
c. \(\dfrac{11\cdot4-11}{2-13}=\dfrac{11\left(4-1\right)}{-11}=\dfrac{1\cdot3}{-1}=-3\)
d. \(\dfrac{49+7\cdot49}{49}=\dfrac{49\cdot\left(1+7\right)}{49}=\dfrac{8}{1}=8\)
*Rút gọn :
`(8.5-8.2)/16=(8.(5-2))/(8.2)=(8.3)/(8.2)=3/2`
`(11.4-11)/(2-13)=(11.(4-1))/(-11)=(11.3)/(-1.11)=3/(-1)=-3`
*Quy đồng :
- Ta có : `-3=(-3)/1`
UCLN(2,1)=2
$\to\begin{cases} \dfrac32=\dfrac32\\\dfrac{-3}1=\dfrac{-6}2\end{cases}$
\(\frac{-1997\cdot1996+1}{-1995\cdot\left(-1997\right)+1996}\)
\(=\frac{-1997\cdot\left(1995+1\right)+1}{1995\cdot1997+1996}\)
\(=\frac{-1997\cdot1995+\left(-1997\right)+1}{1995\cdot1997+1996}\)
\(=\frac{-1997\cdot1995+\left(-1996\right)}{1995\cdot1997+1996}\)
\(=-1\)
\(\frac{-1997.1996+1}{-1995.\left(-1997\right)+1996}\)
\(=\frac{-1997.1996+1}{1995.1997+1996}\)
\(=\frac{-1997.\left(1995+1\right)+1}{1995.1997+1996}\)
\(=\frac{-1997.1995+-1997+1}{1995.1997+1996}\)
\(=\frac{1995.-1997+-1996}{1995.1997+1996}\)
\(=\frac{-\left(1995.1997+1996\right)}{1995.1997+1996}\)
\(=-1\)
a/ \(-\frac{81}{504}=-\frac{9}{56};-\frac{10101}{101010}=\frac{-1}{10}\)
b/ \(\frac{72\cdot5+72\cdot3}{144\cdot2+144\cdot6}=\frac{72\left(5+3\right)}{144\left(2+6\right)}=\frac{8}{2\cdot8}=\frac{1}{2}\)
\(\frac{-81}{504}=-\frac{81:9}{504:9}=-\frac{9}{56}\)
\(\frac{-10101}{101010}=\frac{-10101:10101}{101010:10101}=\frac{-1}{10}\)
\(\frac{72.5+72.3}{144.2+144.6}=\frac{72.\left(5+3\right)}{144\left(2+6\right)}=\frac{72.8}{144.8}=\frac{72.8}{72.2.8}=\frac{1}{2}\)
\(445-52+155+452\\ =\left(445+155\right)+\left(452-52\right)\\ =600+4000\\ =1000\\ 102\cdot55-102\cdot73+51\cdot36\\ =102\cdot\left(55-73\right)+51\cdot36\\ =102\cdot\left(-18\right)+102\cdot18\\ =0\\ 4\cdot5\cdot2\cdot25\cdot5-1000\\ =5000-1000\\ =4000\\ 8782-291\cdot13\\ =4999\\ 254+\left\{38\cdot\left[\left(42-16\right):2\right]\right\}\\ =254+\left[38\cdot\left(26:2\right)\right]\\ =254+\left(38\cdot13\right)\\ =748\\ \left(2791\cdot34+7882:14\right)\cdot0+1510-510:2\\ =1510-255\\ =1255\)
a. \(445-52+155+452=\left(445+155\right)+\left(452-52\right)=600+400=1000\)
b. \(102\cdot55-102\cdot73+51\cdot36=102\left(55-73+18\right)=102\cdot0=0\)
c.\(4\cdot5\cdot2\cdot25\cdot5-1000=4\cdot25\cdot5\cdot5\cdot2-1000=100\cdot50-1000=5000-1000=4000\)
d. \(8782-291\cdot13=8782-3783=4999\)
e. \(254+\left\{38\left[\left(42-16\right)\div2\right]\right\}=254+\left\{30\cdot13\right\}=254+494=748\)
f. \(\left(2791\cdot34+7882\div14\right)\cdot0+1510-510\div2=0+1510-255=1255\)\
3/2=1,5
cách giải thế nào đã rồi mình tick cho