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1253:254
Ta có: 1253 = (53)3 = 59
254 = (52)4 = 58
Vậy: 59:58 =51
Nên: 1253:254=51
1)\(1\frac{3}{4}-1\frac{3}{8}:\left(\frac{3}{8}+\frac{9}{20}\right)=\frac{7}{4}-\frac{11}{8}:\frac{33}{40}=\frac{7}{4}-\frac{5}{3}=\frac{1}{12}\)
a)( \(\dfrac{-8}{40}\)+\(\dfrac{25}{40}\))+(\(\dfrac{-3}{8}\))
=\(\dfrac{17}{40}\)+(\(\dfrac{-3}{8}\))
=\(\dfrac{17}{40}\)+(\(\dfrac{-15}{40}\))
=\(\dfrac{2}{40}\)
=\(\dfrac{1}{20}\)
b)\(\dfrac{-5}{21}\)-(\(\dfrac{16}{21}\)-\(\dfrac{21}{21}\))
=\(\dfrac{-5}{21}\)-\(-5\)
=\(\dfrac{-25}{105}\)-\(\dfrac{-105}{105}\)
=\(\dfrac{16}{21}\)
bằng đây thôi nha mỏi tay quá
c: =3/7(-5/11+6/11)=3/7*1/11=3/77
d:=8/-3(4/3-7/3)=8/3
e: =3,4-5/8-0,4-3/8
=2-1
=1
a, \(\dfrac{x}{2}+\dfrac{3x}{4}=\dfrac{4}{5}\Leftrightarrow\dfrac{10x+15x}{20}=\dfrac{16}{20}\Rightarrow25x=16\Leftrightarrow x=\dfrac{16}{25}\)
b, \(\dfrac{3}{7}.\dfrac{5}{8}-\dfrac{3}{8}.\dfrac{13}{8}+\dfrac{1}{7}=\dfrac{15}{56}-\dfrac{39}{64}+\dfrac{1}{7}\)
\(=\dfrac{120}{448}-\dfrac{273}{448}+\dfrac{64}{448}=-\dfrac{89}{448}\)
1*2*3*...*9-1*2*3*...*8-1*2*3*...8*8=9!-8!-8!*8
=8!*9-8!*1-8!*8
=8!*(9-1)-8!*8
=8!*8-8!*8
=0
1
8/8+8/8-3/3*1/1=1+1-1*1=1+1-1=2-1=1