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b: Ta có: \(1\frac{13}{15}\cdot\left(0,5\right)^2\cdot3+\left(\frac{8}{15}-1\frac{19}{60}\right):1\frac{23}{24}\)
\(=\frac{28}{15}\cdot3\cdot\frac14+\left(\frac{8}{15}-\frac{79}{60}\right):\frac{47}{24}\)
\(=\frac{28}{5}\cdot\frac14+\left(\frac{32}{60}-\frac{79}{60}\right)\cdot\frac{24}{47}=\frac75+\frac{-47}{60}\cdot\frac{24}{47}\)
\(=\frac75-\frac25=\frac55=1\)
c: \(\left(-2\right)^3\cdot\frac{-1}{24}+\left(\frac45-1,2\right):\frac{2}{15}\)
\(=\left(-8\right)\cdot\frac{-1}{24}+\left(\frac45-\frac65\right)\cdot\frac{15}{2}\)
\(=\frac13+\frac{-2}{5}\cdot\frac{15}{2}=\frac13-3=-\frac83\)
d: \(\left(-\frac25\right)^2+\frac12\cdot\left(4,5-2\right)-255\)
\(=\frac{4}{25}+\frac12\cdot2,5-255\)
=0,16+1,25-255
=-253,59
(-2)3 . (-\(\dfrac{1}{24}\)) + ( \(\dfrac{4}{5}\) - 1,2): \(\dfrac{2}{15}\)
= -8 .(-\(\dfrac{1}{24}\)) + ( \(\dfrac{4}{5}\) - \(\dfrac{6}{5}\)) \(\times\) \(\dfrac{15}{2}\)
= \(\dfrac{1}{3}\) - \(\dfrac{2}{5}\) \(\times\) \(\dfrac{15}{2}\)
= \(\dfrac{1}{3}\) - 3
= -\(\dfrac{8}{3}\)
2) 132 - 52 = ( 13 - 5 )( 13 + 5 ) = 8 x 18 = 4 x 4 x 3 x 3 = 122
`a)5/8x+2/5=1/5`
`=>5/8x=1/5-2/5`
`=>5/8x=-1/5`
`=>x=-1/5:5/8=-8/25`
`b)5/7:x+11/7=18/7`
`=>5/7:x=18/7-11/7`
`=>5/7:x=1`
`=>x=5/7`
`c)(-1,2).(-3/24)+(0,4-1 4/15):1 2/3`
`=(-6/5).(-1/8)+(2/5-19/15):5/3`
`=3/20+(-13/15)*3/5`
`=3/20-13/25=-37/100`
a)5/8.x+2/5=1/5
5/8.x=1/5 - 2/5
5/8.x=-1/5
x=(-1/5):5/8
x=(-1/5).8/5
x=-8/25. Vậy x=-8/25
b)5/7:x +11/7=18/7
5/7:x=1
x=5/7:1
x=5/7. Vậy x=5/7
\(24\cdot5^5+5^2\cdot5^3\)
\(=24\cdot5^5+5^{2+3}\)
\(=24\cdot5^5+5^5\)
\(=5^5\cdot\left(24+1\right)\)
\(=5^5\cdot25\)
\(=5^5\cdot5^2\)
\(=5^7\)
`@` `\text {Ans}`
`\downarrow`
`24 * 5^5 + 5^2 * 5^3`
`= 24*5^5 +`\(5^{2+3}\)
`= 24*5^5 + 5^5`
`= 5^5* (24+1)`
`= 5^5*25`
`= 5^5*5^2`
`= 5^7`