Chứng minh rằng
(5^7-5^6+5^5) chia hết cho 21
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(x - 45) x 27 = 0
=> x - 45 = 0 : 27
=> x - 45 = 0
=> x = 0 + 45
=> x = 45
Vậy: ..
`(x - 45) . 27 = 0`
`=> x - 45 = 0 : 27`
`=> x - 45 = 0`
`=> x = 45`
Vậy `x = 45`
Bài 3:
\(a,x-\dfrac{2}{5}=0,24\\ =>x-\dfrac{2}{5}=\dfrac{6}{25}\\ =>x=\dfrac{6}{25}+\dfrac{2}{5}\\ =>x=\dfrac{16}{25}\\ b,\left(\dfrac{7}{3}x-0,6\right):3\dfrac{2}{5}=1\\ =>\left(\dfrac{7}{3}x-\dfrac{3}{5}\right):\dfrac{17}{5}=1\\ =>\dfrac{7}{3}x-\dfrac{3}{5}=\dfrac{17}{5}\\ =>\dfrac{7}{3}x=\dfrac{17}{5}+\dfrac{3}{5}=4\\ =>x=4:\dfrac{7}{3}=\dfrac{12}{7}\\ c,\left(2\dfrac{4}{5}x-50\right):\dfrac{2}{3}=51\\ =>\dfrac{14}{5}x-50=\dfrac{2}{3}\cdot51=34\\ =>\dfrac{14}{5}x=34+50=84\\ =>x=84:\dfrac{14}{5}=30\)
Bài 4:
a: \(5\dfrac{4}{7}:x=13\)
=>\(\dfrac{39}{7}:x=13\)
=>\(x=\dfrac{39}{7}:13=\dfrac{3}{7}\)
b: \(6\dfrac{2}{9}x+3\dfrac{10}{27}=22\dfrac{1}{7}\)
=>\(\dfrac{56}{9}x=22+\dfrac{1}{7}-3-\dfrac{10}{27}=19+\dfrac{-43}{189}=\dfrac{3548}{189}\)
=>\(x=\dfrac{3548}{189}:\dfrac{56}{9}=\dfrac{887}{294}\)
c: \(\left(\dfrac{7}{3}x-0,6\right):3\dfrac{2}{5}=1\)
=>\(\left(\dfrac{7}{3}x-0,6\right)=1\cdot3\dfrac{2}{5}=3,4\)
=>\(\dfrac{7}{3}x=3,4+0,6=4\)
=>\(x=4:\dfrac{7}{3}=\dfrac{12}{7}\)
d: \(\left(2\dfrac{4}{5}x-50\right):\dfrac{2}{3}=51\)
=>\(\left(2,8x-50\right)=51\cdot\dfrac{2}{3}=34\)
=>2,8x=34+50=84
=>\(x=\dfrac{84}{2,8}=30\)
e:
\(\left(4\dfrac{1}{2}-2x\right)\cdot3\dfrac{2}{3}=\dfrac{11}{15}\)
=>\(\left(\dfrac{9}{2}-2x\right)\cdot\dfrac{11}{3}=\dfrac{11}{15}\)
=>\(\dfrac{9}{2}-2x=\dfrac{11}{15}:\dfrac{11}{3}=\dfrac{3}{15}=\dfrac{1}{5}\)
=>\(2x=\dfrac{9}{2}-\dfrac{1}{5}=\dfrac{45}{10}-\dfrac{2}{10}=\dfrac{43}{10}\)
=>\(x=\dfrac{43}{20}\)
a: \(5^{56}=\left(5^7\right)^8=78125^8\)
\(11^{24}=\left(11^3\right)^8=1331^8\)
mà 78125>1331
nên \(5^{56}>11^{24}\)
b: \(5^{23}=5\cdot5^{22}< 6\cdot5^{22}\)
c:
\(7\cdot2^{13}< 8\cdot2^{13}=2^3\cdot2^{13}=2^{16}\)
a: \(\dfrac{21}{25}\cdot\dfrac{11}{9}\cdot\dfrac{5}{7}=\dfrac{21}{7}\cdot\dfrac{5}{25}\cdot\dfrac{11}{9}\)
\(=3\cdot\dfrac{11}{9}\cdot\dfrac{1}{5}=\dfrac{11}{3}\cdot\dfrac{1}{5}=\dfrac{11}{15}\)
b: \(\dfrac{5}{23}\cdot\dfrac{17}{26}+\dfrac{5}{23}\cdot\dfrac{9}{26}\)
\(=\dfrac{5}{23}\left(\dfrac{17}{26}+\dfrac{9}{26}\right)\)
\(=\dfrac{5}{23}\cdot1=\dfrac{5}{23}\)
c: \(\dfrac{7}{13}\cdot\dfrac{5}{19}+\dfrac{7}{19}\cdot\dfrac{8}{13}-3\dfrac{7}{19}\)
\(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}\right)-3-\dfrac{7}{19}\)
\(=\dfrac{7}{19}-3-\dfrac{7}{19}=-3\)
\(a,\dfrac{21}{25}.\dfrac{11}{9}.\dfrac{5}{7}\)
\(=\dfrac{21.11.5}{25.9.7}\)
\(=\dfrac{1.11.1}{5.3.1}\)
\(=\dfrac{11}{15}\)
\(b,\dfrac{5}{23}.\dfrac{17}{26}+\dfrac{5}{23}.\dfrac{9}{26}\)
\(=\dfrac{5}{23}.\left(\dfrac{17}{26}+\dfrac{9}{26}\right)\)
\(=\dfrac{5}{23}.1\)
\(=\dfrac{5}{23}\)
\(c,\dfrac{7}{13}.\dfrac{5}{19}+\dfrac{7}{19}.\dfrac{8}{13}\)
\(=\dfrac{7}{19}.\dfrac{5}{13}+\dfrac{7}{19}.\dfrac{8}{13}\)
\(=\dfrac{7}{19}.\left(\dfrac{5}{13}+\dfrac{8}{13}\right)\)
\(=\dfrac{7}{19}.1\)
\(=\dfrac{7}{19}\)
Ngày thứ nhất đội đào được: \(28\cdot\dfrac{1}{4}=7\left(m^3\right)\)
Sau ngày thứ nhất còn lại 28-7=21(m3)
Ngày thứ hai đội đào được: \(21\cdot\dfrac{3}{7}=9\left(m^3\right)\)
Ngày thứ ba đội đào được: 21-9=12(m3)
Số phần đất đội đào trong ngày thứ hai là:
\(\dfrac{3}{7}.\left(1-\dfrac{1}{4}\right)=\dfrac{9}{28}\)
Số phần đất còn lại là:
\(1-\left(\dfrac{1}{4}+\dfrac{9}{28}\right)=\dfrac{3}{7}\)
Ngày thứ 3 đội phải đào là:
\(28.\dfrac{3}{7}=12\left(m^3\right)\)
\(7\cdot4^{x-1}+4^{x+1}=23\)
=>\(7\cdot4^x\cdot\dfrac{1}{4}+4^x\cdot4=23\)
=>\(4^x\left(\dfrac{7}{4}+4\right)=23\)
=>\(4^x=23:\dfrac{23}{4}=4\)
=>x=1
\(119\times24-53\times23-24\times66\)
\(=\left(119-66\right)\times24-53\times23\)
\(=53\times24-53\times23\)
\(=53\times\left(24-23\right)\)
\(=53\times1\)
\(=53\)
\(119\cdot24-53\cdot23-24\cdot66\)
\(=24\left(119-66\right)-53\cdot23\)
\(=53\cdot24-53\cdot23=53\)
\(3^{x+2}-5.3^x\\ =3^x\left(3^2-5\right)\\ =3^x.\left(9-5\right)\\ =4.3^x\)
\(5^7-5^6+5^5\\ =5^5\cdot5^2-5^5\cdot5+5^5\\ =5^5\cdot\left(5^2-5+1\right)\\ =5^5\cdot\left(25-5+1\right)\\ =5^5\cdot21⋮21\)
=> `5^7-5^6+5^5` chia hết cho 21
`5^7 - 5^6 + 5^5`
`= 5^5 . (5^2 - 5 + 1)`
`= 5^5 . (25 - 5 + 1)`
`= 5^5 . 21 vdots 21 (đpcm)`