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\(\frac{3}{5}.\frac{8}{27}.\frac{5}{3}=1.\frac{8}{27}.1=\frac{8}{27}\)
\(\frac{7}{19}.\frac{1}{3}+\frac{7}{19}.\frac{2}{3}=\frac{7}{19}.\left(\frac{1}{3}+\frac{2}{3}\right)=\frac{7}{19}.1=\frac{7}{19}\)
\(\frac{12}{5}.4-4.\frac{7}{5}=4\left(\frac{12}{5}-\frac{7}{5}\right)=4.1=4\)
\(\frac{3}{5}x\frac{8}{27}x\frac{5}{3}\)
\(=\frac{3}{5}x\frac{5}{3}x\frac{8}{27}\)
\(=1x\frac{8}{27}\)
\(=\frac{8}{27}\)
\(\frac{7}{19}x\frac{1}{3}+\frac{7}{19}x\frac{2}{3}\)
\(=\frac{7}{19}x\left(\frac{1}{3}+\frac{2}{3}\right)\)
\(=\frac{7}{19}x1=\frac{7}{19}\)
\(\frac{12}{5}x4-4x\frac{7}{5}\)
\(=4x\left(\frac{12}{5}-\frac{7}{5}\right)\)
\(=4x1=4\)
Đúng luôn nên các bn nhớ k mk nhé
a) \(x-\frac{10}{3}=\frac{7}{15}\cdot\frac{3}{5}\) b) \(x+\frac{3}{22}=\frac{27}{121}\cdot\frac{11}{9}\)
\(\Leftrightarrow x-\frac{10}{3}=\frac{7}{25}\) \(\Leftrightarrow x+\frac{3}{22}=\frac{3}{11}\)
\(\Rightarrow x=\frac{7}{25}+\frac{10}{3}\) \(\Rightarrow x=\frac{3}{11}-\frac{3}{22}\)
\(x=\frac{271}{75}\) \(x=\frac{3}{22}\)
c) \(\frac{8}{23}.\frac{46}{24}-x=\frac{1}{3}\) d) \(1-x=\frac{49}{65}.\frac{5}{7}\)
\(\Leftrightarrow\frac{2}{3}-x=\frac{1}{3}\) \(\Leftrightarrow1-x=\frac{7}{13}\)
\(\Rightarrow x=\frac{2}{3}-\frac{1}{3}\) \(\Rightarrow x=1-\frac{7}{13}\)
\(x=\frac{1}{3}\) \(x=\frac{6}{13}\)
45/7 : 3/5 + 1 = 82/7
3/5 x 8/9 = 8/15
55 - 54 + 53
= 53 ( 25 - 5 + 1 )
= 53. 21
Mà 21 ⋮ 7 ⇒ 55 - 54 + 53 ⋮ 7
Tìm số tự nhiên n , sao cho :
a, \(n+4⋮n\)
b, \(3\times n+7⋮n\)
c, \(27-5\times n⋮n\)
d, \(n+6⋮n+2\)
a: =>4 chia hết cho n
=>\(n\in\left\{1;2;4\right\}\)
b: =>3n+7 chia hết cho n
=>7 chia hết cho n
=>\(n\in\left\{1;7\right\}\)
c: 27-5n chia hết cho n
=>27 chia hết cho n
=>\(n\in\left\{1;3;9;27\right\}\)
d: =>n+2+4 chia hết cho n+2
mà n là số tự nhiên
nên \(n+2\in\left\{2;4\right\}\)
hay \(n\in\left\{0;2\right\}\)
a) Có: \(3+3^2+3^3+3^4+...+3^{99}\\ =\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\\ =\left(3+3^2+3^3\right)+3^3\left(3+3^2+3^3\right)+...+3^{97}\left(3+3^2+3^3\right)\\ =39+3^3\cdot39+...+3^{97}\cdot39\\ =13\cdot3+3^3\cdot13\cdot3+...+3^{97}\cdot13\cdot3\\ =13\left(3+3^4+...+3^{98}\right)⋮13\left(đpcm\right)\)
b) Có: \(81^7-27^9-9^{13}\\ =\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\\ =3^{28}-3^{27}-3^{26}\\ =3^{26}\left(3^2-3-1\right)\\ =3^{24}\cdot\left(3^2\cdot5\right)\\ =3^{24}\cdot45⋮45\left(đpcm\right)\)
c) Có: \(24^{54}\cdot54^{24}\cdot2^{10}\\ =\left(2^3\cdot3\right)^{54}\cdot\left(2\cdot3^3\right)^{24}\cdot2^{10}\\ =2^{162}\cdot3^{54}\cdot2^{24}\cdot3^{72}\cdot2^{10}\\ =2^{196}\cdot3^{126}\\ =2^7\cdot\left(2^{189}\cdot3^{126}\right)\\ =2^7\cdot\left[\left(2^3\right)^{63}\cdot\left(3^2\right)^{63}\right]\\ =2^7\left(8^{63}\cdot9^{63}\right)\\ =2^7\cdot72^{63}⋮72^{63}\left(đpcm\right)\)
a) ta có: 3 + 32 + 33 + 34 + ... + 399
= (3 + 32 + 33) + (34 + 35 +36) + ... + (397 + 398 + 399)
= 3(1 + 3 + 32) + 34(1 + 3 + 3) + ... + 396(1 + 3 + 3)
= 3.13 + 34.13 + ... + 396.13
= 13(3 + 34 + ... + 396) ⋮ 13
vậy (3 + 32 + 33 + 34 + ... + 399) ⋮ 13
b) ta có: 817 - 279 - 913
= (34)7 - (33)9 - (32)13
= 328 - 327 - 326
= 326(32 - 3 - 1)
= 326 . 5 = 324 (9.5) = 324 . 45 ⋮ 45
Vậy (817 - 279 - 913) ⋮ 45
c) ta có: 2454.5424.210
= (23.3)54 . (2.33)24 . 210
= 2162 . 354 . 224 . 372 . 210
= 2196 . 3126
= (2193.3124).(23.32)
= (2193.3124).72 ⋮ 72
vậy (2454.5424.210) ⋮ 72
273 x 345= (33)3 x 345 = 33.3 x 345 = 39 x 345 = 39+45 = 354.