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\(B=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{51}\right)=\)\(\frac{250}{51}\)
\(B=5\left(\frac{1}{1}-\frac{1}{51}\right)=\frac{250}{51}\)
\(B=5\left(\frac{51}{51}-\frac{1}{51}\right)=\frac{250}{51}\)
\(B=5.\frac{50}{51}=\frac{250}{51}\)
\(B=\frac{250}{51}=\frac{250}{51}\)
\(\dfrac{13}{4}\times\dfrac{2}{3}\times\dfrac{4}{13}\times\dfrac{3}{2}\\ =\left(\dfrac{13}{4}\times\dfrac{4}{13}\right)\times\left(\dfrac{2}{3}\times\dfrac{3}{2}\right)\\ =1\times1\\ =1\)
Giải:
\(\dfrac{13}{4}.\dfrac{2}{3}.\dfrac{4}{13}.\dfrac{3}{2}\)
\(=\left(\dfrac{13}{4}.\dfrac{4}{13}\right)\left(\dfrac{2}{3}.\dfrac{3}{2}\right)\)
\(=\left(\dfrac{13.4}{4.13}\right)\left(\dfrac{2.3}{3.2}\right)\)
\(=1.1=1\)
Vậy giá trị của biểu thức trên là 1.
Chúc bạn học tốt!
a, \(\left(2\dfrac{3}{5}-3\dfrac{5}{9}\right):\left(3\dfrac{10}{21}-1\dfrac{3}{7}\right)\)
\(=\dfrac{-43}{45}:\dfrac{43}{21}=\dfrac{-43}{45}.\dfrac{21}{43}=\dfrac{-7}{15}\)
b, \(5\dfrac{1}{2}-14\dfrac{3}{7}:\dfrac{9}{13}-3\dfrac{4}{7}:\dfrac{9}{13}\)
\(=5\dfrac{1}{2}-14\dfrac{3}{7}.\dfrac{13}{9}-3\dfrac{4}{7}.\dfrac{13}{9}\)
\(=5\dfrac{1}{2}-\dfrac{13}{9}.\left(14\dfrac{3}{7}+\dfrac{4}{7}\right)\)
\(=5\dfrac{1}{2}-\dfrac{13}{9}.15=5\dfrac{1}{2}-\dfrac{65}{3}\)
\(=\dfrac{-97}{6}\)
Chúc bạn học tốt!!!
a) \(2^{x+2}-2^x=96\)\(\Leftrightarrow2^x.4-2^x=96\)
\(\Leftrightarrow2^x\left(4-1\right)=96\)\(\Leftrightarrow2^x.3=96\)\(\Leftrightarrow2^x=32=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(10^6-5^7=\left(2.5\right)^6-5^{6+1}=2^6.5^6-5^6.5=5^6\left(2^6-5\right)=5^6\left(64-5\right)=5^6.59⋮59\)
c) \(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}\)
\(=3^{24}\left(3^4-3^3-3^2\right)=3^{24}\left(81-27-9\right)=3^{24}.45⋮45\)
1. Ta có: 2x + 2 - 2x = 96
=> 2x.4 - 2x = 96
=> 2x(4 - 1) = 96
=> 2x = 96 : 3
=> 2x = 32
=> 2x = 25
=> x = 5
2. Ta có: 106 - 57 = (2.5)6 - 57 = 26.56 - 56.5 = 56(64 - 5) = 56 . 59 \(⋮\)59
817 - 279 - 913 = (34)7 - (33)9 - (32)13 = 328 - 327 - 326 = 324.(34 - 33 - 32) = 224. 45 \(⋮\)45
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
a) 378
b) 3
c) 2
d) 2
e) \(\frac{8748}{1715}\)
Mình thấy bài e) bạn có ghi thiếu ko vậy.81^2 x;: hay là cộng trừ vậy?
4x 52 + 81:32 - (13-4)2
= 208 + 81:9- 92
= 208 + 9 - 81
= 217 - 81
=136
tinh nhanh nhe