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a) A = 1 + ( - 2 ) + 3 + ... + ( - 98 ) + 99
A = 1 + [ ( - 2 ) + 3 ] + ... + [ ( - 98 ) + 99 ]
A = 1 + 1 + ... + 1
A = 1 . 50
A = 50
miucool:
A = 1 + (-2) + 3 + ... + (-98) + 99
A = 1 + [(-2) + 3)] + ... + [(-98) + 99)]
A = 1 + 1 + ... + 1
Có 50 cặp số:
A = 1 . 50
A = 50
Vậy A =50
(^_^)
a: \(=\left(\dfrac{2}{18}-\dfrac{15}{18}-\dfrac{72}{18}\right):\left(\dfrac{21}{36}-\dfrac{1}{36}-\dfrac{360}{36}\right)\)
\(=\dfrac{-85}{18}:\dfrac{-170}{18}\)
\(=\dfrac{85}{170}=\dfrac{1}{2}\)
b: \(=\left(\dfrac{5}{8}-\dfrac{5}{6}-\dfrac{5}{32}+\dfrac{5}{64}\right):\left(1-\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}\right)\)
\(=\dfrac{-55}{192}:\dfrac{3}{8}=\dfrac{-55}{192}\cdot\dfrac{8}{3}=-\dfrac{55}{72}\)
a)
6x+70 =570-440
6x+70 =130
6x =130-70
6x =60
x =60:6
x =10
b)^ là mũ nhé!
(2^x+1).2^x=1000+24
(2^x+1).2^x=1024
2^x.2.2^x=1024
2^x.2^x=1024:2=512
2.2^x=512
2^x=512:2=256
2^x=2^8
x=8
\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=2.\dfrac{3}{16}\)
\(A=\dfrac{3}{8}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{111}\)
\(B=\dfrac{12}{37}\)
\(a,4^{21}:16^5.\)
\(=4^{21}:\left(4^2\right)^5.\)
\(=4^{21}:4^{10}.\)
\(=4^{21-10}=4^{11}.\)
Vậy.....
\(b,32^8:4^{19}.\)
\(=\left(2^5\right)^8:\left(2^2\right)^{19}.\)
\(=2^{40}:2^{38}.\)
\(=2^{40-38}.\)
\(=2^2=4.\)
Vậy.....
\(c,27^{15}:9^{22}.\)
\(=\left(3^3\right)^{15}:\left(3^2\right)^{22}.\)
\(=3^{45}:3^{44}.\)
\(=3^{45-44}.\)
\(=3^1=3.\)
Vậy.....
\(d,25^{10}:125^6.\)
\(=\left(5^2\right)^{10}:\left(5^3\right)^6.\)
\(=5^{20}:5^{18}.\)
\(=5^{20-18}.\)
\(=5^2=25.\)
Vậy.....
~ Hok tốt!!! ~ :))
a, 421 : 165
= 421 : (42 )5
= 421 : 410
= 411
b, 328 : 419
= (25)8 : (22 )19
= 240 : 238
= 22
c, 2715 : 922
= (33 ) 15 : (32 )22
= 345 : 344
= 3
d, 2510 : 1256
= (52)10 : (53)6
= 520 : 518
= 52
Cho e hỏi cái này. Ở câu 1 ý, cuối đề là \(-\frac{1}{7}\) sao xuống dưới phải đổi thành -1 thế ạ ? E chưa hiểu lắm :<
Bài 1 :
a ) ( 2637 - n ) - ( 2\(^{10}\) - 7 ) = 15\(^2\) - 20
( 2637 - n ) - 1024 = 205
2637 - n = 205 + 1024
2637 - n =1229
n = 2637 - 1229
n =1408
b) n\(^3\) = n\(^9\)
<=> n = 1 hoặc n = 0
Vì nếu n > 1 => n khi nâng nên luỹ thừa 9 sẽ lớn hơn khi nâng lên luỹ thừa 3
Nếu n < 0 => n khi nâng nên luỹ thừa 3 sẽ lớn hơn hơn khi nâng lên luỹ thừa 9 .
Bài 2 : So sánh
a) 2\(^{15}\) và 3\(^{10}\)
2\(^{15}\) = \(\left(2^3\right)^5\) = 8\(^5\)
3\(^{10}\) = \(\left(3^2\right)^5\) = 9\(^5\)
Vì 9\(^5\) > 8\(^5\) nên \(3^{10}>2^{15}\)
b) 7 x 2\(^{2017}\) và 2\(^{2020}\)
Ta có : 7 x 2\(^{2017}\) < 8 x 2\(^{2017}\)
Mà 8 x \(2^{2017}\) = 2\(^3\) x 2\(^{2017}\) = 2\(^{2020}\)
Vậy : 7 x 2\(^{2017}\) < 2\(^{2020}\)
c) 21\(^{15}\) và 27\(^5\) x 49\(^8\)
21\(^{15}\) = 7\(^{15}\) x 3\(^{15}\)
27\(^5\) x 49\(^8\) = \(\left(3^3\right)^5\) x \(\left(7^2\right)^8\) = 3\(^{15}\) x 7\(^{16}\)
So sánh : 7\(^{15}\) x 3\(^{15}\) và 7\(^{16}\) x 3\(^{15}\)
=> 7\(^{16}\) x 3\(^{15}\) > \(7^{15}\) x 3\(^{15}\) . Vì 3\(^{15}\) = 3\(^{15}\) mà 7\(^{16}\) > 7\(^{15}\) => 7\(^{16}\) x 3\(^{15}\) > 7\(^{15}\) x 3\(^{15}\)
Vậy : 21\(^{15}\) < 27\(^5\) x 49\(^8\)
a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)
=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)
=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)
=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)
b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)
=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)
c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)
=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)
= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)
d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)
=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)
=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)
e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
b)Có \(63^7< 64^7\)
\(64^7=\left(2^6\right)^7=2^{42}\)
\(16^{12}=\left(2^4\right)^{12}=2^{48}\)
Mà \(2^{42}< 2^{48}\Rightarrow63^7< 64^7< 16^{12}\Rightarrow63^7< 16^{12}\)
B = 1 + ( -4 ) + 7 + ( -10 ) + ... + 319 + ( -322 ) + 32
B - 32 = [ 1 + ( -4 )] + [ 7 + ( -10 )] + ... + [ 319 + ( -322 )] có 54 nhóm
= 54 ( -3 ) = -162
B = ( -162 ) + 32 = -130
Vậy B = -130
B = 1 + ( -4 ) + 7 + ( -10 ) + ... + 319 + ( -322 ) + 32
B - 32 = [ 1 + ( -4 )] + [ 7 + ( -10 )] + ... + [ 319 + ( -322 )] có 54 nhóm
= 54 ( -3 ) = -162
B = ( -162 ) + 32 = -130
Vậy B = -130
BÀI NÀY MÌNH THAM KHẢO TRÊN MẠNG NHOOO