B= 2^27 : 4^12
K
Khách
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TK
3
23 tháng 2 2019
\(b)4\cdot(-5)^2+2\cdot(-5)-20\)
\(=4\cdot25+2\cdot(-5)-20\)
\(=100+(-10)-20=70\)
\(c)27\cdot(15-12)-15\cdot(27-12)\)
\(=27\cdot15-27\cdot12-15\cdot27-15\cdot12\)
\(=(27\cdot15-15\cdot27)-27\cdot12-15\cdot12\)
\(=0-(27-15)\cdot12=0-12\cdot12=0-144=-144\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
26 tháng 9 2023
\(b=\text{}\dfrac{15.2^{24}.3^{14}-4.3^{15}.2^{24}}{19.2^{24}.3^{14}-6.2^{24}.3^{15}}=\)
\(=\dfrac{2^{24}.3^{14}\left(15-4.3\right)}{2^{24}.3^{14}\left(19-6.3\right)}=3\)
H
1
VN
3 tháng 3 2020
a) 27.(15-12)-15.(27-12)
= 27.15-27.12-15.27-15.12
=405 - 324 - 405 - 180
= -504
NH
22 tháng 8 2020
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)
22 tháng 8 2020
\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)
\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)
\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)
\(C=\frac{4}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 10 2023
a) \(4^8\cdot4^4=\left(2^2\right)^8\cdot\left(2^2\right)^4=2^{16}\cdot2^8=2^{16+8}=2^{24}\)
b) \(5^{12}\cdot7-5^{11}\cdot10\)
\(=5^{11}\cdot\left(5\cdot7-10\right)=5^{11}\cdot\left(35-10\right)=5^{11}\cdot25\)
\(=5^{11}\cdot5^2=5^{11+2}=5^{13}\)
d) \(27^{16}:9^{10}\)
\(=\left(3^3\right)^{16}:\left(3^2\right)^{10}=3^{48}:3^{20}=3^{48-20}=3^{28}\)
e) \(125^3:25^4=\left(5^3\right)^3:\left(5^2\right)^4=5^9:5^8=5^{9-8}=5\)
f) \(24^4:3^4-32^{12}:16^{12}\)
\(=\left(24:4\right)^4-\left(32:16\right)^{12}\)
\(=6^4-2^{12}\)
\(=2^4\cdot\left(3^4-2^8\right)=2^4\cdot-175=-2800\)
16 tháng 3 2020
a) Giải ko đc. Kiểm tra lại đề
b) 27.15 + 85.27 - 16.12
= 27(15 + 85) - 16.(10 + 2)
= 27.100 - [(16 . 10)+(16 . 2)
= 2700 - [160 + 32]
= 2700 - 192 = 2508
c) (-12) + 83 + (-48) + 17
= [(-12) + (-48)] + (83 + 17)
= [-70] + 100 = 100 - 70 = 30
MAY MẮN NHÉ!
DT
Dang Tung
CTVHS
20 tháng 12 2023
a) \(x+546=46\\ x=46-546\\ x=-500\)
b) \(2x-19\times3=27\\ 2x-57=27\\ 2x=27+57\\ 2x=84\\ x=84:2\\ x=42\)
c) \(x+12=23+3\times3^4\\ x+12=23+3\times81\\ x=23+243-12\\ x=254\)
DT
Dang Tung
CTVHS
20 tháng 12 2023
d) \(x-12=3-3\times2^4\\ x-12=3-3\times16\\ x=3-48+12\\ x=-33\)
e) \(\left(27-x\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}27-x=0\\x+9=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=27\\x=-9\end{matrix}\right.\)
f) \(\left(-x\right)\left(x-43\right)=0\\ \Rightarrow\left[{}\begin{matrix}-x=0\\x-43=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=43\end{matrix}\right.\)
B = 2^27 : 4^12 = 2^27 : (2^2)^12
= 2^27 : 2^(2x12)
=2^27 : 2^24
=2^3
=8