*Mình cần gấp ạ !
Tính giá trị biểu thức
a) ( 1000-13) . ( 1000-23) : ( 1000-33) ...( 1000 -503)
b) (1/125-1/13) . (1/125-1/23).( 1/125-1/33)...( 1/125-1/253)
K
Khách
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Những câu hỏi liên quan
NN
1
NN
21 tháng 8 2020
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
PN
1
28 tháng 10 2021
a)[(-15).8]:4
=-120:4
=30
b)[(-125):(-5)].(-13)
=25.(-13)
=325
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1
\(log_5125=log_55^3=3\)
\(log_6216=log_66^3=3\)
\(log_{10}\dfrac{1}{10000}=log_{10}10^{-4}=-4\)
\(log\sqrt{1000}=log_{10}10^{\dfrac{3}{2}}=\dfrac{3}{2}\)
\(81^{log_35}=3^{3log_35}=3^{log_3125}=125\)
\(125^{log_52}=5^{3log_52}=5^{log_58}=8\)
\(\left(\dfrac{1}{49}\right)^{log_7\dfrac{1}{8}}=7^{-2log_7\dfrac{1}{8}}=7^{log_764}=64\)
\(\left(\dfrac{1}{625}\right)^{log_52}=5^{-4log_52}=5^{log_5\dfrac{1}{16}}=\dfrac{1}{16}\)
NT
6
28 tháng 5 2020
Bài làm:
a) \(\frac{x}{3}=\frac{2}{3}+\left(-\frac{1}{7}\right)\)
<=>\(\frac{x}{3}=\frac{2.7-1.3}{3.7}\)
<=>\(\frac{x}{3}=\frac{11}{21}\)
<=> \(21.x=33\)
<=> \(x=\frac{11}{7}\)
b) \(x-\frac{1}{4}=\frac{2}{13}\)
<=> \(x=\frac{1}{4}+\frac{2}{13}\)
<=> \(x=\frac{13.1+2.4}{4.13}\)
<=> \(x=\frac{21}{52}\)
c)\(\frac{4}{7}.x=\frac{9}{8}-\frac{125}{1000}\)
<=> \(\frac{4}{7}.x=\frac{9}{8}-\frac{1}{8}\)
<=> \(\frac{4}{7}.x=\frac{8}{8}\)
<=> \(x=1:\frac{4}{7}\)
<=> \(x=\frac{7}{4}\)
Học tốt!!!!
30 tháng 5 2020
7 + 67 + 56 - 76 + 368 - 93 + 45 -88 =
=? ? ? ? ? ? ?
= ? ? ? ? ?
= ?
T
13 tháng 7 2019
#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 3 2023
em muốn nhanh thì lần sau em tách câu hỏi ra chứ đừng hỏi nhiều trong một câu em nhé
\(\dfrac{12}{48}\) = \(\dfrac{12:12}{48:12}=\) \(\dfrac{1}{2}\) \(\dfrac{49}{140}\) = \(\dfrac{49:7}{140:7}\) = \(\dfrac{7}{20}\)
\(\dfrac{125}{1000}\) = \(\dfrac{125:125}{1000:125}\) = \(\dfrac{1}{8}\) \(\dfrac{352}{253}\) = \(\dfrac{352:11}{253:11}\)= \(\dfrac{32}{23}\)
\(\dfrac{75}{300}=\) \(\dfrac{75:75}{300:75}\) = \(\dfrac{1}{4}\) \(\dfrac{561}{132}\) = \(\dfrac{561:33}{132:33}\) = \(\dfrac{17}{4}\)
24 tháng 7 2017
\(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\)
\(A=\left(-1\right)^{2n+n+n+1}\)
\(A=\left(-1\right)^{4n+1}\)
\(B=\left(10000-1^2\right).\left(10000-2^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-100^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-10000\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...0\left(10000-1000^2\right)\)
\(B=0\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...0....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=0\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-10^3\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-1000\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...0}\)
\(D=1999^0\)
\(D=1\)
_Minh ngụy_
a) ( 1000-13) . ( 1000-23) . ( 1000-33) ...( 1000 -503)
\(=\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot...\cdot\left(1000-10^3\right)\cdot.....\cdot\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\cdot\left(100-2^3\right)\cdot...\cdot\left(1000-1000\right)\cdot...\cdot\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot......\cdot0\cdot......\left(1000-50^3\right)\)
\(=0\)
b) (1/125-1/13) . (1/125-1/23).( 1/125-1/33)...( 1/125-1/253)
\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{5^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{125}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot....\cdot0\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=0\)