\(\frac{1}{5}\))5 . 55

b) (0,125) 

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21 tháng 9 2017

a)1

b)1

c)4

21 tháng 9 2017

A) (1/5)^5 . 5^5 = 1/5^5 . 5^5 = 5^5 / 5^5 = 1.

B)(0,125)^3 . 512 = (1/8)^3 . 512 = 1/8^3 . 512 = 1/512 . 512 = 1.

C) (0,25)^4 . 1024 = (1/4)^4 . 1024 = 1/4^4 . 1024 = 1/256 . 1024 = 4.

4 tháng 7 2018

\(\left(\dfrac{1}{7}\right)^7\cdot7^7=\left(\dfrac{1}{7}\cdot7\right)^7=1^7=1\\ \left(0,125\right)^3\cdot512=\left(0,125\right)^3\cdot8^3=\left(0,125\cdot8\right)^3=1^3=1\\ \left(0,25\right)^4\cdot1024=\left(0,25\right)^4\cdot256\cdot4=\left(0,25\right)^4\cdot4^4\cdot4=\left(0,25\cdot4\right)^4\cdot4=1^4\cdot4=4\)

5 tháng 7 2018

a) \(\left(\dfrac{1}{7}\right)^7.7^7=\left(\dfrac{1}{7}.7\right)^7=1^7=1\)

b) \(\left(0.125\right)^3.512=\left(0.125\right)^3.8^3=\left(0.125\cdot8\right)^3=1^3=1\)

c) \(\left(0.25\right)^4.1024=\left(0.25\right)^4.4^5=\left(0.25\right)^4.4^4.4=\left(0.25.4\right)^4.4=1^4.4=1.4=4\)

11 tháng 9 2016

a) \(\left(\frac{1}{5}\right)^5.5^5=\frac{1^5}{5^5}.5^5=1^5=1\)

b) \(\left(0,25\right)^4.1024=\left(\frac{1}{4}\right)\text{ }^4.2^{10}=\frac{1^4}{4^4}.2^{10}=\frac{1}{\left(2^2\right)^4}.2^{10}=\frac{1}{2^8}.2^{10}=2^2=4\)

c) \(\left(0,125\right)^3.512=\left(\frac{1}{8}\right)^3.2^9=\frac{1^3}{8^3}.2^9=\frac{1}{\left(2^3\right)^3}.2^9=\frac{1}{2^9}.2^9=1\)

22 tháng 8 2017

Bạn giỏi quá !ngaingung

21 tháng 10 2019

a) Câu này thiếu đề nhé bạn.

b) \(\frac{25}{5^n}=5\)

\(\Rightarrow5^n=25:5\)

\(\Rightarrow5^n=5\)

\(\Rightarrow5^n=5^1\)

\(\Rightarrow n=1\)

Vậy \(n=1.\)

c) \(\frac{81}{\left(-3\right)^n}=-243\)

\(\Rightarrow\left(-3\right)^n=81:\left(-243\right)\)

\(\Rightarrow\left(-3\right)^n=-\frac{1}{3}\)

\(\Rightarrow\left(-3\right)^n=\left(-3\right)^{-1}\)

\(\Rightarrow n=-1\)

Vậy \(n=-1.\)

e) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)

\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow n=4\)

Vậy \(n=4.\)

f) \(\left(-\frac{3}{4}\right)^n=\frac{81}{256}\)

\(\Rightarrow\left(-\frac{3}{4}\right)^n=\left(-\frac{3}{4}\right)^4\)

\(\Rightarrow n=4\)

Vậy \(n=4.\)

Chúc bạn học tốt!

22 tháng 10 2019

d) \(\frac{1}{2}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n.\left(\frac{1}{2}+4\right)=288\)

\(\Rightarrow2^n.\frac{9}{2}=288\)

\(\Rightarrow2^n=288:\frac{9}{2}\)

\(\Rightarrow2^n=64\)

\(\Rightarrow2^n=2^6\)

\(\Rightarrow n=6\)

Vậy \(n=6.\)

g) \(-\frac{512}{343}=\left(-\frac{8}{7}\right)^n\)

\(\Rightarrow\left(-\frac{8}{7}\right)^n=\left(-\frac{8}{7}\right)^3\)

\(\Rightarrow n=3\)

Vậy \(n=3.\)

h) \(5^{-1}.25^n=125\)

\(\Rightarrow5^{-1}.5^{2n}=5^3\)

\(\Rightarrow5^{-1+2n}=5^3\)

\(\Rightarrow-1+2n=3\)

\(\Rightarrow2n=3+1\)

\(\Rightarrow2n=4\)

\(\Rightarrow n=4:2\)

\(\Rightarrow n=2\)

Vậy \(n=2.\)

k) \(3^{-1}.3^n+6.3^{n-1}=7.3^6\)

\(\Rightarrow3^{n-1}+6.3^{n-1}=7.3^6\)

\(\Rightarrow3^{n-1}.\left(1+6\right)=7.3^6\)

\(\Rightarrow3^{n-1}.7=7.3^6\)

\(\Rightarrow n-1=6\)

\(\Rightarrow n=6+1\)

\(\Rightarrow n=7\)

Vậy \(n=7.\)

Chúc bạn học tốt!

a) \(\left(0,25\right)^3\cdot32=0,015625\cdot32=0,5\)

b) \(\left(-0,125\right)^3\cdot80^4=\dfrac{-1}{512}\cdot40960000=80000\)

c) \(\dfrac{8^2\cdot4^5}{2^{20}}=\dfrac{2^{3^2}\cdot2^{2^5}}{2^{20}}=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{2^{16}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\)

d) \(\dfrac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{3^{4^{11}}\cdot3^{17}}{3^{3^{10}}\cdot3^{2^{15}}}=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)

4 tháng 4 2020
https://i.imgur.com/rAuF0P4.jpg
AH
Akai Haruma
Giáo viên
19 tháng 3 2019

1.

\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)

\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)

\(=2x^5y^4-4x^2y^3\)

2.

\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)

\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)

\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)

3.

\(5x-7xy^2+3x-\frac{1}{2}xy^2\)

\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)

\(=8x-\frac{15}{2}xy^2\)

AH
Akai Haruma
Giáo viên
19 tháng 3 2019

4.

\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)

\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)

\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)

5.

\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)

\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)

\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)

6.

\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)

\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)