Tìm số nguyên N biết:
\(a,\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(b,\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(c,\left(\frac{-3}{4}\right)^n=\frac{81}{256}\)
\(d,\left(2x+3\right)^2=\frac{9}{121}^n\)
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tìm các số nguyên n biết
a. \(\left(\frac{1}{3}\right)^n=\frac{1}{18}\)
b. \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
c. \(\left(\frac{-3}{4}\right)^n=\frac{81}{256}\)
d. 27n : 3n=9
e. \(\frac{1}{2}.2^n+4.2^n=9.2^5\)
Tìm số tự nhiên n biết:
a) \(\left[\left(0,5\right)^3\right]^n\)=\(\frac{1}{64}\)
b) \(\frac{64}{\left(-2\right)^{n+1}}=4\)
c) \(\left(\frac{1}{3}\right)^{n+1}=\frac{1}{81}\)
d) \(\left(\frac{3}{4}\right)^n.\frac{1}{2}=\frac{81}{512}\)
\(a,\left[\left(0,5\right)^3\right]^n=\frac{1}{64}\Rightarrow\left(0,125\right)^n=0,125^2\Rightarrow n=2\)
\(b,\frac{64}{\left(-2\right)^{n+1}}=4\Rightarrow\left(-2\right)^{n+1}=\frac{64}{4}\Rightarrow\left(-2\right)^{n+1}=16\Rightarrow\left(-2\right)^{n+1}=\left(-2\right)^4\)
\(\Rightarrow n+1=4\Rightarrow n=3\)
\(c,\left(\frac{1}{3}\right)^{n+1}=\frac{1}{81}\Rightarrow\left(\frac{1}{3}\right)^{n+1}=\left(\frac{1}{3}\right)^4\Rightarrow n+1=4\Rightarrow n=3\)
\(d,\left(\frac{3}{4}\right)^n.\frac{1}{2}=\frac{81}{512}\Rightarrow\left(\frac{3}{4}\right)^n=\frac{81}{512}:\frac{1}{2}=\frac{81}{256}\Rightarrow\left(\frac{3}{4}\right)^n=\left(\frac{3}{4}\right)^4\Rightarrow n=4\)
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Tìm n∈Z biết :
a,27n/3n
b,frac{25}{5^n}5
c,frac{81}{left(-3right)^n}-243
d,frac{1}{2}cdot2^n+4cdot2^n9cdot2^5
e,(frac{1}{3})nfrac{1}{81}
f,left(frac{-3}{4}right)^nfrac{81}{256}
g,frac{-512}{343}left(frac{-8}{7}right)^n
h,5-1*25n125
k,3-1*3n+6*3n-17*36
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Tìm n∈Z biết :
a,27n/3n
b,\(\frac{25}{5^n}\)=5
c,\(\frac{81}{\left(-3\right)^n}=-243\)
d,\(\frac{1}{2}\cdot2^n+4\cdot2^n=9\cdot2^5\)
e,(\(\frac{1}{3}\))n=\(\frac{1}{81}\)
f,\(\left(\frac{-3}{4}\right)^n=\frac{81}{256}\)
g,\(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
h,5-1*25n=125
k,3-1*3n+6*3n-1=7*36
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!
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!
b)\(\frac{25}{5^n}\)=5
\(5^n\)=25/5
\(5^n\)=\(5^1\)
⇒ n = 1
c) \(\frac{81}{\left(-3\right)^n}\)=-243
\(\left(-3\right)^n\)=81/-243
\(\left(-3\right)^n\)=\(\frac{-1}{3}\)
\(\left(-3\right)^n\)=\(\left(-3\right)^{-1}\)
⇒n=-1
a)\(27^x:3^x=9\)
b)\(\frac{125}{5^x}=25\)
c)\(\frac{-243}{\left(-3\right)}x=-245\)
d)\(\left(\frac{1}{3}\right)x=\frac{1}{81}\)
e)\(\frac{-512}{343}=\left(\frac{-8}{7}\right)^x\)
g)\(\left(\frac{-3}{4}\right)^x=\frac{81}{256}\)
a) x=1
b) x=1
c) x= -(245/81)
d) x= 1/27
e) x=3
g) x=4
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Tính m, n, p biết
a) \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
b) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
c) \(\left(-0,25\right)^p=\frac{1}{256}\)
a ) \(\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow m=4\)
b ) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow\left(\frac{3}{5}^2\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow\left(\frac{9}{25}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow n=5\)
c ) \(\left(-0,25\right)^p=\frac{1}{256}\)
\(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)
\(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\left(-\frac{1}{4}\right)^4\)
\(\Leftrightarrow p=4\)
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\(a.\)
\(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow m=4\)
Vậy : \(m=4\)
\(b.\)
\(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{15}\)
\(\Rightarrow n=5\)
Vậy : \(n=5\)
\(c.\)
\(\left(-0,25\right)^p=\frac{1}{256}\)
\(\Rightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)
\(\Rightarrow\left(-\frac{1}{4}\right)^p=\left(\frac{1}{4}\right)^4\)
\(\Rightarrow p=4\)
Vậy : \(p=4\)
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1) Tính:
a) \(\frac{6^3-3.6^2+3^2}{-13}\)
b) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
2) Tìm n \(\in\)Z:
a) 27n : 3n = 9
b) \(\frac{25}{5^n}=5\)
c) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
d) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
ko bt làm thì xuống lớp 6 hocj đi
Bạn 12345678901 xuống lớp 1 học đạo đức làm người nhé bạn. Lịch sự tí đi
a)Tìm số nguyên dương n thỏa mãn:
\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)=\frac{2013}{2014}\)
b)tìm a sao cho
\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)=11.a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
bài 1: Tìm x,y biết rằng: x+(-frac{31}{12})^2left(frac{49}{12}right)^2-xy^2 bài 2: tìm x biết:a.5^x.left(5^3right)^2625 b.left(frac{12}{25}right)^xleft(frac{5}{3}right)^{-2}-left(-frac{3}{5}right)^4 c.left(-frac{3}{4}right)^{3x-1}frac{256}{81} d.172x^2-7^9:98^32^{-3}Bài 3: Tìm x varepsilonN biết:a.8 2^xle2^9times2^{-5} b.27 81^3:3^x 243 left(frac{2}{5}right)^xleft(frac{5}{2}right)^{-3}timesleft(-frac{2}{5}right)^2c.
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bài 1: Tìm x,y biết rằng:
\(x+(-\frac{31}{12})^2=\left(\frac{49}{12}\right)^2-x=y^2\)
bài 2: tìm x biết:
a.\(5^x.\left(5^3\right)^2=625\) b.\(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\) c.\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\)
d.\(172x^2-7^9:98^3=2^{-3}\)
Bài 3: Tìm x \(\varepsilon\)N biết:
a.\(8< 2^x\le2^9\times2^{-5}\) b.\(27< 81^3:3^x< 243\) \(\left(\frac{2}{5}\right)^x>\left(\frac{5}{2}\right)^{-3}\times\left(-\frac{2}{5}\right)^2\)c.
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
1 Tính m,n,p biết :
a) \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
b) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
c) \(\left(-0,25\right)^p=\frac{1}{256}\)
Mình cần gấp bạn nào nhanh mình tick ạ
a/ \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
\(\Leftrightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Leftrightarrow m=4\left(tm\right)\)
b/ \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow n=10\)
\(\Leftrightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)
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a)(1/3)^m=(1/3)^4
b)(3/5)^n=(3/5)^10
c)(-0,25)^p=(-0,25)^4
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a)\(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
\(\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow m=4\)
vay \(m=4\)
b) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)
\(\Rightarrow n=10\)
vay \(n=10\)
c) \(\left(-0,25\right)^p=\frac{1}{256}\)
\(\left(\frac{-1}{4}\right)^p=\left(\frac{1}{4}\right)^4\)
\(\Rightarrow p=4\)
voi mu duong thi \(\left(\frac{-1}{4}\right)^4=\left(\frac{1}{4}\right)^4\)
vay \(p=4\)
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