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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!
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1/vì (1,782x-2-1,78x):1,78x=0
nên 1,78x2-2-1,78x=0
=>1,782x-2=1,78x
=>2x-2=x
2x=x+2
=>x=2
2/vì cơ số bằng nhau nên ta có
x-2=1;-1;0
ta có: x-2=1 => x=3
x-2=-1 => x=1
x-2=0 => x=2
3/ta có
(x+2)3=33 =>x+2=3 =>x=1
mik mệt rồi bạn cứ gải tiếp đi
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a: \(=3^2\cdot3^3\cdot3^{-4}\cdot3^2=3^{2+3-4+2}=3^3\)
b: \(=2^2\cdot2^5:\left(2^3\cdot\dfrac{1}{2^4}\right)=2^7:\dfrac{1}{2}=2^8\)
c: \(=9\cdot32\cdot\dfrac{4}{9}=128=2^7\)
d: \(=\dfrac{1}{27}\cdot3^4=3^1\)
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a) \(9\cdot3^3\cdot\frac{1}{81}\cdot3^2=3^2\cdot3^3\cdot\left(\frac{1}{3}\right)^43^2=3^7\cdot\frac{1}{3^4}=3^3\)
b) \(4\cdot2^5:\left(2^3\cdot\frac{1}{16}\right)=2^2\cdot2^5:\left(2^3\cdot\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^8\)
c) \(3^2\cdot2^5\cdot\left(\frac{2}{3}\right)^2=3^2\cdot2^5\cdot\frac{2^2}{3^2}=2^7\)
d) \(\left(\frac{1}{3}\right)^2\cdot\frac{1}{3}\cdot9^2=\frac{1}{3^2}\cdot\frac{1}{3}\cdot3^4=\frac{1}{3^3}\cdot3^4=3\)
a)9.33.\(\frac{1}{81}\).32
=32.33.34.\(\frac{1}{3^4}\).32
=311.\(\frac{1}{3^4}\)
=37
b)4.25:(\(2^3.\frac{1}{16}\))
=22.25:(\(2^3.\frac{1}{2^4}\))
=27:\(\frac{2^3}{2^4}\)
=27.\(\frac{2^4}{2^3}\)
=\(\frac{2^{11}}{2^3}\)
=28
c)32.25.\(\left(\frac{2}{3}\right)^2\)
=32.25.\(\frac{2^2}{3^2}\)
=\(\frac{3^2.2^5.2^2}{3^2}\)
=27
d)\(\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2\)
=\(\frac{1^2}{3^2}.\frac{1}{3}.\left(3^2\right)^2\)
=\(\frac{1^2}{3^2}.\frac{1}{3}.3^4\)
=\(\frac{1^2}{3^2}.\frac{3^4}{3}\)
=\(\frac{1^2}{3^2}.3^3\)
=3
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a) \(\frac{1}{81}\times\left(\frac{1}{3}\right)^{-2}\times9\times3^3\)
\(=\frac{3^7}{3^4}\)
\(=3^3\)
b) \(\left(2^5\times4\right)\div\left(2^3\times\frac{1}{16}\right)\)
\(=2^7\div\frac{2^3}{2^{\text{4}}}\)
\(=2^7\div\frac{1}{2}\)
=\(2^6\)
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linhpham linh
\(\left(x+\frac{3}{4}\right)^2=\frac{49}{16}\)
\(\Rightarrow\left(x+\frac{3}{4}\right)^2=\frac{7^2}{4^2}\)
\(\Rightarrow\left(x+\frac{3}{4}\right)^2=\left(\frac{7}{4}\right)^2\)
\(\Rightarrow x+\frac{3}{4}=\frac{7}{4}\)
\(\Rightarrow x=\frac{7}{4}-\frac{3}{4}\)
\(\Rightarrow x=1\)
a) \(\left(x+\frac{3}{4}\right)^2=\frac{49}{16}\)
=> x + \(\frac{3}{4}=\frac{7}{4}\)
=> x = \(\frac{7}{4}-\frac{3}{4}=\frac{4}{4}=1\)
c) (3x - 1)2 = 81
=> 3x - 1 = 9
=> 3x = 10
=> x = \(\frac{10}{3}\)
Ta có: \(9.3^3.\frac{1}{81}.3^2\)
\(=3^2.3^3.\frac{1}{3^4}.3^2\)
\(=3^3=27\)
Bài làm :
Ta có :
\(9.3^3.\frac{1}{81}.3^2=3^2.3^3.\frac{1}{3^4}.3^2=\frac{3^7}{3^4}=3^3=27\)
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!