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Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
\(\frac{1}{9}.3^4.3^x=3^7\)
\(\Leftrightarrow3^x=3^7:\frac{1}{9}:3^4=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
a. x = {3;-3}
b. x thuộc rỗng
c. x2-4=0
x2 = 4
x={2;-2}
d. x2+1=82
x2 =83
x thuộc rỗng
e. (2x)2=6
x thuộc rỗng
f. (x-1)2=9
TH1: x-1=3=>x=4
TH2: x-1=-3=>x=-2
Vậy x={4;-2}
g.(2x+3)2=25
TH1: 2x+3=5=> x=1
Th2: 2x+3=-5=>x=-4
VẬY X={1;-4}
a, x^2= 9
=>\(\sqrt{9}=3\)
b,\(x^2=5=>x=\sqrt{5}\)
c, x^2-4=0
=>x^2=4
=>x=2
d, x^2+1=82
=>x^2=81 =>\(\sqrt{81}=9\)
3, 2x^2=6
=>x= \(\sqrt{6}\)
f, {x-1} ^2=9
=> x-1=3
=>x=2
g{ 2x+3}^2=25
=> 2x+3=5
=>2x=2
=>x=1
a)3^x+1=9^x
3^x+1=3.3^x
3^x+1=3^x+1
=>x thuộc TH Z
b)2^3.x+2=4^x+5
2^3x+2=2^2.(x+5)
2^3x+2=2^2x+10
2^3x=2^2x+8
3x-2x=8
=>x=8
c)3^2x-1=243
3^2x=243.3
3^2x=729
3^2x=3^6
=>2x=6
x=6:2=3
chúc bạn học tốt nha
a) \(\left(\frac{1}{4}\right)^x:\frac{1}{16}=\frac{1}{4}\)
\(\left(\frac{1}{4}\right)^x=\frac{1}{4}\cdot\frac{1}{16}\)
\(\left(\frac{1}{4}\right)^x=\frac{1}{64}\)
\(\left(\frac{1}{4}\right)^x=\left(\frac{1}{4}\right)^3\)
\(\Rightarrow x=3\)
b) \(81^x:9^x=729\)
\(\left(81:9\right)^x=729\)
\(9^x=9^3\)
\(\Rightarrow x=3\)
a) \(\left(\frac{1}{4}\right)^x:\frac{1}{16}=\frac{1}{4}\)
\(\Rightarrow\left(\frac{1}{4}\right)^x=\frac{1}{4}\times\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{4}\right)^x=\frac{1}{64}\)
\(\Rightarrow\left(\frac{1}{4}\right)^x=\left(\frac{1}{4}\right)^3\)
\(\Rightarrow x=3\)
Vậy x = 3
b) \(81^x:9^x=729\)
\(\Rightarrow\left(81:9\right)^x=729\)
\(\Rightarrow9^x=9^3\)
\(\Rightarrow x=3\)
Vậy x = 3
_Chúc bạn học tốt_
a, (\(\frac{1}{2}\))x = \(\frac{1}{64}\)
<=> (\(\frac{1}{2}\))x = (\(\frac{1}{2}\))6
<=> x = 6
b, 35 :3x = 9
<=> 35 :3x = 32
<=> 3x = 35 :32
<=> 3x = 33
<=> x = 3
Chúc bạn học tốt
Ta có \(3^{x+1}=9^x\)
\(\Rightarrow3^x.3=3^{2x}\)
\(\Rightarrow3^{2x}-3^x.3=0\)
\(\Rightarrow3^x.9-3^x.3=0\)
\(\Rightarrow3^x.\left(9-3\right)=0\)
\(\Rightarrow3^x.6=0\)
Suy ra không tìm được x thỏa mãn
Ta có \(3^{x+1}=9^x\)
<=>\(3^{x+1}=\left(3^2\right)^x\)
<=>\(3^{x+1}=3^{2x}\)
<=>x+1=2x
<=>x-2x=1
<=>x.(1-2)=1
<=>x.(-1)=1
<=>x =1:(-1)
<=>x=-1
Vậy x=-1