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A, (2x+1)3=343
=> (2x+1)3=73
=> 2x + 1 = 7
=> 2x = 6
=> x = 3
B, 2x+2x+3=144
=> 2x+2x . 23 =144
=> 2x ( 1 + 23 ) =144
=> 2x ( 1 + 8 ) =144
=> 2x . 9 =144
=> 2 x = 16
=> 2 x = 2 4
=> x = 4
C, 3x+3x+2=2430
=> 3x+3x . 32 =2430
=> 3x . ( 1 + 32 ) =2430
=> 3x . ( 1 + 9 ) =2430
=> 3x . 10 =2430
=> 3x = 243
=> 3x = 3 5
=> x = 5
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\(2^x+2^{x+1}=48\)
\(2^x.1+2^x.2^1=48\)
\(2^x\left(1+2\right)=48\)
\(2^x.3=48\)
\(2^x=16\)
\(2^x=2^4\)
\(\Rightarrow x=4\)
Vậy x = 4
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a) 3x.3x+2 =81
3x.3x.32 =81
3x.3x.9 =81
3x.3x =81/9
3x.3x =9
31.31 =9
x =1
b)2(x+15)+3(x+25)=2050
2(x+15)+3(x+15+10)=2050
2(x+15)+3(x+15)+3.10=2050
(x+15)(2+5)+30=2050
(x+15)5=2050-30
(x+15)5=2020
x+15=2020/5
x+15=404
x =404-15
x =389
c)2x+2x+1+2x+2+2x+3=960
2x+2x.2+2x.22+2x.23=960
2x(1+2+22+23)=960
2x(1+2+4+8)=960
2x.15=960
2x =960/15
2x =64
26 =64
x =6
bài d làm tương tự c,e làm tương tự a
f)3x.32x+3=729
33x.33=729
33x.27 =729
33x =729/27
33x =27
33.1 =27
x =1
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Câu 2.
b) Gọi tổng trên là A.
Số số hạng của A là :
(2012-1):1+1=2012(số hạng)
Nhóm 4 số hạng với nhau, ta được số nhóm là:
2012:4=503(nhóm)
Ta có:
A= \(5+5^2+5^3+...+5^{2012}\)
A= ( \(5+5^2+5^3+5^4\)) + ... + ( \(5^{2009}+5^{2010}+5^{2011}+5^{2012}\))
A= 65.12 + ... + 65.12.\(5^{2008}\)
Vậy A chia hết cho 65.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a) Ta có: \(2^x\cdot4=128\)
\(\Leftrightarrow2^x=32\)
hay x=5
Vậy: x=5
b) Ta có: \(2^x-26=6\)
\(\Leftrightarrow2^x=32\)
hay x=5
Vậy: x=5
c) Ta có: \(27\cdot3^x=3^7\)
\(\Leftrightarrow3^x=\frac{3^7}{27}=\frac{3^7}{3^3}=3^4\)
hay x=4
Vậy: x=4
d) Ta có: \(3^x=81\)
\(\Leftrightarrow3^x=3^4\)
hay x=4
Vậy: x=4
e) Ta có: \(64\cdot4^x=4^5\)
\(\Leftrightarrow4^3\cdot4^x=4^5\)
\(\Leftrightarrow4^{x+3}=4^5\)
\(\Leftrightarrow x+3=5\)
hay x=2
Vậy: x=2
g) Ta có: \(49\cdot7^x=2401\)
\(\Leftrightarrow7^2\cdot7^x=7^4\)
\(\Leftrightarrow7^{x+2}=7^4\)
\(\Leftrightarrow x+2=4\)
hay x=2
Vậy: x=2
h) Ta có: \(3^4\cdot3^x=3^7\)
\(\Leftrightarrow3^{x+4}=3^7\)
\(\Leftrightarrow x+4=7\)
hay x=3
Vậy: x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
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a,3x+1 +3x+3x-1=30.(34)3
3x+1+3x+3x-1=20726199
3x.3+3x.1+3x:3=20726199
3x.3+3x.1+3x.\(\frac{1}{3}\)=20726199
3x(3+1+\(\frac{1}{3}\))=20726199
3x.\(\frac{13}{3}\)=20726199
3x=20726199:\(\frac{13}{3}\)
3x=4782969
3x=314
Vậy x:=14
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2^{x-1}+2^{x+1}+2^{x+2}=104\)
=> \(2^{x-1}+2^x\cdot2+2^x\cdot2^2=104\)
=> \(2^x:2+2^x\cdot\left(2+2^2\right)=104\)
=> \(2^x\cdot\frac{1}{2}+2^x\cdot6=104\)
=> \(2^x\cdot\left(\frac{1}{2}+6\right)=104\Rightarrow2^x=104:\left(\frac{1}{2}+6\right)=104:\frac{13}{2}=16\)
=> \(x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(3.2^{x+1}=12\)
\(2^{x+1}=12:3\)
\(2^{x+1}=4\)
\(2^{x+1}=2^2\)
\(x+1=2\)
\(x=2-1\)
\(x=1\)
Vậy \(x=1\)
c) \(2^{x-1}=2^3+2^4-2^3\)
\(2^{x-1}=8+16-8\)
\(2^{x-1}=16\)
\(2^{x-1}=2^4\)
\(x-1=4\)
\(x=5\)
Vậy \(x=5\)
d) \(x^{50}=x\)
\(x^{50}-x=0\)
\(\Rightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left\{0;1\right\}\)
\(b.3.2^{x+1}=12\\ \Rightarrow2^{x+1}=4\\ \Rightarrow2^{x+1}=2^2\\ \Rightarrow x=1\\ \)
c) \(2^{x-1}=2^3-2^3+2^4\\ \Rightarrow2^{x-1}=0+16\\ \Rightarrow2^{x-1}=16\\ \Rightarrow2^{x-1}=2^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
d) \(x^{50}=x\\ \Rightarrow x=0;1\)
e) \(2\left(2x-1\right)^4=32\\ \Rightarrow\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\frac{3}{2}\)
g) Bí
\(3^{x+2}+3^x=2430\)
\(\Rightarrow3^x\cdot\left(9+1\right)=2430\)
\(\Rightarrow3^x=2430:10=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
ti ck nha
²⁴ʱ๖ۣۜSυρɾεмε...
3^{x+2}+3^x=24303x+2+3x=2430
\Rightarrow3^x\cdot\left(9+1\right)=2430⇒3x⋅(9+1)=2430
\Rightarrow3^x=2430:10=243⇒3x=2430:10=243
\Rightarrow3^x=3^5⇒3x=35
\Rightarrow x=5⇒x=5