2.3 mũ x +1 –3 mũ x=135
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21 tháng 5 2022
a: \(7\cdot3^x=5\cdot3^7+2\cdot3^7\)
\(\Leftrightarrow7\cdot3^x=7\cdot3^7\)
=>3x=37
hay x=7
b: \(4^{x+3}-3\cdot4^{x+1}=13\cdot4^{11}\)
\(\Leftrightarrow4^{x+1}\left(4^2-3\right)=13\cdot4^{11}\)
=>x+1=11
hay x=10
d: \(\left(x-1\right)^{13}=\left(x-1\right)^{12}\)
\(\Leftrightarrow\left(x-1\right)^{12}\left(x-2\right)=0\)
hay \(x\in\left\{1;2\right\}\)
ZS
0
CG
1
A
16 tháng 11 2023
a)
\(\text{( 25 – 2x )³ : 5 – 3^2 = 4^2}\)
\(\text{( 25 – 2x )³ : 5 – 9 = 16}\)
\(\text{( 25 – 2x )³ : 5 = 16 + 9}\)
\(\text{( 25 – 2x )³ : 5 = 25}\)
\(\text{( 25 – 2x )³ = 25 . 5}\)
\(\text{( 25 – 2x )³ = 125}\)
\(\text{( 25 – 2x )³ = 5³}\)
\(\text{25 – 2x = 5}\)
\(\text{2x = 25 – 5}\)
\(\text{2x = 20}\)
\(\text{x = 10}\)
\(\text{________________________________________}\)
b)
\(\text{2.3^x = 10.3^12 + 8.27^4}\)
\(\text{2.3^x = 10.3^12 + 8.(3^3)^4}\)
\(\text{2.3^x = 3^12 . (10+8)}\)
\(\text{2.3^x = 3^12 . 18}\)
\(\text{3^x = 3^12 . 18:2}\)
\(\text{3^x = 3^12 . 9}\)
\(\text{3^x = 3^12 . 3^2}\)
\(\text{3^x = 3^14}\)
\(\text{=> x=14}\)
QD
2
24 tháng 10 2020
2 . 3x + 5 . 3x+1 = 153
3x. ( 2 + 5.3 ) = 153
3x . 17 = 153
3x = 9 = 32
x = 2
Vậy x = 2
HT
24 tháng 10 2020
\(2.3^x+5.3^{x+1}=153\)
\(\Leftrightarrow2.3^x+5.3^x.3=153\)
\(\Leftrightarrow3^x.\left(2+5.3\right)=153\)
\(\Leftrightarrow3^x.17=153\)
\(\Leftrightarrow3^x=153:17\)
\(\Leftrightarrow3^x=9\)
\(\Leftrightarrow3^x=3^2\)
\(\Rightarrow x=2\)
6 tháng 5 2019
a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
NT
16 tháng 10 2016
\(2.3^x=10.3^{12}+8.27^4\\ \Rightarrow2.3^x=10.3^{12}+8.\left(3^3\right)^4\\ \Rightarrow2.3^x=10.3^{12}+8.3^{12}\\ \Rightarrow2.3^x=3^{12}\left(10+8\right)\\ \Rightarrow2.3^x=3^{12}.18\)
\(=>3^x=3^{12}.18:2\\ \Rightarrow3^x=3^{12}.3^2\\ \Rightarrow3^x=3^{10}\)
Ta có: 2.3x+1-3x=135
3x+1=135
3x=134
3x=Vô lý
Bn xem lại đề bài đi bn
2.3 mũ x.3-3 mũ x=135
3 mũx.(2+3)=135
3mũx.5=135
3mũx=135:5
3 mũ x=3 mũ 3