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a)3x-1+5.3x-1=162
3x-1.1+5.3x-1=162
3x-1.(5+1)=162
3x-1.6=162
3x-1=162/6
3x-1=27
3x-1=33
Suy ra x-1=3
x=3+1=4
b)2x+3+2x=144
2x.8+2x=144
2x.8+2x.1=144
2x.(8+1)=144
2x.9=144
2x=144/9
2x=16
2x=24
Suy ra x=4
a) Ta có 3x - 1 + 5.3x - 1 = 162
<=> 6.3x - 1 = 162
<=> 3x - 1 = 27
<=> 3x - 1 = 33
<=> x - 1 = 3
<=> x = 4
Vậy x = 4 là giá trị cần tìm
b) 2x + 3 + 2x = 144
<=> 2x.23 + 2x = 144
<=> 2x.8 + 2x = 144
<=> 2x.9 = 144
<=> 2x = 16
<=> 2x = 24
<=> x = 4
Vậy x = 4 là giá trị cần tìm
`#3107`
b)
`2.3^x = 162`
`\Rightarrow 3^x = 162 \div 2`
`\Rightarrow 3^x = 81`
`\Rightarrow 3^x = 3^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
c)
`(2x - 15)^5 = (2 - 15)^3`
\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`
\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)
`d)`
\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!
`e)`
\(7\cdot4^{x-1}+4^{x-1}=23?\)
\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)
Bạn xem lại đề!
`f)`
\(2\cdot2^{2x}+4^3\cdot4^x=1056\)
\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
_____
\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)
\(\Rightarrow\left(x\div3+17\right)\div10=2\)
\(\Rightarrow x\div3+17=20\)
\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)
Vậy, `x = 9.`
b, 2 + 4 + 6 +.....+2x =210
=>2*(1+2+3+....+x) =210
=> 1+2+3+....+x = 105
=> (x+1)*x : 2 = 105
=> (x+1)*x = 210
=> (x+1)*x = 15*14
=>x=14
c) [(x - 300)^2 - 280 ] * 2 = 19440
=> (x - 300)^2 - 280 = 9720
=> (x - 300)^2 = 10000 =1002
=> x - 300 =100
=> x=400
a, 3x-1+ 5* 3x-1 = 162
3x-1 * (5 + 1) = 162
3x-1 * 6 = 162
3x-1 = 27 = 33
=> x-1=3
=> x=4