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Ta có: 3x+2+4.3x+1=7.36
=> 3x + 1.(3 + 4) = 7.36
=> 3x + 1 . 7 = 7.36
=> 3x + 1 = 36
=> x + 1 = 6
=> x = 5
Lời giải:
1.
$3^{x+2}+4.3^{x+1}=7.3^6$
$3^{x+1}.3+4.3^{x+1}=7.3^6$
$3^{x+1}(3+4)=7.3^6$
$3^{x+1}.7=7.3^6$
$\Rightarrow 3^{x+1}=3^6$
$\Rightarrow x+1=6$
$\Rightarrow x=5$
2.
$5^{x+4}-3.5^{x+3}=2.5^{11}$
$5^{x+3}.5-3.5^{x+3}=2.5^{11}$
$5^{x+3}(5-3)=2.5^{11}$
$2.5^{x+3}=2.5^{11}$
$\Rightarrow 5^{x+3}=5^{11}$
$\Rightarrow x+3=11$
$\Rightarrow x=8$
3.
$4^{x+3}-3.4^{x+1}=13.4^{11}$
$4^{x+1}.4^2-3.4^{x+1}=13.4^{11}$
$4^{x+1}.16-3.4^{x+1}=13.4^{11}$
$13.4^{x+1}=13.4^{11}$
$\Rightarrow 4^{x+1}=4^{11}$
$\Rightarrow x+1=11$
$\Rightarrow x=10$
\(3^{x+2}+4.3^{x+1}=7.3\)
\(\Leftrightarrow3^x\left(3^2+4.3\right)=21\)
\(\Leftrightarrow3^x.21=21\)
\(\Leftrightarrow3^x=1\Leftrightarrow x=0\)
`#3107.\text {DN}`
\(3^{x+2}+4\cdot3^{x+1}+3^{x-1}=6^6\)
`=> 3^x*3^2 + 4*3^x*3 + 3^x * 1/3 = 6^6`
`=>3^x*(3^2 + 12 + 1/3) = 6^6`
`=> 3^x * 64/3 = 6^6`
`=> 3^x = 6^6 \div 64/3`
`=> 3^x = 2187`
`=> 3^x = 3^7`
`=> x = 7`
Vậy, `x = 7.`
\(3^{x+2}+4.3^{x+1}+3^{x-1}=6^6\)
\(3^{x-1}.3^3+4.3^{x-1}.3^2+3^{x-1}=6^6\)
\(3^{x-1}.\left(27+9.4+1\right)=6^6\)
\(3^{x-1}.\left(27+36+1\right)=2^6.3^6\)
\(3^{x-1}.64=3^{x-1}.2^6=3^6.2^6\)
\(\Rightarrow\)\(3^6=3^{x-1}\Rightarrow x=7\)
3^x.3^2+4.3^x.3+3^x:3=6^6
3^x.9+4.3^x.3+3^x/3=36
nhóm lại mà làm nhé.
k nha
a) (2x - 3)2 = 16
=> (2x - 3)2 = 42
=> \(\orbr{\begin{cases}2x-3=4\\2x-3=-4\end{cases}}\)
=> \(\orbr{\begin{cases}2x=7\\2x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)
Vậy ...
b) 2.3x + 2 + 4. 3x + 1 = 10.36
=> 2.3x + 1. 3 + 4.3x + 1 = 10 . 36
=> 6.3x + 1 + 4.3x + 1 = 10.36
=> (6 + 4).3x + 1= 10.36
=> 10.3x + 1= 10.36
=> 3x + 1= 36
=> x + 1 = 6
=> x = 6 - 1
=> x = 5
\(a,\left(2x-3\right)^2=16\)
\(\Rightarrow\left(2x-3\right)^2=4^2\)
\(\Rightarrow2x-3=4\)
\(2x=4+3\)
\(2x=7\)
\(x=7:2\)
\(x=\frac{7}{2}\)
1:
\(\Leftrightarrow4\cdot3^x\cdot\dfrac{1}{9}+2\cdot3^x\cdot3=4\cdot3^4+2\cdot3^7\)
\(\Leftrightarrow3^x\cdot\left(\dfrac{4}{9}+6\right)=3^4\cdot\left(4+2\cdot3^3\right)\)
\(\Leftrightarrow3^x=729\)
hay x=6
2: \(\Leftrightarrow3^x\cdot4\cdot\dfrac{1}{3}+3^x\cdot2\cdot9=4\cdot3^6+2\cdot3^9\)
\(\Leftrightarrow3^x\cdot\dfrac{58}{3}=42282\)
=>3x=2187
hay x=7
3x+2 + 4.3x+1 = 7.36
3x+1.(3 + 4) = 7.36
3x+1.7 = 7.36
=> 3x+1 = 36
=> x + 1 = 6
=> x = 5