3x+3x+1=108
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e) \(2^x+2^{x+3}=144\)
\(=>2^x+2^x.2^3=144\)
\(=>2^x.\left(1+2^3\right)=144\)
\(=>2^x.9=144\)
\(=>2^x=144:9\)
\(=>2^x=16=2^4\)
\(=>x=4\)
__________
f) \(3^x+3^{x+1}=108\)
\(=>3^x+3^x.3=108\)
\(=>3^x.\left(1+3\right)=108\)
\(=>3^x.4=108\)
\(=>3^x=108:4\)
\(=>3^x=27=3^3\)
\(=>x=3\)
\(#Wendy.Dang\)
23x+1 = 16
23x+1 = 24
=> 3x + 1 = 4
=> 3x = 4 - 1
3x = 3
x = 3 : 3
x = 1
Lời giải:
$45-5(x+4)=10$
$5(x+4)=45-10=35$
$x+4=35:5=7$
$x=7-4=3$
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$(7x-15):3-6=17$
$(7x-15):3=17+6=23$
$7x-15=23\times 3=69$
$7x=69+15=84$
$x=84:7=12$
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$3^{x+2}-5.3^x=108$
$3^x(3^2-5)=108$
$3^x.4=108$
$3^x=108:4=27=3^3$
$\Rightarrow x=3$
\(45-5\left(x+4\right)=10\)
\(\Rightarrow5\left(x+4\right)=45-10\)
\(\Rightarrow5\left(x+4\right)=35\)
\(\Rightarrow x+4=35:5\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=7-4\)
\(\Rightarrow x=3\)
______________
\(\left(7x-15\right):3-6=17\)
\(\Rightarrow\left(7x-15\right):3=17+6\)
\(\Rightarrow\left(7x-15\right):3=23\)
\(\Rightarrow7x-15=23\cdot3\)
\(\Rightarrow7x-15=69\)
\(\Rightarrow7x=69+15\)
\(\Rightarrow7x=84\)
\(\Rightarrow x=12\)
______________
\(3^{x+2}-5\cdot3^x=108\)
\(\Rightarrow3^x\cdot\left(3^2-5\right)=108\)
\(\Rightarrow3^x\cdot\left(9-5\right)=108\)
\(\Rightarrow3^x\cdot4=108\)
\(\Rightarrow3^x=108:4\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
( 3x - 21 ) : 4 + 108 = 114
=> ( 3x - 21 ) : 4 = 114 - 108
=> ( 3x - 21 ) : 4 = 6
=> 3x - 21 = 6 x 4
=> 3x - 21 = 24
=> 3x = 24 + 21
=> 3x = 45
=> x = 45 : 3
=> x = 15
X / y = 2/3
3x = 2y
thay thế
3x + 7Y = 108
2y + 7Y = 108
9Y = 108
y = 12
3x = 2 * 12
3x = 24
x = 8
Lớn hơn là 12
\(a,5x-5^2=3x+25\)
\(5x-3x=25+5^2\)
\(2x=50\)
\(x=50:2\)
\(x=25\)
\(b,3x+69:23=2x-108:36\)
\(3x+3=2x-3\)
\(3x-2x=-3-3\)
\(x=-6\)
Học tốt
Ta có:
\(\frac{-84}{44}< 3x< \frac{108}{9}\)
\(\Rightarrow-1,90909...< 3x< 12\)
\(\Rightarrow x\in2;3\)
Có: \(3x=4y\Leftrightarrow\frac{x}{4}=\frac{y}{3}\)
Đặt \(\frac{x}{4}=\frac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=3k\end{matrix}\right.\)
Thay \(x=4k,y=3k\) vào \(x.y=108\), có:
\(4k.3k=108\\ \Leftrightarrow k\in\left\{3;-3\right\}\)
+Khi \(k=3\Rightarrow\left\{{}\begin{matrix}x=3.4=12\\y=3.3=9\end{matrix}\right.\)
+Khi \(k=-3\Rightarrow\left\{{}\begin{matrix}x=-3.4=-12\\y=-3.3=-9\end{matrix}\right.\)
Vậy...
\(\Leftrightarrow3^x\left(1+3^1\right)=108\\ \Leftrightarrow3^x=\dfrac{108}{4}=27=3^3\\ \Leftrightarrow x=3\)