\(3.4^{x-2}+4^{x-1}+4^x=92\)
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30 tháng 10 2015
3.4^x-2+4^x-1+4^x=92
12^x-2+4^x-1+4^x=92
4(3^x-2+1^x-1+1^x)=92
4(3^x-2+2)=92
3^x-2+2=23
3^x-2=21
3^x-2=3.7
=>x-2=1
x=3
PQ
1
3 tháng 2 2015
3.4x-2+4x-1+4x=92
4x(3+1+1)-2-1=92
4x.5-2-1 =92
4x.5 =92+1+2
4x.5 =95
4x =95:5
4x =19
AH
Akai Haruma
Giáo viên
23 tháng 7 2023
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.
$4^{x+3}-3.4^{x+1}=13.4^{11}$
$4^{x+1}.4^2-3.4^{x+1}=13.4^{11}$
$4^{x+1}.16-3.4^{x+1}=13.4^{11}$
$13.4^{x+1}=13.4^{11}$
$\Rightarrow 4^{x+1}=4^{11}$
$\Rightarrow x+1=11$
$\Rightarrow x=10$
DT
1
21 tháng 10 2023
\(\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2013}{2015}\)
=>\(\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+...+\dfrac{2}{x}-\dfrac{2}{x+1}=\dfrac{2013}{2015}\)
=>\(1-\dfrac{2}{x+1}=\dfrac{2013}{2015}\)
=>\(\dfrac{2}{x+1}=\dfrac{2}{2015}\)
=>x+1=2015
=>x=2014
31 tháng 7 2022
1: \(\Leftrightarrow\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+4}{96}+1\right)=\left(\dfrac{x+6}{94}+1\right)+\left(\dfrac{x+8}{92}+1\right)\)
=>x+100=0
hay x=-100
2: \(\Leftrightarrow x\cdot\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{3}x-\dfrac{2}{3}\)
=>3/4x+5/4=-1/3x+7/3
=>13/12x=13/12
hay x=1
21 tháng 10 2021
1) \(\dfrac{1}{4}x-\dfrac{1}{3}=\dfrac{-5}{9}\)
\(\Rightarrow\dfrac{1}{4}x=-\dfrac{2}{9}\Rightarrow x=-\dfrac{8}{9}\)
2) \(2^{x-3}-3.2^x=-92\)
\(\Rightarrow2^x\left(2^{-3}-3\right)=-92\)
\(\Rightarrow2^x.\dfrac{-23}{9}=-92\)
\(\Rightarrow2^x=32\Rightarrow x=5\)
T
0
H9
HT.Phong (9A5)
CTVHS
6 tháng 10 2023
Bài 1:
a) \(4^{x+2}+4^x=68\)
\(\Rightarrow4^x\cdot\left(4^2+1\right)=68\)
\(\Rightarrow4^x\cdot17=68\)
\(\Rightarrow4^x=\dfrac{68}{17}\)
\(\Rightarrow4^x=4\)
\(\Rightarrow4^x=4^1\)
\(\Rightarrow x=1\)
b) \(5\cdot2^{x+4}-3\cdot2^x=308\)
\(\Rightarrow2^x\cdot\left(5\cdot2^4-3\right)=308\)
\(\Rightarrow2^x\cdot\left(5\cdot16-3\right)=308\)
\(\Rightarrow2^x\cdot77=308\)
\(\Rightarrow2^x=\dfrac{308}{77}\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
c) \(4\cdot3^{x+1}+7\cdot3^x=513\)
\(\Rightarrow3^x\cdot\left(4\cdot3+7\right)=513\)
\(\Rightarrow3^x\cdot19=513\)
\(\Rightarrow3^x=\dfrac{513}{19}\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
d) \(5^{x+4}-5^x=3120\)
\(\Rightarrow5^x\cdot\left(5^4-1\right)=3120\)
\(\Rightarrow5^x\cdot\left(625-1\right)=3120\)
\(\Rightarrow5^x\cdot624=3120\)
\(\Rightarrow5^x\cdot\dfrac{3120}{624}\)
\(\Rightarrow5^x=5\)
\(\Rightarrow5^x=5^1\)
\(\Rightarrow x=1\)
f) \(3\cdot4^{2x+1}-16^x=2816\)
\(\Rightarrow3\cdot4^{2x+1}-\left(4^2\right)^x=2816\)
\(\Rightarrow3\cdot4^{2x+1}-4^{2x}=2816\)
\(\Rightarrow4^{2x}\cdot\left(3\cdot4-1\right)=2816\)
\(\Rightarrow4^{2x}\cdot11=2816\)
\(\Rightarrow4^{2x}=\dfrac{2816}{11}\)
\(\Rightarrow4^{2x}=256\)
\(\Rightarrow\left(2^2\right)^{2x}=2^8\)
\(\Rightarrow2^{4x}=2^8\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=2\)
Bài 2:
\(2^x+124=5^y\)
\(\Rightarrow5^y-2^x=124\)
\(\Rightarrow5^y-2^x=125-1\)
\(\Rightarrow\left\{{}\begin{matrix}5^y=125\\2^x=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5^y=5^3\\2^x=2^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=3\\x=0\end{matrix}\right.\)
Vậy: ....
3.4x-2+4x-1+4x=92
3.4x-2+4x-2+1+4x--2+2=92
3.4x-2+4x-2.4+4x-2.16=92
4x-2.(3+4+16)=92
4x-2.23=92
4x-2=92:23
4x-2=4
=>x-2=1
x=3
vay...
\(3.4^{x-2}+4^{x-1}+4^x=92\)\
<=> \(\frac{3}{16}.4^x+\frac{1}{4}.4^x+4^x=92\)
<=> \(\left(\frac{3}{16}+\frac{1}{4}+1\right).4^x=92\)
<=> \(\frac{23}{16}.4^x=92\)
<=> \(4^x=64\)
<=> x = 3