2x+3 - 2x+1 = 128
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a) 2x . 4 = 128
<=> 2x = 32
<=> 2x = 25
<=> x = 5
b) x15 = x1
<=> x15 - x = 0
<=> x(x14 - 1) = 0
<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
c) (2x + 1)3 = 125
<=> (2x + 1)3 = 53
<=> 2x + 1 = 5
<=> 2x = 4
<=> x = 2
d) (x - 5)4 = (x - 5)6
<=> (x - 5)6 - (x - 5)4 = 0
<=> (x - 5)4[(x - 5)2 - 1] = 0
<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)
Khi (x - 5)4 = 0 => x - 5 = 0 => x = 5
Khi (x - 5)2 - 1 = 0 <=> (x - 5)2 = 12 <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
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a.23x +1=128
= 23x x 2=128
=128:2=64=2 mũ 6
vậy x=2
b.(7x-11)3=25+52+200
(7x-11)3=257=6,3579 mũ 3
7x=17,3579
x=2,4797
c.(2x+1)+(2x+2)+(2x+3)+...+(2x+101)=5757
vế trái có 101 số hạng
VT =(2x +1+2x+101).101:2=(4x+102).101:2=5757
(4x+102).101 =5757.2=11514
(4x+102)=11514:101=114
4x=114-102=12
x=12:4=3
vậy x=3
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\(a,2^x.4=128\\2^x.2^2=2^7\\ 2^x=\dfrac{2^7}{2^2}=2^{7-2}=2^5\\ Vậy:x=5\\ ----\\ b,\left(2x+1\right)^3=125=5^3\\ \Rightarrow 2x+1=5\\ 2x=5-1=4\\ x=\dfrac{4}{2}=2\\ ----\\ c,2x-2^6=6\\ 2x=6+2^6=6+64\\ 2x=70\\ x=\dfrac{70}{2}=35\\ ----\\ d,49.7^x=2401\\ 7^x=\dfrac{2401}{49}=49=7^2\\ Vậy:x=2\)
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52\(x+1\) + 31 = 128
5\(2x+1\) + 3 = 128
5\(^{2x+1}\) = 128 - 3
5\(^{2x+1}\) = 125
5\(^{2x+1}\) = 53
2\(x\) + 1 = 3
2\(x\) = 3 - 1
2\(x\) = 2
\(x\) = 2: 2
\(x\) = 1
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720:[41-(2x-5)]=2^3.5
720:[41-2x+5]=40
[41+5-2x]=720:40
46-2x=18
2x=46-18
2x=28
x=14
a, 2x.4=128
2x=128:4
2x=32
2x=25
x=5
Vay x=5
b, x15=x
x=1
Vay x=1
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( 2x + 1 )3 = 96
( 2x + 1 )3 = 63
2x + 1 = 6
2x = 6 - 1
2x = 5
x = 5 : 2
x = 2,5
16x < 1284
x = 0;1;2;3;4;........
\(2^{x+3}-2^{x+1}=128\)
\(2^{x+1}\cdot2^2-2^{x+1}=128\)
\(2^{x+1}\cdot\left(4-1\right)=128\)
\(2^{x+1}=\frac{128}{3}\)
Ủa đề sai ko bạn xem lại nhé ^^
\(x=4,415037499\)
mk nói thật đấy thử đi