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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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Ta có : (x - 5)4 = (x - 5)6
=> (x - 5)4 - (x - 5)6 = 0
=> (x - 5)4 (1 - (x - 5)2) = 0
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)=0\\\left(x-5\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x-5=1;-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=4;6\end{cases}}\)
Vậy x = {4;5;6}
Tìm x thuộc N:
2x.4=128
2x =128:4
2x =32
x =32:2
x =16
b.x.15=x
x thỏa mãn điều kiện:0,1
c.(2x+1)3=125
(2x+1)3=53
==>2x+1=5
2x =5-1
2x =4
x =4:2=2
phần d mk chưa hiểu lắm
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a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
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Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2^x.4=128\)
\(\Rightarrow2^x=128:4=32\)
Mà \(32=2^5\)
\(\Rightarrow2^x=2^5\)
Vậy x = 5
b) \(x^{15}=x\)
\(\Rightarrow x=\left\{0;1;-1\right\}\)
c) Ta có: \(125=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1=4\)
\(\Rightarrow x=4:2=2\)
Vậy x = 2
d) Ta có: \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow x=5\)
Ủng hộ tớ nha?
\(2^x.4=128\)
\(\Rightarrow2^x.2^2=2^7\)
\(\Rightarrow x+2=7\)
\(\Rightarrow x=5\)
\(x^{15}=x\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=+-1\end{cases}}\)
\(\left(2x+1\right)^3=125\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x . 4 = 128
2x = 128 : 4
2x = 32
2x = 2 . 2 . 2 . 2 . 2
2x = 25
x = 5
(2x + 1)3 = 125
(2x + 1)3 = 5 . 5 . 5
(2x + 1)3 = 53
2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2
x15 = x
x = 1
(x - 5)4 = (x - 5)6
x = 6
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Mình giải phần 1 ) thôi
\(1)\)
\(a)\frac{3}{2}x-\frac{1}{3}=1-x\)
\(\Rightarrow\frac{3}{2}x+x=1-\frac{1}{3}\)
\(\Rightarrow\frac{5}{2}x=\frac{2}{3}\)
\(\Rightarrow x=\frac{2}{3}:\frac{5}{2}\)
\(\Rightarrow x=\frac{2}{3}.\frac{2}{5}\)
\(\Rightarrow x=\frac{4}{15}\)
b ) \(\left(\frac{1}{3}+x\right)^3=27\)
\(\Rightarrow\frac{1}{3}+x=3\)
\(\Rightarrow x=3-\frac{1}{3}\)
\(\Rightarrow x=\frac{9}{3}-\frac{1}{3}\)
\(\Rightarrow x=\frac{8}{3}\)
Chúc bạn học tốt !!!
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\(a,2^x.4=128\\2^x.2^2=2^7\\ 2^x=\dfrac{2^7}{2^2}=2^{7-2}=2^5\\ Vậy:x=5\\ ----\\ b,\left(2x+1\right)^3=125=5^3\\ \Rightarrow 2x+1=5\\ 2x=5-1=4\\ x=\dfrac{4}{2}=2\\ ----\\ c,2x-2^6=6\\ 2x=6+2^6=6+64\\ 2x=70\\ x=\dfrac{70}{2}=35\\ ----\\ d,49.7^x=2401\\ 7^x=\dfrac{2401}{49}=49=7^2\\ Vậy:x=2\)
2^x . 4=128 x^15=x (2x+1)^3=125 (x-5)^4=(x-5)^6
2^x=128 : 4 =>x=1 (2x+1)^3=5^3 =>x=5 hoặc 6
2^x=32 => 2x+1=5
2^x=2^5 2x=5-1
=>x=5 2x=4
x=4 : 2
x=2