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\(2^x.4=128\)
\(2^x=128:4\)
\(2^x=32\)
\(\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
\(x^{15}=x\Leftrightarrow x\in\left\{-1;0;1\right\}\)
\(\left(2x+1\right)^3=125\)
\(\Leftrightarrow\left(2x+1\right)^3=5^3\)
\(\Leftrightarrow2x+1=5\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
\(\left(x-5\right)^6=\left(x-5\right)^4\)
\(\Leftrightarrow\hept{\begin{cases}x-5=-1\\x-5=0\\x-5=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\x=5\\x=6\end{cases}}\)
\(\text{Vậy:}\)\(x\in\left\{4;5;6\right\}\)
\(2^x.4=128\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5.\)
\(x^{15}=x\Rightarrow\orbr{\begin{cases}x=\pm1\\x=0\end{cases}}\)
\(\left(2x+1\right)^3=125\)
<=> \(\left(2x+1\right)^3=5^3\)
<=> \(2x+1=5\)
<=> \(x=2\)
\(\left(x-5\right)^6=\left(x-5\right)^4\)
<=> \(\left(x-5\right)^6-\left(x-5\right)^4=0\)
<=> \(\left(x-5\right)^4.\left[\left(x-5\right)^2-1\right]=0\)
<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)
Giải ra được x = 5 ; x = 6 ; x = 4 .
\(\left(2x+1\right).y=5\)
\(\Rightarrow2x+1;y\inƯ\left(5\right)=\left\{1;5\right\}\)
\(TH1:\hept{\begin{cases}2x+1=1\\y=5\end{cases}\Rightarrow\hept{\begin{cases}x=0\\y=5\end{cases}}}\)
\(TH2:\hept{\begin{cases}2x+1=5\\y=1\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}}\)
Vậy....................
a: =>15-(x-2)=-13-27=-40
=>x-2=15+40=55
hay x=57
b: =>5-x=-114+12=-102
=>x=107
c: \(\Leftrightarrow\left|x\right|=-1-5=-6\)(vô lý)
d: \(\Leftrightarrow\left|x-3\right|=3\)
=>x-3=3 hoặc x-3=-3
=>x=6 hoặc x=0
a) 2^x.2^4=128
=>2^x.2^2=2^7
=>2^x=2^7:2^2
=>2^x=2^5
=>x=5
b)x^15=x
=>x^15-x=0
=>x(x^16-x)=0
=>2 trượng hợp:x=0 và x^16-1=0(x^16-1=0 cx 2 th nha)
b),d),e) như nhau nha!
c) dễ rồi
\(a)2^x\cdot4=128\)
\(\Rightarrow2^x=\frac{128}{4}\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(b)x^{15}=x\)
\(\Rightarrow x^{15}-x=0\)
\(\Rightarrow x(x^{14}-1)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x^{14}=1\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=1\end{cases}}\)
\(c)(2x+1)^3=125\)
\(\Rightarrow(2x+1)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2=2\)
\(d)(x-5)^4=(x-5)^6\)
\(\Rightarrow(x-5)^6-(x-5)^4=0\)
\(\Rightarrow(x-5)^4\cdot\left[(x-5)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}(x-5)^4=0\\(x-5)^2-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=6\end{cases}}\)
\(e)(2x-15)^5=(2x-15)^3\)
\(\Rightarrow(2x-15)^5-(2x-15)^3=0\)
\(\Rightarrow(2x-15)^3-\left[(2x-15)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}(2x-15)^3=0\\(2x-15)^2-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\varnothing\\x=8\end{cases}}\)
Chúc bạn hoc tốt :>
\(a.2^x.4=128\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(b.x^{15}=x\)
\(\Rightarrow x^{15}-x=0\)
\(\Rightarrow x.\left(x^{14}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^{14}=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(c.\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow\left(2x+1\right)=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=\frac{4}{2}\)
\(\Rightarrow x=2\)
\(d.\left(x-5\right)^4=\left(x-5^6\right)\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4.\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(e.\left(2x-15\right)^5=\left(2x-15\right)^4\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^4=0\)
\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7,5\\x=8\end{cases}}\)
a)
2x . 4 = 128
2x + 2 = 27
x + 2 = 7
x = 7 - 2
x = 5
b)
x15 = x
x15 - x = 0
x . (x14 - 1) = 0
Có 2 trường hợp:
TH1:
_ x = 0
TH2:
_ x14 - 1 = 0
_ x có hai trường hợp:
+ TH1:
+ x = -1
+ TH2:
+ x = 1
Đáp số:
x1 = 0
x2 = -1
x3 = 1
c)
(2x + 1)3 = 125
(2x + 1)3 = 53
2x + 1 = 5
2x = 5 - 1
2x = 4
x = 2
d)
(x - 5)4 = (x - 5)6
(x - 5)4 - (x - 5)6 = 0
(x - 5)4 . [1 - (x - 5)2] = 0
Có 2 trường hợp:
TH1:
_ (x - 5)4 = 0
_ x = 5
TH2:
_ 1 - (x - 5)2 = 0
_Có 2 trường hợp:
+ TH1:
+ x = 4
+ TH2:
+ x = 6
Đáp số:
x1 = 5
x2 = 4
x3 = 6
e)
(2x - 15)5 = (2x - 15)3
(2x - 15)5 - (2x - 15)3 = 0
(2x - 15)3 . [(2x - 15)2 - 1] = 0
Có 2 trường hợp:
_ TH1:
_ (2x - 15)3 = 0
_ x =\(\frac{15}{2}\)
_TH2:
_ (2x - 15)2 - 1 = 0
_ Có 2 trường hợp:
+ x = 7
+ x = 8
Đáp số:
x1 = \(\frac{15}{2}\)
x2 = 7
x3 = 8