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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 <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)
b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)
c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)
d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)
e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)
g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)
h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)
i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)
k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)
Bài 1L
a) \(\left(x-7\right)\left(x+3\right)< 0\)
TH1:
\(\hept{\begin{cases}x-7>0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>7\\x< -3\end{cases}}}\)( loại )
TH2:
\(\hept{\begin{cases}x-7< 0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 7\\x>-3\end{cases}\Leftrightarrow}-3< x< 7}\)( chọn )
Vậy \(-3< x< 7\)
Bài 2:
a) \(\left(5x+8\right)-\left(2x-15\right)+21=2x-5\)
\(\Leftrightarrow5x+8-2x+15+21=2x-5\)
\(\Leftrightarrow5x-2x-2x=-5-21-8-15\)
\(\Leftrightarrow x=-49\)
Vậy ...
a, 3 - 2x = 3 . (5 - x) + 4
3 - 2x = 15 - 3x + 4
-2x + 3x = 15 + 4 - 3
x = 16
b, 4 - (7x + 2017) = 6 . (5 - x) - 2017
4 - 7x - 2017 = 30 - 6x - 2017
-7x + 6x = 30 - 2017 - 4 + 2017
-x = 26
x = -26
c, 15 - x . (x + 1) = 4 - x^2 + 2x
15 - x^2 - x = 4 - x^2 + 2x
-x^2 - x + x^2 - 2x = 4 - 15
-3x = -11
x = 11/3
d, -4 . (x - 5) + 2016 = 3 . (8 - x) - (2x - 2016)
-4x + 20 + 2016 = 24 - 3x - 2x + 2016
-4x + 3x +2x = 24 + 2016 - 20 - 2016
x = 4
đúng 100%
a/(x-5)^4=9x-5)^6\(\Rightarrow\) TH1:x-5=1\(\rightarrow\) x=6
\(\Rightarrow\) TH2:x-5=-1\(\rightarrow\) x=4
\(\Rightarrow\) TH3:x-5=0\(\rightarrow\) x=5
vậy....................
b/(2x-15)^5=(2x-15)^6\(\Rightarrow\) 2x-15=1\(\rightarrow\) 2x=16\(\rightarrow\)x=8
\(\Rightarrow\) 2x-15=0\(\rightarrow\)2x=15\(\rightarrow\)x=7.5
vậy................
a) Có : ( x-5 ) \(^4\) = ( x - 5 ) \(^6\)
\(\Rightarrow\left(x-5\right)^6:\left(x-5\right)^4=1\Rightarrow\left(x-5\right)^2=1\)
\(\Rightarrow x-5=1\) hoặc \(x-5=-1\)
\(\Rightarrow x=1+5\) hoặc \(x=-1+5\)
\(\Rightarrow x=6\) hoặc \(x=4\)
b) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5:\left(2x-15\right)^3=1\Rightarrow\left(2x-15\right)^2=1\)
\(\Rightarrow2x-15=1\) hoặc \(2x-15=-1\)
\(\Rightarrow2x=16\) hoặc \(2x=14\)
\(\Rightarrow x=8\) hoặc \(x=7\)
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
=> 2x-15 = 1 hoặc 0
=> x=8 (x là số tự nhiên)
b) => x-5 = 0 ; 1 hoặc -1
=> x = 5 ; 6 hoặc 4
a) \(x^{15}=x\)
\(\Leftrightarrow x\left(x^{14}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^{14}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
b) \(\left(x-5\right)^6=\left(x-5\right)^4\)
\(\Leftrightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Leftrightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\\x-5=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=1\\2x-15=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
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