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a: x^3=7^3
=>x^3=343
=>\(x=\sqrt[3]{343}=7\)
b: x^3=27
=>x^3=3^3
=>x=3
c: x^3=125
=>x^3=5^3
=>x=5
d: (x+1)^3=125
=>x+1=5
=>x=4
e: (x-2)^3=2^3
=>x-2=2
=>x=4
f: (x-2)^3=8
=>x-2=2
=>x=4
h: (x+2)^2=64
=>x+2=8 hoặc x+2=-8
=>x=6 hoặc x=-10
j: =>x-3=2 hoặc x-3=-2
=>x=1 hoặc x=5
k:
9x^2=36
=>x^2=36/9
=>x^2=4
=>x=2 hoặc x=-2
l:
(x-1)^4=16
=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3x-1\right)^3=64\)
\(\left(3x-1\right)^3=4^3\)
3x - 1 = 4
3x = 4 + 1
3x = 5
x = 5/3
\(\left(3x-1\right)^3=64\)
\(\Leftrightarrow3x-1=4\)
\(\Leftrightarrow3x=5\Leftrightarrow x=\frac{5}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9^7.81:9^5=9^7.9^2:9^5=9^{7+2-5}=9^4\\ x^{12}:x.x^8=x^{12-1+8}=x^{19}\\ 16.2^4:8=2^4.2^4:2^3=2^{4+4-3}=2^5\\ 64.4^5:16=4^3.4^5:4^2=4^{3+5-2}=4^6\\ 3^{12}.3:3^8=3^{12+1-8}=3^5\\ 7^9.7^{12}:2015^0=7^{9+12}:1=7^{19}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2. 2(x-1) +3 2-x) =- 1
\(\Leftrightarrow2x-2+6x-3=-1\)
\(\Leftrightarrow8x-5=-1\Leftrightarrow8x=4\Leftrightarrow x=\frac{1}{2}\)
Vậy x = 1/2
3. ( n2 + 3 ) chia hết cho ( n - 1)
\(\Leftrightarrow n^2-1+4⋮n-1\Leftrightarrow\left(n-1\right)\left(n+1\right)+4⋮n-1\)
Vì n thuộc Z => ( n-1) ( n+1) thuộc Z
\(\Rightarrow\left(n-1\right)\left(n+1\right)+4⋮n-1\Leftrightarrow4⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Phần còn lại bn tự làm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)
=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)
=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)
b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)
c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)
d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)
=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)
=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)
e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)
=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)
=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)
f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)
=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)
=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)
=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1+\left(x-\frac{2}{3}\right)^3=\frac{37}{64}\)
\(\left(x-\frac{2}{3}\right)^3=\frac{37}{64}-1\)
\(\left(x-\frac{2}{3}\right)^3=\frac{-27}{64}\)
\(\left(x-\frac{2}{3}\right)^3=\left(\frac{-3}{4}\right)^3\)
\(\Rightarrow x-\frac{2}{3}=\frac{-3}{4}\)
\(\Rightarrow x=\frac{-3}{4}+\frac{2}{3}\)
\(\Rightarrow x=\frac{-1}{12}\)
\(1+\left(x-\frac{2}{3}\right)^3=\frac{37}{64}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{37}{64}-1\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{-27}{64}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\left(\frac{-3}{4}\right)^3\)
\(\Rightarrow x-\frac{2}{3}=\frac{-3}{4}\)
\(\Rightarrow x=\frac{-3}{4}+\frac{2}{3}\)
\(\Rightarrow x=\frac{-9}{12}+\frac{8}{12}\)
\(\Rightarrow\frac{-1}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: ( x - l ) 3 = 3 3 nên x -1 = 3. Do đó x = 4.
b) Ta có: ( x + 1 ) 3 = 4 3 nên x +1 = 4. Do đó x = 3.