Tìm x biết :
a) x2 - 3 = 13
b) (x - 1)3 -1=26
c) 3x . 27= 96
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TM
2
17 tháng 2 2021
a) 9x-1=32
( 32 )x-1 = 32
32x-2 = 32
⇒ 2x-2 = 2
2x = 2+2
2x = 4
x = 4 : 2
x = 2
b) 5x+2=625
5x+2= 54
⇒ x+2 = 4
x = 4-2
x = 2
c) 2x: 25= 2
2x:25 = 21
2x = 21 . 25
2x = 26
⇒ x = 6
d) 3x:27=3
3x:33 = 31
3x = 31.33
3x = 34
⇒ x = 4
17 tháng 2 2021
a) Ta có: \(9^{x-1}=3^2\)
\(\Leftrightarrow3^{2x-2}=3^2\)
\(\Leftrightarrow2x-2=2\)
\(\Leftrightarrow2x=4\)
hay x=2
Vậy: x=2
b) Ta có: \(5^{x+2}=625\)
\(\Leftrightarrow5^{x+2}=5^4\)
\(\Leftrightarrow x+2=4\)
hay x=2
Vậy: x=2
c) Ta có: \(2^x:2^5=2\)
\(\Leftrightarrow2^{x-5}=2^1\)
\(\Leftrightarrow x-5=1\)
hay x=6
Vậy: x=6
d) Ta có: \(3^x:27=3\)
\(\Leftrightarrow3^x:3^3=3\)
\(\Leftrightarrow3^{x-3}=3^1\)
\(\Leftrightarrow x-3=1\)
hay x=4
Vậy: x=4
CM
11 tháng 9 2019
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
17 tháng 10 2021
\(\left(2x+x^2\right)\left(x^2-3x+2\right)=0\Leftrightarrow x\left(x+2\right)\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=2\end{matrix}\right.\\ A=\left\{-2;0;1;2\right\}\)
\(3\le x^3\le27\Leftrightarrow x\in\left\{2;3\right\}\\ B=\left\{2;3\right\}\)
\(\Leftrightarrow A\cup B=\left\{-2;0;1;2;3\right\}\)
29 tháng 10 2020
a) x2 - 9 = 3( x - 3 )
⇔ ( x - 3 )( x + 3 ) - 3( x - 3 ) = 0
⇔ ( x - 3 )( x + 3 - 3 ) = 0
⇔ ( x - 3 ).x = 0
⇔ x - 3 = 0 hoặc x = 0
⇔ x = 3 hoặc x = 0
b) 3( 3x2 + 1 ) = 6 - 2( 3x + 2 )
⇔ 9x2 + 3 = 6 - 6x - 4
⇔ 9x2 + 6x + 3 - 6 + 4 = 0
⇔ 9x2 + 6x + 1 = 0
⇔ ( 3x + 1 )2 = 0
⇔ 3x + 1 = 0
⇔ x = -1/3
6 tháng 3 2022
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
16 tháng 11 2021
\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)
\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)
a) \(\Rightarrow x^2=16\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
b) \(\Rightarrow\left(x-1\right)^3=27\Rightarrow x-1=3\Rightarrow x=4\)
c) \(\Rightarrow3^x.3^3=3^{12}\)
\(\Rightarrow3^x=3^9\Rightarrow x=9\)