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a, \(3^{2x+2}=9^{10}\\ 3^{2x+2}=\left(3^2\right)^{10}\\ 3^{2x+2}=3^{20}\\ \Rightarrow2x+2=20\\ \Rightarrow2x=18\\ \Rightarrow x=9\)Vậy x = 9
b, \(3^{3x}=27^{13}\\ 3^{3x}=\left(3^3\right)^{13}\\ 3^{3x}=3^{39}\\ \Rightarrow3x=39\\ \Rightarrow x=13\)Vậy x = 13
c, \(2^x=4^6\cdot16^3\\ 2^x=\left(2^2\right)^6\cdot\left(2^4\right)^3\\ 2^x=2^{12}\cdot2^{12}\\ 2^x=2^{24}\\ \Rightarrow x=24\)Vậy x = 24
d, \(2^x=32^5\cdot64^6\\ 2^x=\left(2^5\right)^5\cdot\left(2^6\right)^6\\ 2^x=2^{25}\cdot2^{36}\\ 2^x=2^{61}\\ \Rightarrow x=61\)Vậy x = 61
32 < 2x < 22x-3 . 28-2x
=> 25 < 2x < 22x - 3 . 28 - 2x
=> 25 < 2x < 25
=> 2x = 25
=> x = 5
a) 2x.(1 + 23) = 144
2x . 9 = 144
2x = 16
=> x = 4
b) (2x - 1)10 = (2x - 1)100
(2x - 1)100 - (2x - 1)10 = 0
(2x - 1)10.[ (2x - 1)90 - 1] = 0
=> (2x - 1)10 = 0 hoặc (2x - 1)90 - 1 = 0
=> 2x = 1 hoặc (2x - 1)90 = 1
=> x = \(\frac{1}{2}\) hoặc \(2x-1=\orbr{\begin{cases}1\\-1\end{cases}}\)
=> \(2x=\orbr{\begin{cases}2\\0\end{cases}}\)
=> x = {\(\frac{1}{2};1;0\)}
2.Tìm x, biết
a,(x-1)3=3 (x-1)3=343
=> (x-1)3=73
=> x -1 = 7
=> x = 8
B, (X-2)4=4096
(X-2)4= 84
=> x - 2 = 8
=> x = 10
C,(2x2-13)4=(-5)4
=> 2x2-13 = -5
=> 2x2 = 8
=> x2 = 4
=> x = 2
Study well
a. (x-1)3 = 343
=> (x-1)3 = 73
=> x - 1 = 7
=> x = 7 + 1
=> x = 8
33 + 2x = 2713
=> 33 . 32x = (33)13
=> 32x = 33.13 : 33
=> 32x = 313
=> 2x = 13
=> x = 13 : 2
=> x = 13/2
a) 3x+1=9x
3x+1=(32)x
3x+1=32x
=>x+1=2x
x-2x=-1
-x=-1
=>x=1
b)32x-1=243
32x-1=35
=>2x-1=5
2x=6
x=3
Hằng đẳng thức đó bn:
\(\left(a+b\right)\left(a^2-ab+b^2\right)\)
Thay vào thì: \(-\left(x-3\right)\left(x^2-3x+9\right)=-\left[\left(x-3\right)\left(x^2-3x+3^2\right)\right]\)
\(=-\left(x^3-27\right)=-x^3+27\)
Bài làm:
Ta có: \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)=\left(x-3\right)^3+3\left(2x+1\right)^2-\left(x^3-5x+1\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3+27=x^3-9x^2+27x-27+12x^2+12x+3-x^3+5x-1\)
\(\Leftrightarrow6x^2+41x-51=0\)
\(\Leftrightarrow6\left(x^2+\frac{41}{6}x+\frac{1681}{144}\right)-\frac{2905}{24}=0\)
\(\Leftrightarrow\left(x+\frac{41}{12}\right)^2-\frac{\left(\sqrt{2905}\right)^2}{12^2}=0\)
\(\Leftrightarrow\left(x+\frac{41}{12}-\frac{\sqrt{2905}}{12}\right)\left(x+\frac{41}{12}+\frac{\sqrt{2905}}{12}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{2905}-41}{12}\\x=\frac{-\sqrt{2905}-41}{12}\end{cases}}\)
332 - 2x = 913
=> 332 : 32x = (32)13
=> 332 : 32x = 326
=> 32x = 332 : 326
=> 32x = 36
=>2x = 6
=> x = 3
\(3^{32-2X}=9^{13}\)
(=) \(3^{32-2X}=3^{26}\)
=>32-2X=26
(=)2X=6
(=)X=3