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\(l,\\ 2^x=1=2^0\\ Vậy:x=0\\ m,\\ 3^x=81=3^4\\ Vậy:x=4\\ n,\\ 3^x=37=3^3\\ Vậy:x=3\\ o,\\ 9^x=3^4=\left(3^2\right)^2=9^2\\ Vậy:x=2\)
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\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\)
\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}\)
Số nào xuất hiện 2 lần thì thay thế những số đó bằng số 1.
\(B=\frac{1}{2020}\)
B = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2019}\right).\left(1-\frac{1}{2020}\right)\)
= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2018}{2019}.\frac{2019}{2020}\)
= \(\frac{1.2.3...2019}{2.3.4..2020}\)(Nếu có 2 thừa số giống nhau lặp lại ở tử số và mẫu số thì rút gọn coi như triệt tiêu hết và không có gì)
= \(\frac{1}{2020}\)
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2^n.4=128
2^n=128:4
2^n=32
2^n=2^5
Suy ra n=5
3^n+1.9=81
3^n+1=81:9
3^n+1=9
3^n+1=3^2
Suy ra n=1
15^n-2=9^2:3^4
15^n-2=3^4:3^4
15^n-2=3^0
15^n-2=1
15^n-2=15^0
Suy ra n=2
co gi ko hieu thi hoi nha
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a) \(6⋮\left(x-1\right)\left(đkxđ:x\ne1;x\inℕ\right)\)
\(\Rightarrow x-1\in U\left(6\right)=\left\{1;2;3;6\right\}\)
\(\Rightarrow x\in\left\{2;3;4;7\right\}\)
b) \(14⋮\left(2x+3\right)\left(đkxđ:x\ne-\dfrac{3}{2};x\inℕ\right)\)
\(\Rightarrow2x+3\in U\left(14\right)=\left\{1;2;7;14\right\}\)
\(\Rightarrow x\in\left\{-1;-\dfrac{1}{2};2;\dfrac{9}{2}\right\}\)
\(\Rightarrow x\in\left\{-2\right\}\)
\(a,6⋮\left(x-1\right)\\ \Rightarrow\left(x-1\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\\ Ta.có:x-1=-6\Rightarrow x=-5\left(loại\right)\\ x-1=-3\Rightarrow x=-2\left(loại\right)\\ x-1=-2\Rightarrow x=-1\left(loại\right)\\ x-1=-1\Rightarrow x=0\left(nhận\right)\\ x-1=1\Rightarrow x=2\left(nhận\right)\\ x-1=2\Rightarrow x=3\left(nhận\right)\\ x-1=3\Rightarrow x=4\left(nhận\right)\\ x-1=6\Rightarrow x=7\left(nhận\right)\\ Vậy:x\in\left\{0;2;3;4;7\right\}\)
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a, \(x^2\) = \(x^3\)
\(x^3\) - \(x^2\) = 0
\(x^2\)( \(x\) -1) = 0
\(\left[{}\begin{matrix}x^2=0\\x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(x\) \(\in\) { 0; 1}
e, 32\(x+1\) = 27
\(3^{2x}\)+1 = 33
2\(x\) + 1 = 3
2\(x\) = 2
\(x\) = 1
g, 62 = 6\(x-3\)
2 = \(x-3\)
\(x\) = 3 + 2
\(x\) = 5
\(a,x^2=x^3\\ \Rightarrow x^2-x^3=0\\ \Rightarrow x^2\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\1-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(b,3^{2x+1}=27\\ \Rightarrow3^{2x+1}=3^3\\ \Rightarrow2x+1=3\\ \Rightarrow2x=3-1\\ \Rightarrow2x=2\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)
\(c,6^2=6^{x-3}\\ \Rightarrow6^{x-3}=6^2\\ \Rightarrow x-3=2\\ \Rightarrow x=2+3\\ \Rightarrow x=5\)
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a) x = 1
b) x = 2
c) ko có Đ/Á
d) x = 3
a) 5 . x3 - 5 = 0
=> 5 . ( x3 - 1 ) = 0
=> x3 - 1 = 0
=> x3 = 1
=> x3 = 13
=> x = 1
b) ( x + 1 )3 = 27
=> ( x +1 )3 = 33
=> x + 1 = 3
=> x = 2
c) Sửa đề :
( x + 1 )4 = 16
=> ( x + 1 )4 = 16
=> ( x + 1 )4 = 24
=> x + 1 = 4
=> x = 2
h) ( 2x - 1 )3 = 125
=> ( 2x - 1 )3 = 53
=> 2x - 1 = 5
=> 2x = 6
=> x = 3
\(3^4.3^{x+1}=3^{20}\)
\(3^{x+5}=3^{15+5}\)
\(=>x=15\)
Vậy x=15
\(81\cdot3^{x+1}=3^{20}\)
\(3^4\cdot3^x\cdot3=3^{20}\)
\(3^5\cdot3^x=3^{20}\)
\(3^x=3^{20}:3^5=3^{15}\)
\(\Rightarrow x=15\)