Tìm các số tự nhiên \(a\ne b\ne0\)sao cho \(\hept{\begin{cases}a+b=n\\ab⋮n\end{cases}}\)(n thuộc N*)
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Tìm các số tự nhiên \(a\ne b\ne0\)sao cho \(\hept{\begin{cases}a+b=n\\ab⋮n\end{cases}}\)(n thuộc N*)
\(=\left(2.\left(2^3\right)^4.\left(3^3\right)^2+2^2.\left(2.3\right)^9\right):\left(2^{12}.\left(3^2\right)^3.31\right)\)
\(=\left(2^{13}.3^6+2^{11}.3^9\right):\left(2^{12}.3^6.31\right)\)
\(=\left[2^{11}.3^6\left(2^2+3^3\right)\right]:\left(2^{12}.3^6.31\right)\)
\(=\frac{2^{11}.3^6.31}{2^{12}.3^6.31}=\frac{1}{2}\)
Đưa về phân số:
\(=\frac{2.8^4.27^2+4.6^9}{2^{12}.9^3.31}\)
\(=\frac{2.\left(2^3\right)^4.\left(3^3\right)^2+2^2.\left(2.3\right)^9}{2^{12}.\left(3^2\right)^3.31}\)
\(=\frac{2.2^{3.4}.3^{3.2}+2^2.2^9.3^9}{2^{12}.3^{2.3}.31}\)
\(=\frac{2.2^{12}.3^6+2^{2+9}.3^9}{2^{12}.3^6.31}\)
\(=\frac{2^{1+12}.3^6+2^{11}.3^9}{2^{12}.3^6.31}\)
\(=\frac{2^{13}.3^6+2^{11}.3^9}{2^{12}.3^6.31}\)
\(=\frac{2^{11}.3^6\left(2^2+3^3\right)}{2^{12}.3^6.31}\)
\(=\frac{2^{11}.3^6.31}{2^{12}.3^6.31}=\frac{1}{2}\)
Em hiểu hơn ko?
a) 25 . 8 3 - 22 . 83
\(=8^3.\left(25-22\right)=8^3.3=1536\)
b) 54 - 2 . 53
\(=5^3\left(5-2\right)=5^3.3=375\)
\(c,\frac{\left(3^4+3^3\right)\left(3^4+3\right)}{9^2}=\frac{108.84}{81}=112\)
d) 33. 22 - 33. 19
\(=3^3.\left(22-19\right)=3^3.3=3^4=81\)
e) 24 . 5 - [ 131 - ( 13- 42) ]
\(=80-\left(131+3\right)\)
\(=-54\)
2) Tìm x
a) 30- 5 . ( x - 3 ) = 5
\(5\left(x-3\right)=25\)
\(x-3=5\)
\(x=8\)
b) 9 + 3 . x = 37 : 32
\(9+3x=243\)
\(3x=234\)
x=78
C2 :
\(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^{2019}\)
\(2\cdot3\cdot3^{x+1}+4\cdot3^{x+1}=10\cdot3^{2019}\)
\(6\cdot3^{x+1}+4\cdot3^{x+1}=10\cdot3^{2019}\)
\(\left(6+4\right)\cdot3^{x+1}=10\cdot3^{2019}\)
\(10\cdot3^{x+1}=10\cdot3^{2019}\)
\(\Rightarrow x+1=2019\)
\(x=2019-1\)
\(x=2018\)
Vậy x = 2018
Chắc sai =))
\(\frac{1313}{1212}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=1\)
\(x=\frac{1313}{1212}:1\)
\(x=\frac{13}{12}\)
Lời giải
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)
\(x=\frac{1313}{1212}:1\)
\(x=\frac{13}{12}\)
49 - 3 . ( x + 6 ) = 13
3 . ( x + 6 ) = 49 - 13
3 . ( x + 6 ) = 36
x + 6 = 36 : 3
x + 6 = 12
x = 12 - 6
x = 6
49 - 3.( x + 6 ) = 13
3.( x + 6 ) = 36
x + 6 = 12
x = 6
Vậy x = 6
#)Giải :
\(9^8:\left(27^3.81^2\right)=\left(3^2\right)^8:\left[\left(3^3\right)^3.\left(3^4\right)^2\right]=3^{16}:\left[3^9.3^8\right]\)
Đến đây ez rùi ^^
\(9^8:\left(27^3.81^2\right)=3^{16}:\left(3^9.3^8\right)=3^{16}:\left(3^{9+8}\right)=3^{16}:3^{17}=3^{-1}=\frac{1}{3}\)
\(9^8:\left(27^3.81^2\right)\)
\(=3^{16}:\left(3^9.3^8\right)\)
\(=3^{16}:3^{17}\)
\(=\frac{1}{3}\)