5! + 5! + 5! = 120 + 120 + 120 = 120.3 = 360
Ta có: 5! = 1*2*3*4*5 => 5! + 5! + 5! = 1*2*3*4*5 + 1*2*3*4*5 + 1*2*3*4*5 = 3*(1*2*3*4*5)
= 3*120
= 360
Tìm số tự nhiên n, biết rằng:
\(\dfrac {4^{5} + {4^{5}} +{4^{5}} + {4^{5}}}{{3^{5}} + {3^{5}} + {3^{5}}}\) . \(\dfrac{6^{5} + {6^{5}} + {6^{5}} + {6^{5}} + {6^{5}} + {6^{5}} }{2^{5} + 2^{5}} = 2^{n}\)
Số tự nhiên n thỏa mãn:
\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^n\)
4^5+4^5+4^5+4^5/3^5+3^5+3^5.6^5+6^5+6^5+6^5+6^5+6^5/2^5+2^5=8^x Tìm x
Tìm x biết \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^4}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)=2\(2^x\)
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\)*\(\frac{6^5+6^5+^{_{6^5}}+6^5+6^5+6^5}{2^5+2^5}\)=\(^{2^x}\)
Tìm x biết
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^{\left|2x+6\right|}\)
tìm n biết:
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^n\)
tìm số nguyên dương n biết:\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^n\)
số nguyên dương n thỏa mãn đẳng thức ?
4^5+4^5+4^5+4^5/3^5+3^5+3^5.6^5+6^5+6^5+6^5+6^5+6^5/2^5+2^5=2^n là n = ?
