Cho S= 1+5+5 mũ 2+5 mũ 3+......+ 5 mũ 20.TÌm n thỏa mãn 4S + 1= 5 mũ n
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3 tháng 9 2023
a) \(S=1+5+5^2+5^3+...+5^{28}\)
\(S=\left(1+5\right)+\left(5^2+5^3\right)+...+\left(5^{27}+5^{28}\right)\)
\(S=1\left(1+5\right)+5^2\left(1+5\right)+...+5^{27}\left(1+5\right)\)
\(S=\left(1+5^2+...+5^{27}\right).6⋮3\left(dpcm\right)\)
b) \(S=1+5+5^2+5^3+...+5^{28}\)
\(\Rightarrow5S=5+5^2+5^3+5^4+...+5^{29}\)
\(\Rightarrow5S-S=\left(5+5^2+5^3+5^4+...+5^{29}\right)-\left(1+5+5^2+5^3+...+5^{28}\right)\)
\(\Rightarrow4S=5^{29}-1\)
\(\Rightarrow4S+1=5^{29}-1+1\)
\(\Rightarrow4S=5^{29}=5^n\)
\(\Rightarrow n=29\)
3 tháng 9 2023
a) \(S=1+5+5^2+5^3+...+5^{28}\)
\(\Rightarrow S=\left(1+5\right)+5^2\left(1+5\right)+...+5^{27}\left(1+5\right)\)
\(\Rightarrow S=6+5^2.6+...+5^{27}.6\)
\(\Rightarrow S=6\left(1+5^2+...+5^{27}\right)⋮6\)
\(\Rightarrow S=6\left(1+5^2+...+5^{27}\right)⋮3\)
\(\Rightarrow dpcm\)
b) Bạn xem lại đề
AH
Akai Haruma
Giáo viên
7 tháng 10 2023
Lời giải:
$A=5^{50}-5^{48}+5^{46}-5^{44}+....-5^4+5^2-1$
$5^2A=5^{52}-5^{50}+5^{48}-5^{46}+...-5^6+5^4-5^2$
$\Rightarrow A+5^2A=5^{52}-1$
$\Rightarrow 26A=5^{52}-1$
$\Rightarrow 5^{52}-1+1=5^n$
$\Rightarrow 5^{52}=5^n$
$\Rightarrow n=52$
16 tháng 3 2020
b1
ta có : n+4 = (n+1)+3
=>n+1+3 chia hết cho n+1
vì n+1 chia hết cho n+1
=>3 chia hết cho n+1
=> n+1 chia hết cho 3
=> n+1 thuộc Ư 3 =[1;3]
=> n+1=1 n+1=3
n =1-1 n =3-1
n =0 n =2
vậy n thuộc [0;2]
VM
13 tháng 8 2019
3.
a) \(\left(x-1\right)^3=125\)
=> \(\left(x-1\right)^3=5^3\)
=> \(x-1=5\)
=> \(x=5+1\)
=> \(x=6\)
Vậy \(x=6.\)
b) \(2^{x+2}-2^x=96\)
=> \(2^x.\left(2^2-1\right)=96\)
=> \(2^x.3=96\)
=> \(2^x=96:3\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> \(x=5\)
Vậy \(x=5.\)
c) \(\left(2x+1\right)^3=343\)
=> \(\left(2x+1\right)^3=7^3\)
=> \(2x+1=7\)
=> \(2x=7-1\)
=> \(2x=6\)
=> \(x=6:2\)
=> \(x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
27 tháng 7 2023
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
27 tháng 7 2023
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
22 tháng 12 2017
a,\(5^3.2-100:4+2^3.5\)
= 125 . 2 - 25 + 8 . 5
= 250 - 25 + 40
= 265
b, \(6^2:9+50.2-3^3.3\)
= 36 : 9 + 100 - 27 . 3
= 4 + 100 - 81
= 23
21 tháng 10 2019
b) \(5^3\cdot2-100:4+2^3\cdot5\)
\(=125\cdot2-25+8\cdot5\)
\(=250-25+40\)
\(=225+40=265\)
c) \(6^2:9+50\cdot2+3^3-3\)
\(=36:9+100+27-3\)
\(=4+100+27-3\)
\(=104+27-3=131-3=128\)
d) \(3^2\cdot5+2^3\cdot10-81:3\)
\(=9\cdot5+8\cdot10-27\)
\(=45+80-27\)
\(=125-27=98\)
e) \(5^{13}:5^{10}-25\cdot2^2\)
\(=5^{13-10}-5^2\cdot2^2\)
\(=5^3-\left(5\cdot2\right)^2\)
\(=125-10^2\)
\(=125-100=25\)
f) \(20:2^2+5^9:5^8\)
\(=20:4+5^{9-8}\)
\(=5+5^1=5+5=10\)
g) \(100:5^2+7\cdot3^2\)
\(=10^2:5^2+7\cdot9\)
\(=\left(10:5\right)^2+63\)
\(=2^2+63=4+63=67\)
h) \(84:4+3^9:3^7+5^0\)
\(=21+3^{9-7}+1\)
\(=21+3^2+1\)
\(=21+9+1=30+1=31\)
i) \(29-\left[16+3\cdot\left(51-49\right)\right]\)
\(=29-\left[16+3\cdot2\right]\)
\(=29-\left[16+6\right]\)
\(=29-22=7\)
j) \(\left(15^{19}:5^{17}+3\right)\cdot0:7\)
\(=\left[\left(3\cdot5\right)^{19}:5^{17}+3\right]\cdot0\)
Vì số nào nhân cho 0 cũng bằng 0 nên giá trị biểu thức trên bằng 0
k) \(7^9:7^7-3^2+2^3\cdot5\)
\(=7^{9-7}-9+8\cdot5\)
\(=7^2-9+40\)
\(=49-9+40=40+40=80\)
l) \(1200:2+6^2\cdot2^1+18\)
\(=600+36\cdot2+18\)
\(=600+72+18\)
\(=600+\left(72+18\right)=600+90=690\)
m) \(5^9:5^7+70:14-20\)
\(=5^{9-7}+5-20\)
\(=5^2+5-20\)
\(25+5-20=30-20=10\)
Những câu sau mình làm sau nhé bạn!!!!!!!
\(S=1+5+5^2+5^3+......+5^{20}\)
\(\Rightarrow5S=5+5^2+5^3+5^4+........+5^{21}\)
\(\Rightarrow5S-S=5^{21}-1\)
\(\Rightarrow4S=5^{21}-1\)
Ta có: \(4S+1=5^n\)
\(\Leftrightarrow5^{21}-1+1=5^n\)
\(\Leftrightarrow5^n=5^{21}\)\(\Leftrightarrow n=21\)
Vậy \(n=21\)