D=15(4^2+1)(4^4+1)...(4^64+1))
15(4^2+1)(4^4+1)...(4^64+1)
15(4^2+1)(4^4+1)...(4^64+1)
⇒D=(42−1)(42+1)(44+1)(464+1)
=>D=(44−1)...(464+1)
=>D=(464−1)(464+1)
=>D=4128−1
- Vậy D=4128−1
\(15\cdot\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)....\left(4^{64}+1\right)\)
\(=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)....\left(4^{64}+1\right)\)
\(=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)...\left(4^{64}+1\right)\)
\(=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)
\(=\left(4^{32}-1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)
\(=\left(4^{64}-1\right)\left(4^{64}+1\right)\)
\(=4^{128}+1\)
lỗi nên đăng lại nhe
15(4^2+1)(4^4+1)...(4^64+1)
⇒D=(42−1)(42+1)(44+1)(464+1)
=>D=(44−1)...(464+1)
=>D=(464−1)(464+1)
=>D=4128−1
- Vậy D=4128−1
Tính A=(4^2+1)(4^4+1)(4^8+1)(4^16+1)(4^32+1)-1/15 . 4^64
a,|x|-7/6=9/15
b,|x-4/3|=1/6
c,|x-4/3|-1/3=1/2
d,8/3-|7/9-x|=-1/5
e,|x-1/4^2|-25/64=0
f,(x-1/4)^2+17/64=21/32
a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)
=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)
=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)
b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)
c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)
d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)
=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)
=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)
e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)
=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)
=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)
f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)
=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)
=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)
=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)
A=(4^2+1)(4^4+1)(4^8+1)(4^16+1)(4^32+1)−1/15 . 4^64
Ta có (42 - 1)(42 + 1) = 44 - 1
Ta có 15A = (42 - 1)(42 + 1)(44 + 1)(48 + 1)(416 + 1)(432 + 1) - 464 = 464 - 1 - 464 = -1
=> A = \(\frac{-1}{15}\)
Ghi lại cái đề cho rõ hơn đi t giải cho
Cái đề đọc không ra.Ghi lại cái đề cho rõ hơn đi t giải cho
Tính thuận tiện
a) A=(15 - 6 13/18) : 11 1/27 - 2 1/8 : 1 11/40
b) B=(-3,2) . -15/64 + (0,8 - 2 4/15): 3 2/3
c) C= -7/9 . 4/11 + -7/9. 7/11 + 5 7/9
d) D= 50% .1 1/3 . 10 . 7/35 . 0,75
e) E= (2/5)^2 + 5 1/2 (4,5 - 2) + 23/-4
a: \(=\left(15-6-\dfrac{13}{18}\right):\dfrac{298}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)
\(=\dfrac{149}{18}\cdot\dfrac{27}{298}-\dfrac{5}{3}=\dfrac{3}{2}-\dfrac{5}{3}=\dfrac{9-10}{6}=\dfrac{-1}{6}\)
b: \(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\dfrac{-22}{15}:\dfrac{11}{2}\)
\(=\dfrac{3}{4}-\dfrac{4}{15}=\dfrac{29}{60}\)
c: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=\dfrac{-7}{9}+\dfrac{7}{9}+5=5\)
d: \(=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot10\cdot\dfrac{1}{5}\cdot\dfrac{3}{4}=1\)
e: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+\dfrac{-23}{4}=\dfrac{204}{25}\)
a,3/2-5/6:(1/2)2+\(\sqrt{0,25-\sqrt{\dfrac{1}{4}}}\)
b,-4/3:2/9+13/12:-13/8
c,(-1/2)2-[-1/6:|-1+5|-\(\sqrt{64}.\left(\dfrac{1}{4}\right)^2\)
d,15^11.5^7.9^2/5^18.27^6
\(a,=\dfrac{3}{2}-\dfrac{5}{6}:\dfrac{1}{4}+\sqrt{\dfrac{1}{4}-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{10}{3}+\sqrt{\dfrac{1}{2}}=-\dfrac{11}{6}+\dfrac{\sqrt{2}}{2}=\dfrac{-33+3\sqrt{2}}{6}\)
\(b,=-\dfrac{4}{3}\cdot\dfrac{9}{2}+\dfrac{13}{12}\cdot\left(-\dfrac{8}{13}\right)=6-\dfrac{2}{3}=\dfrac{16}{3}\\ c,=\dfrac{1}{4}-\left(-\dfrac{1}{6}:4-8\cdot\dfrac{1}{16}\right)=\dfrac{1}{4}-\left(-\dfrac{1}{24}-\dfrac{1}{2}\right)\\ =\dfrac{1}{4}-\dfrac{13}{24}=-\dfrac{7}{24}\\ d,=\dfrac{3^{11}\cdot5^{11}\cdot5^7\cdot3^4}{5^{18}\cdot3^{18}}=\dfrac{1}{3^3}=\dfrac{1}{27}\)
BT7: Tính
\(3,C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{16}+1\right)\)
\(4,D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(5,E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
3: =(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1
4:
D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)
=(4^8-1)(4^8+1)*...*(4^64+1)
=...
=4^128-1
5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)
=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1
=5^256-1+5^256-1
=2*5^256-2
3, \(C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)
\(C=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)
\(C=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)
\(C=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(C=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(C=5^{32}-1\)
4, \(D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^4-1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^8-1\right)\left(4^8+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^{16}-1\right)\left(4^{16}+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^{32}-1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)
\(D=\left(4^{64}-1\right)\left(4^{64}+1\right)\)
\(D=4^{128}-1\)
5, \(E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{256}+1\right)\)
\(E=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{128}+1\right)\left(5^{256}+1\right)\)
\(E=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{256}+1\right)\)
....
\(E=\left(5^{128}-1\right)\left(5^{128}+1\right)\left(5^{256}+1\right)\)
\(E=\left(5^{256}-1\right)\left(5^{256}+1\right)\)
\(E=5^{512}-1\)
chứng minh
a) (a+a^2+a^3+a^4+................+a^29+a^30) chia hết (a+1)
b)15/460+15/498+15/754+.............+15/6160<1
c)1/2-1/4+1/8-1/16++1/32-1/64<1/3
d)1/3-2/3^2+3/3^3-4/3^4+...........+99/3^99+100/3^100
giúp mk vs nha
bài 1
(4^2+1)(4^4+1)(4^8+1)(4^16+1)(4^32+1)-1/15*4^64
Bài 2:Cho x+1/x=10. Tính S=x^5+1/x^5
Bài 2:
Ta có: \(\frac{x+1}{x}=10\) hay \(\frac{x^1+1^1}{x^1}=10^1\)
Nên suy ra : \(\frac{x^5+1}{x^5}=10^5\)
= 100000 ( do 15 cũng sẽ =1 nên không viết mũ 5 cũng chả sao)