( 1/16 )^3 : (1/8 )^2
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a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)
b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)
\(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-81^{16}\)
\(A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-81\)
\(A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-81^{16}\)
\(A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)-81^{16}\)
\(A=\left(3^{16}-1\right)\left(3^{16}+1\right)-81^{16}\)
\(A=3^{32}-1-81^{16}\)
A = 8.( 32 + 1 ).( 34 + 1 ).( 38 + 1).( 316 + 1 ) - 8116
A = ( 32 - 1).( 32 + 1 ).( 34 + 1 ).( 38 + 1).( 316 + 1 ) - 8116
A = ( 34 - 1 ).( 34 + 1 ).( 38 + 1).( 316 + 1 ) - 8116
A = ( 38 - 1 ).( 38 + 1).( 316 + 1 ) - 8116
A = ( 316 - 1 ).( 316 + 1 ) - 8116
A = ( 332 - 1 ) - 8116
A = -364
\(1+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
\(=\dfrac{5}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
\(=\dfrac{11}{8}+\dfrac{1}{16}\)
\(=\dfrac{23}{16}\)
______
\(2-\dfrac{1}{8}-\dfrac{1}{12}-\dfrac{1}{16}\)
\(=\dfrac{15}{8}-\dfrac{1}{12}-\dfrac{1}{16}\)
\(=\dfrac{43}{24}-\dfrac{1}{16}\)
\(=\dfrac{83}{48}\)
_________
\(\dfrac{4}{99}\times\dfrac{18}{5}:\dfrac{12}{11}+\dfrac{3}{5}\)
\(=\dfrac{8}{55}:\dfrac{12}{11}+\dfrac{3}{5}\)
\(=\dfrac{8}{55}\times\dfrac{11}{12}+\dfrac{3}{5}\)
\(=\dfrac{2}{15}+\dfrac{3}{5}\)
\(=\dfrac{11}{15}\)
__________
\(\left(1-\dfrac{3}{4}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{3}\right)\)
\(=\dfrac{1}{4}\times\dfrac{4}{3}\times\dfrac{2}{3}\)
\(=\dfrac{4\times2}{4\times3\times3}\)
\(=\dfrac{2}{3\times3}\)
\(=\dfrac{2}{9}\)
a) 1 + 1/4 + 1/8 + 1/16
= 16/16 + 4/16 + 2/16 + 1/16
= 23/16
b) 2 - 1/8 - 1/12 - 1/16
= 96/48 - 6/46 - 4/48 - 3/48
= 83/48
c) 4/99 × 18/5 : 12/11 + 3/5
= 8/55 : 12/11 + 3/5
= 2/15 + 3/5
= 2/15 + 9/15
= 11/15
d) (1 - 3/4) × (1 + 1/3) : (1 - 1/3)
= 1/4 × 4/3 : 2/3
= 1/3 : 2/3
= 2
c;=(50-49)(50+49)+(48-47)(48+47)+.............+(2+1)(2-1)
=50+49+48+............+1
=(50+1)50=2550:2=1275
d;=(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)(2^16+1)
=2^32-1
e;=(3-1)(3+1)(3^2+1)...........(3^16+1)
=(3^2-1)(3^2+1)..............(3^16+1)
=(3^16-1)(3^16+1)=3^32-1
tu tinh ket qua luy thua tao khong thua hoi dau
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
\(\frac{1}{16}^3:\frac{1}{8}^2=\frac{1}{4096}:\frac{1}{64}=\frac{1}{4096}.64=\frac{1}{64}\)
Bài làm:
Ta có: \(\left(\frac{1}{16}\right)^3\div\left(\frac{1}{8}\right)^2\)
\(=\left(\frac{1}{2^4}\right)^3\div\left(\frac{1}{2^3}\right)^2\)
\(=\frac{1}{2^{12}}\div\frac{1}{2^6}\)
\(=\frac{1}{2^6}\)