A=(3^2+1).(3^4+1).(3^8+1).(3^16+1)-3^32/2 (3^32/2 là phân số nha)
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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
Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\).Ta có :
\(=>\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=>2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=>2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
...............................................................................
Cuối cùng \(=>2A=3^{64}-1\).
\(=>A=\frac{3^{64}-1}{2}\)
Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=...........................................\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
a)\(\frac{32}{64}-\frac{16}{64}+\frac{8}{64}-\frac{4}{64}+\frac{2}{64}-\frac{1}{64}\le\frac{1}{3}\)
\(\Rightarrow\frac{32-16+8-4+2-1}{64}=\frac{23}{64}\)\
\(\Rightarrow\frac{23}{64}=0,359375;\frac{1}{3}=0,33333...\)
đề sao lạ vậy
1/ \(=-\frac{64}{27}.\frac{243}{32}\)
\(=-\frac{243}{16}\)
2/ \(=\frac{1}{81}.\frac{5361441}{64}\)
\(=\frac{6561}{64}\)
3/ \(=-\frac{2197}{512}.36,71356045\)
\(=-\frac{2048}{13}\)
tíc mình nha
1\(\left(-\frac{4}{3}\right)^3.\left(\frac{9}{16}\right)^5=-\frac{2187}{16384}\)
2\(\left(\frac{1}{3}\right)^4.\left(-\frac{9}{2}\right)^6=\frac{6561}{64}\)
3\(\left(-\frac{13}{8}\right)^3.\left(-\frac{32}{13}\right)^4=-41,00457607\)
a ) 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
Đạt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
A x 2 = 2 x ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)
A x 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
Lấy A x 2 - A ta có :
A x 2 - A = 1 + 1/2 + ..... + 1/128 - 1/2 + 1/4 + ........ + 1/256
A x ( 2 - 1 ) = 1 - 1/ 256
A = 255/256
b) 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
A x 3 = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/ 3 + 1/9 + 1/27 + 1/81 + 1/243
Lấy A x 3 - A ta có :
A x 3 - A = 1 + 1/3 + 1/9 +..... + 1/243 - 1/3 + 1/9 +........+ 1/243 + 1/29
A x ( 3 - 1 ) = 1 - 1/29
A x2 = 28/29
A = 28/29 : 2 ( tự tính
Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{256}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{128}\)
\(\Rightarrow2A-A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2B=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2B=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(B=\dfrac{3^{32}-1}{2}< A=3^{32}-1\)
\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =>2B=2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)=3^{32}-1\\ =>A=\dfrac{3^{32}-1}{2}< B\)
Mình ghi nhầm đề bài 1 tí đề bài là :
So sánh 2 số A và B biết :
A = (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) và B = 3^32 - 1
Lời giải:
8A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=[(3^2-1)(3^2+1)](3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=(3^8-1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=(3^{16}-1)(3^{16}+1)-4.3^{32}$
$=3^{32}-1-4.3^{32}$
$=-3.3^{32}-1=-3^{33}-1$
$\Rightarrow A=\frac{-3^{33}-1}{8}$