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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
(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
B=[(45.79+45.21)]:90-5^2]:5+2^3 B=[(45.79+45.21):90-25]:5+8 B=[(45.(79+21):65]:13 B=[(45.100):65]:13 B=[4500:65]:13 B=4500:65:13
Chứng minh rằng:
a) 1/2-1/4+1/8-1/16+1/32-1/64<1/3
b) 1/3-2/3^2+3/3^3-3/3^4+...+99/3^99-100/3^100<3/16
Ta có: \(\frac{2.4+2.4.8+4.8.16+8.16.32}{3.4+2.6.8+4.12.16+8.24.32}\)
\(=\frac{4\left(2+2.8+8.16+2.16.32\right)}{4\left(3+3.8+12.16+2.24.32\right)}\)
\(=\frac{2+2.8+8.16+2.16.32}{3+3.8+12.16+2.24.32}\)
\(=\frac{2\left(1+8+64+16.32\right)}{3\left(1+8+64+16.32\right)}=\frac{2}{3}\)
Lời giải:
8A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=[(3^2-1)(3^2+1)](3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=(3^8-1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=(3^{16}-1)(3^{16}+1)-4.3^{32}$
$=3^{32}-1-4.3^{32}$
$=-3.3^{32}-1=-3^{33}-1$
$\Rightarrow A=\frac{-3^{33}-1}{8}$