Tính: (16^3−64^2):8^3
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Ta có :
\(A=\frac{3}{2}+\frac{3}{4}+...+\frac{3}{128}\)
\(2A=3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}\)
\(2A-A=\left(3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}\right)-\left(\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}+\frac{3}{128}\right)\)
\(A=3-\frac{3}{128}\)
\(A=\frac{381}{128}\)
Ủng hộ mk nha !!! ^_^
b:3=1/2+1/4+1/8+1/16+1/32+1/64+1/128
[b:3]x2=1+1/2+1/4+1/8+1/16+1/32+1/64
[b:3]x2-[b:3]=1+1/2+1/4+1/8+1/16+1/32+1/64-1/2+1/4+1/8+1/16+1/32+1/64+1/128
b:3=1-1/128
b:3=127/128
b=127/128x3
b=381/128
B.2=\(3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}\)\(+\frac{3}{128}\)
B.2-B=\(\left(3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}+\frac{3}{128}\right)\).\(\left(\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}+\frac{3}{128}\right)\)
B=3-\(\frac{3}{128}\)
B=\(\frac{381}{128}\)
3(1/2+1/4+1/8+1/16.....1/128)
A=(1/2+1/4+1/8+1/16....1/128)
2A-A=A
2A=(1/2+1/4+1/8....1/128)*2
=1+1/2+1/4....1/64
2A-A=1+1/2+1/4+1/8....1/64-(1/2+1/4+1/8......1/128)
A=1-1/128=127/128
Đáp số:3*127/128=381/128
( 163 - 642 ) : 83
= [ ( 24 ) 3 - ( 26 ) 2 ] : ( 23 ) 3
= ( 212 - 212 ) : 29
= 0 : 29
= 0
\(\Rightarrow H=2^4.2^6.2^6:\left(2^6.2^5.2^4\right)\)
\(\Rightarrow H=2^{16}:2^{15}\)
\(\Rightarrow H=2^1=2\)
tíc mình nha
\(\dfrac{16^3-64^2}{8^3}=\dfrac{2^{12}-2^{12}}{2^9}=\dfrac{0}{2^9}=0\)