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a)
\(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(\frac{2.3\left(1.2\right)+2.3\left(2.4\right)+2.3\left(3.6\right)+2.3\left(4.8\right)+2.3\left(5.10\right)}{3.4\left(3.4+6.8+9.12+12.16+15.20\right)}\)
\(=\frac{\left(3.4+6.8+9.12+12.16+15.20\right)}{2.3\left(3.4+6.8+9.12+12.16+15.20\right)}=\frac{1}{2.3}=\frac{1}{6}\)
a) 153^2+99.153+47^2
= 153^2+2.47.153+47^2
= (153+47)^2
=200^2
=40000
b) 126^2-152.126+5776
= 126^2-2.76.126+76^2
= (126-76)^2
= 50^2
= 2500
c)3^8.5^8-(15^4-1).(15^4+1)
= 15^8-[(15^4)^2-1^2]
= 15^8-15^8+1
=1
d) (2+1).(2^2+1).(2^4+1)...(2^32+1)+1
= 1.(2+1).(2^2+1).(2^4+1)...(2^32+1)+1
= (2-1).(2+1).(2+1).(2^4+1)...(2^32+1)+1
= (2^2-1).(2^2+1).(2^4+1)...(2^32+1)+1
= (2^4-1).(2^4+1)...(2^32+1)+1
= (2^8-1)...(2^32+1)+1
= (2^32-1).(2^32+1)+1
= 2^64-1+1
= 2^64
`1/2+2/4+3/6+4/8+5/10+6/12`
`=1/2+1/2+1/2+1/2+1/2+1/2`
`=1/2*6=3`
`1/3+1/4+1/5+8/10+20/15+20/30`
`=(1/3+1/4)+(1/5+4/5)+(4/3+2/3)`
`=7/12+1+2`
`=7/12+3=43/12`
\(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{6}+\dfrac{4}{8}+\dfrac{5}{10}+\dfrac{6}{12}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\)
\(=\dfrac{1}{2}\times6=3\)
\(------\)
\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{8}{10}+\dfrac{20}{15}+\dfrac{20}{30}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{4}{5}+\dfrac{4}{3}+\dfrac{2}{3}\)
\(=\left(\dfrac{1}{3}+\dfrac{4}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)+\dfrac{1}{4}\)
\(=\dfrac{7}{3}+1+\dfrac{1}{4}\)
\(=\dfrac{28}{12}+\dfrac{12}{12}+\dfrac{3}{12}\)
\(=\dfrac{43}{12}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{18}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)
\(A=1-\dfrac{1}{64}\)
\(A=\dfrac{63}{64}\)
giúp em với ạ , em cấn gấp , 4h em học rồi ạ . Cảm ơn những ai giúp em ạ , em tick luôn ạ
đặt
\(A=15.\left(4^2+1\right)\left(4^4+1\right)..\left(4^{32}+1\right)\)
ta có
\(A=\left(4^2-1\right)\left(4^2+1\right)..\left(4^{32}+1\right)\)
\(=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)..\left(4^{32}+1\right)=..=4^{64}-1\)
Đặt \(A=15\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{32}+1\right)\)
\(=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{32}+1\right)\)
\(=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{32}+1\right)\)
\(=\left(4^{32}-1\right)\left(4^{32}+1\right)=4^{64}-1\)