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A=\(\frac{72^3.54^2}{108^4}=\frac{\left(2^3.3^2\right)^3.\left(2.3^3\right)^2}{\left(2^2.3^3\right)^4}=\frac{2^9.3^6.2^2.3^6}{2^8.3^{12}}=\frac{2^{11}.3^{12}}{2^8.3^{12}}=2^3=8\)
B= \(\frac{4^6.3^4.9^5}{6^{12}}=\frac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\frac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)
c) \(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)
1.
\(\frac{72^3\times54^2}{108^4}=\frac{\left(8\times9\right)^3\times\left(27\times2\right)^2}{\left(27\times4\right)^4}=\frac{\left(2^3\times3^2\right)^3\times\left(3^3\times2\right)^2}{\left(3^3\times2^2\right)^4}=\frac{\left(2^3\right)^3\times\left(3^2\right)^3\times\left(3^3\right)^2\times2^2}{\left(3^3\right)^4\times\left(2^2\right)^4}=\frac{2^9\times3^6\times3^6\times2^2}{3^{12}\times2^8}=2^3=8\)
2.
\(\frac{4^6\times3^4\times9^5}{6^{12}}=\frac{\left(2^2\right)^6\times3^4\times\left(3^2\right)^5}{\left(2\times3\right)^{12}}=\frac{2^{12}\times3^4\times3^{10}}{2^{12}\times3^{12}}=3^2=9\)
3.
\(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\times\left(2^8+1\right)}{2^2\times\left(2^8+1\right)}=2^3=8\)
\(b\)
\(\frac{2^{13+5}}{2^{10+2}}\)=\(\frac{2^{18}}{2^{12}}\)
a) \(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{2^{10}.\left(13.4+65.4\right)}{2^{10}.104}+\frac{3^9.\left(3.11+3.5\right)}{3^9.16}\)
\(=\frac{312}{104}+\frac{48}{16}\)
=3+3=6
b) \(\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)
\(=\frac{1.5.6\left(1+2.2.2+4.4.4+9.9.9\right)}{1.3.5\left(1+2.2.2+4.4.4+9.9.9\right)}\)
\(=\frac{1.5.6}{1.3.5}\)
\(=2\)
c) 1+2-3-4+5+6-7-8+...+2009+2010-2011-2012+2013
Nhận xét:Giá trị tuyệt đối của hai số liền nhau hơn kém nhau 1 đơn vị
=> Tổng trên có 2013-1+1=2013(Số hạng)
Nhóm 4 số vào một nhóm, ta được 2013:4=503 nhóm (thừa 1 số)
=>1+2-3-4+5+6-7-8+...+2009+2010-2011-2012+2013
=1+(2-3-4+5)+(6-7-8+9)+...+(2010-2011-2012+2013)
=1+0+0+...+0 (có 503 số 0)
=1+0.503
=1+0
=1
a)38:3^4+2^2.2^3=3^8-4+2^2+3=3^4+2^5=81+32=113
b)3.4^2-2.3^2=3(4^2-2.3)=3(16-6)=3.10=30
c)4^6.3^4.9^5/6^12=(2^2)^6.3^4.(3^2)^5/(2.3)^12=2^12.3^4.3^10/2^12.3^12=2^12.3^14/2^12.3^12=3^14/3^12=3^2=9
e)45^3.20^4.18^2/180^5=(5.3^2)^3.(2^2.5)^4.(2.3^2)^2/(2^2.3^2.5)^5=5^3.3^6.2^8.5^4.2^2.3^4/2^10.3^10.5^5=5^9.3^10.2^10/2^10.3^10.5^5
=5^4/1=5^4=625
d)21^2.14.125/35^3.6=(2.7)^2.2.7.5^3/(5.7)^3.2.3=2^2.7^2.2.7.5^3/5^3.7^3.2.3=2^3.7^3.5^3/5^3.7^3.2.3=2^3/2.3=8/6=4/3
g)2^13+2^5/2^10+2^2=2^5(2^8+1)/2^2(2^8+1)=2^5/2^2=32/4=8
Chúc bạn học tốt !!!!
\(\frac{7256.4375-725}{4375.7255+3650}=\frac{\left(7255+1\right).4375-725}{4375.7255+3650}=\frac{7255.4375+4375-725}{7255.4375+3650}=\frac{7255.4375+3650}{7255.4375+3650}=1\)
\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{16}=3\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^2.78}{26.2^2}=\frac{78}{26}=3\)
\(\left(125^3.7^5-175^5.5\right):2001^{2002}\) ( bạn xem lại đề xem sai đâu ko nhé )
Để Thiên giải câu 3 cho:
(1253.75 -1755;5):20012001
\(=\left[\left(5^3\right)^3.7^5-175^5:5\right]:2001^{2002}\)
\(=\left(5^9.7^5-175:5\right):2001^{2002}\)
\(=\left(5^5.5^4.7^4.7-175^4.175:5\right):2001^{2002}\)
\(=\left(5^5.35^4.7-175^4.35\right):2001^{2002}\)
\(=\left(5^4.35^4.5.7-175^4.35\right):2001^{2002}\)
\(=\left(175^4.35-175^4.35\right):2001^{2002}\)
\(=0:2001^{2002}\)
\(=0\)
Gợi ý
bn thực hiện phép tính tử mẫu bình thường , khi ra nhưng số trùng nhau bn gạch ra nháp cho đến nhưng số tối giản là ra nha
chúc bn
học tốt
A = \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
= \(\frac{3^{10}\left(11+5\right)}{3^9.2^4}\)
= \(\frac{3^{10}.16}{3^9.2^4}\)
= \(\frac{3^{10}.2^4}{3^9.2^4}=3\)
B = \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
= \(\frac{2^{10}\left(13+65\right)}{2^8.104}\)
= \(\frac{2^{10}.78}{2^8.104}\)
= \(\frac{2^{10}.13.2.3}{2^8.2^3.13}\)
= \(\frac{2^{11}.13.3}{2^{11}.13}=3\)