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\(\frac{2^{10}.13.2^{10}.65}{2^8.104}=\frac{1024.13.1024.65}{256.104}\)
\(=\frac{1024.13.65}{256.104}=\frac{865280}{26642}\)
\(=32,5\)
\(\frac{2^{10}.13.2^{10}.65}{2^8.104}=\frac{2^{20}.13.65}{2^8.2^3.13}\)
\(=\frac{2^{20}.13.65}{2^{11}.13}\)
\(=2^9.65\)
\(=512.65\)
\(=33280\)
bài 2
\(2^x\cdot7=224\)
\(\Rightarrow2^x=224\div7\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(\left(3^x+5\right)^2=289\)
\(\Rightarrow\left(3^x+5\right)^2=17^2\)
\(\Rightarrow3^x+5=17\)
\(\Rightarrow3^x=17-5\)
\(\Rightarrow3^x=12\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
Xét : 210 . 13 + 210. 65 = 210 . ( 13 + 65 ) = 210 . 78
28 . 104 = 28 . ( 22 . 26 ) = 28. 22. 26 = 210.26
có : \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)= \(\frac{2^{10}.78}{2^{10}.26}\)= \(\frac{78}{26}\)= 3
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{2^{10}.78}{2^8.104}\)
\(=\frac{2^2.39}{52}=\frac{2^2.39}{2^2.13}\)
\(=\frac{39}{13}\)
\(=3\)
Ta có:
B = \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.2^4}=\frac{3^{10}.16}{3^9.2^4}=\frac{3^{10}.2^4}{3^9.2^4}=3\)
C = \(\frac{2^{10}.39+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(39+65\right)}{2^8.104}=\frac{2^{10}.104}{2^8.104}=4\)
Ta thấy : 3 < 4 => B < C
\(\frac{2^{10}.13+2^{10}.65}{2^{8^{ }}.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^8.2^2.78}{2^8.104}=3\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^8.2^2.78}{2^8.104}=3\)
\(\frac{7256.4375-725}{4375.7255+3650}=\frac{\left(7255+1\right).4375-725}{4375.7255+3650}=\frac{7255.4375+4375-725}{7255.4375+3650}=\frac{7255.4375+3650}{7255.4375+3650}=1\)
\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{16}=3\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^2.78}{26.2^2}=\frac{78}{26}=3\)
\(\left(125^3.7^5-175^5.5\right):2001^{2002}\) ( bạn xem lại đề xem sai đâu ko nhé )
Để Thiên giải câu 3 cho:
(1253.75 -1755;5):20012001
\(=\left[\left(5^3\right)^3.7^5-175^5:5\right]:2001^{2002}\)
\(=\left(5^9.7^5-175:5\right):2001^{2002}\)
\(=\left(5^5.5^4.7^4.7-175^4.175:5\right):2001^{2002}\)
\(=\left(5^5.35^4.7-175^4.35\right):2001^{2002}\)
\(=\left(5^4.35^4.5.7-175^4.35\right):2001^{2002}\)
\(=\left(175^4.35-175^4.35\right):2001^{2002}\)
\(=0:2001^{2002}\)
\(=0\)
jdgfcb
ko biết nha