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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
a)\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{2^4}=\frac{3.2^4}{2^4}=3\)
b)\(B=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.2^3.13}=\frac{2^{10}.78}{2^{11}.13}=3\)
c)\(C=\frac{4^9.36+64^4}{16^4.100}=\frac{2^{18}.2^2.3^2+2^{24}}{2^{16}.2^2.5^2}=\frac{2^{20}\left(3^2+2^4\right)}{2^{18}.5^2}=\frac{2^2.25}{25}=4\)
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 = \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
= \(\frac{2^{10}\left(13+65\right)}{2^8.104}\)
= \(\frac{2^{10}.78}{2^8.104}\)
= \(\frac{2^{10}.13.2.3}{2^8.2^3.13}\)
= \(\frac{2^{11}.13.3}{2^{11}.13}=3\)
\(\frac{7256.4375-725}{4375.7255+3650}=\frac{\left(7255+1\right).4375-725}{4375.7255+3650}=\frac{7255.4375+4375-725}{7255.4375+3650}=\frac{7255.4375+3650}{7255.4375+3650}=1\)
\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{16}=3\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^2.78}{26.2^2}=\frac{78}{26}=3\)
\(\left(125^3.7^5-175^5.5\right):2001^{2002}\) ( bạn xem lại đề xem sai đâu ko nhé )
Để Thiên giải câu 3 cho:
(1253.75 -1755;5):20012001
\(=\left[\left(5^3\right)^3.7^5-175^5:5\right]:2001^{2002}\)
\(=\left(5^9.7^5-175:5\right):2001^{2002}\)
\(=\left(5^5.5^4.7^4.7-175^4.175:5\right):2001^{2002}\)
\(=\left(5^5.35^4.7-175^4.35\right):2001^{2002}\)
\(=\left(5^4.35^4.5.7-175^4.35\right):2001^{2002}\)
\(=\left(175^4.35-175^4.35\right):2001^{2002}\)
\(=0:2001^{2002}\)
\(=0\)
A= 3^10.( 11+5 ) / 3^9. 2^4
A= 3^10. 16 /3 ^9 . 16
A= 3^10/3^9= 3
B= 2^10. ( 13 +65) / 2^8.104
B= 2^10. 78/ 2^8.104
B= 2 ^10.2.39/ 2^8 .2.52
B= 2^11.39/ 2^9.52
B= 2 ^ 2. (39/ 52)
B= 4 . 39/52 = 3
C= (2^3.3^2)^3.( 2.3.3^2 )^2 / (2^2.3^3)^4
C= 2^9.3^6/ 2^2.3^2.3^4
C= 2^9.3^6/ 2^2.3^6
C= 2^9/2^2= 2^5=32
D= 11.3^29-3^30/ 2^2.3^28
D= 3^29.(1- 3)/ 2^2.3^28
D= 3^29.(-2)/ 2^2.3^28
D= 3. (-2/2^2)
D = 3. (-1/2)= -3/2
\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3.16}{16}=3\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{4.78}{104}=\frac{312}{104}=3\)
~ Chúc em học tốt ~
Ta có:
B=\(\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\) = \(\frac{3^{10}.16}{3^9.2^4}\)= \(\frac{3^9.3.16}{3^9.16}\)= 3
C=\(\frac{2^{10}.\left(13+65\right)}{2^8.104}\) =\(\frac{2^{10}.78}{2^8.104}\) = \(\frac{2^8.2^2.78}{2^8.104}\)= \(\frac{4.78}{104}\) = \(\frac{4.78}{4.26}\)=\(\frac{78}{26}\)=3
=>B=C