Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)
\(=\frac{3^{10}.16}{3^9.2^4}\)
\(=\frac{3^{10}.2^4}{3^9.2^4}=3\)
3^10.(11+5) =3.16
3^9 . 2^4 1.16 bỏ số 16 thì được kết quả bằng 3
\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3^{10}.16}{3^9.16}=3\)
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)\(C=\frac{4^9.36+64^4}{16^4.100}=\frac{2^{18}.2^2.3^2+2^{24}}{2^{16}.2^2.5^2}=\frac{2^{20}\left(3^2+2^4\right)}{2^{18}.5^2}=\frac{2^2.25}{25}=4\)
\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3.16}{16}=3\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{4.78}{104}=\frac{312}{104}=3\)
~ Chúc em học tốt ~
Ta có:
B = \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.2^4}=\frac{3^{10}.16}{3^9.2^4}=\frac{3^{10}.2^4}{3^9.2^4}=3\)
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= 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
\(H=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(H=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(H=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(H=\frac{2^{19}.3^9+2^{18}.5.3^9}{2^{19}.3^9+2^{20}.3^{10}}\)
\(H=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9.\left(1+2.3\right)}\)
\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+6\right)}\)
\(H=\frac{2^{18}.3^9.7}{2^{19}.3^9.7}=\frac{1}{2}\)
\(K=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)
\(K=\frac{\left(2^2\right)^7.2^8}{3.2^{15}.\left(2^4\right)^2-5.2^2.2^{20}}\)
\(K=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^{22}}\)
\(K=\frac{2^{22}}{3.2^{23}-5.2^{22}}\)
\(K=\frac{2^{22}}{2^{22}.\left(3.2-5\right)}=\frac{2^{22}}{2^{22}.1}=1\)
Bài 1:
\(A=\frac{3333}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{3333}{101}\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{3333}{101}.\frac{4}{21}=\frac{1111.4}{101.7}=\frac{4444}{707}\)
Bài 2
\(A=\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1+2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(B=\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3+4}{2^{10}-3}=1+\frac{4}{2^{10}-3}\)
Ta thấy \(2^{10}-1>2^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}< \frac{4}{2^{10}-3}\)
Từ đó \(\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{4}{2^{10}-3}\Rightarrow A< B\)
Bài 3\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}\right)+\frac{5}{11}}{\frac{5}{12}+\left(1-\frac{7}{11}\right)}=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}=\frac{\frac{55+60}{11.12}}{\frac{55+48}{12.11}}=\frac{115}{103}\)
Ta có : \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3^{10}.16}{3^9.16}=3\)3
\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\) \(=\frac{3^{10}.16}{3^9.2^4}=\frac{3^{10}.2^4}{3^9.2^4}=3\)