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\(\frac{11.3^{22}.3^7-9^{15}}{2^2.3^{28}}=\frac{11.3^{29}}{2^2.3^{28}}-\frac{3^{30}}{2^2.3^{28}}=\frac{11.3}{4}-\frac{9}{4}=\frac{33-9}{4}=\frac{24}{4}=6\)
\(\frac{27^5.8^2-9^7.4^3}{2^6.9^6}=\frac{3^{15}.2^6}{2^6.3^{12}}-\frac{3^{14}.2^6}{2^6.3^{12}}=27-9=18\)
~ Học tốt ~
\(\frac{8^{14}}{4^4\cdot64^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
\(\frac{2^7\cdot9^3}{6^5\cdot8^2}=\frac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}=\frac{3}{2^4}=\frac{3}{16}\)
\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{6^{12}-6^{11}}\)
\(=\frac{2^{12}3^{10}\left(1+5\right)}{6^{11}\left(6-1\right)}\)
\(=\frac{2^{10}\cdot3^{10}\cdot5\cdot2^2}{6^{10}\cdot6\cdot5}\)
\(=\frac{6^{10}\cdot20}{6^{10}\cdot30}\)
\(=\frac{2}{3}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\) (Sau đó phân tách ra)
=\(\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
=\(\frac{2^{12}.3^{10}+2^9.3^9.2^33.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
=\(\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\) (Gộp và giải như biểu thức thường)
=\(\frac{2^{12}.3^{10}.\left(1+1.5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
=\(\frac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\) (Rút gọn giữa tử và mẫu)
=\(\frac{2.1.6}{1.3.5}=\frac{2.1.2}{1.1.5}=\frac{4}{5}\)
a) Ta có : 2 + 4 + 6 + 8 +... + 16 + 18
SSH : (18 - 2) : 2 + 1 = 9(số)
Tổng : (2 + 18).9 : 2 = 90
\(A=\frac{27\cdot45+27\cdot45}{90}=\frac{27\left(45+45\right)}{90}=\frac{27\cdot90}{90}=27\)
b) Ta có : 135.1420 + 45.780.3 = 135.1420 + 135.780 = 135(1420 + 780) = 135.2200 (*)
3 + 6 + 9 + 12 + ... + 24 + 27
Số số hạng : (27 - 3) : 3 + 1 = 9(số)
Tổng : (3 + 27).9 : 2 = 135 (**)
Từ (*) và (**) suy ra
=> \(B=\frac{135\cdot2200}{135}=2200\)
a)
2 + 4 + 6 + ... + 16 + 18
Số số hạng :
( 18 - 2 ) / 2 + 1 = 9
Tổng :
( 18 + 2 ) x 9 / 2 = 90
\(A=\frac{27\cdot45+27\cdot45}{90}\)
\(=\frac{27\left(45+45\right)}{90}\)
\(=\frac{27\cdot90}{90}\)
\(=27\)
b)
3 + 6 + 9 + 12 + ... + 24 + 27
Số số hạng :
( 27 - 3 ) / 3 + 1 = 9
Tổng :
( 27 + 3 ) x 9 /2 = 135
\(B=\frac{135\cdot1420+45\cdot780\cdot3}{135}\)
\(=\frac{135\cdot1420+135\cdot780}{135}\)
\(=\frac{135\left(1420+780\right)}{135}\)
\(=\frac{135\cdot2200}{135}\)
= 2200
\(A=\frac{2^{12}.27^3}{6^7.16^2}=\frac{2^{12}.\left(3^3\right)^3}{2^7.3^7.\left(2^4\right)^2}=\frac{2^{12}.3^9}{2^7.3^7.2^8}=\frac{2^{12}.3^9}{2^{15}.3^7}=\frac{1.3^2}{2^3.1}=\frac{9}{8}\)
\(A=\frac{2^{12}\cdot27^3}{6^7\cdot16^2}\)
\(A=\frac{4096\cdot19683}{279936\cdot256}\)
\(A=\frac{80621568}{71663616}\)
a=2^12.3^5-2^12.3^4/2^12.3^6+2^12.3^5 - 5^10.7^3-5^10.7^4/5^9.7^3+5^9.7^3.2^3
a=2^12.3^4.(3-1)/2^12.3^5.(3+1)-5^10.7^3.(1-7)/5^9.7^3.(1+8)
a=2/12-30/9
a=1/6-10/3=-19/6
a=
\(A=\frac{13,5.1420+4,5.780.3}{3+6+9+...+24+27}\)
\(=\frac{13,5.1420+\left(4,5.3\right).780}{\left(3+27\right)+\left(6+24\right)+\left(9+21\right)+\left(12+18\right)+\left(15+15\right)}\)
\(=\frac{13,5.\left(1420+780\right)}{30.5}\)
\(=\frac{13,5.2200}{150}\)
\(=\frac{29700}{150}=198\)
Chị làm lại nhé
\(A=\frac{13,5.1420+4,5.780.3}{3+6+9+12+15+18+21+24+27}\)
\(=\frac{13,5.\left(1420+780\right)}{\left(3+27\right)+\left(6+24\right)+\left(9+21\right)+\left(12+18\right)+15}\)
\(=\frac{13,5.2200}{30.4+15}\)
\(=\frac{29700}{135}=220\)
\(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=\frac{1}{9}\)
\(\frac{2^{10}\cdot3^{13}\cdot16^3}{4^{10}\cdot9^6}=\frac{2^{10}\cdot3^{13}\cdot2^{12}}{2^{20}\cdot3^{12}}=\frac{2^{22}\cdot3^{13}}{2^{20}\cdot3^{12}}=2^2\cdot3=12\)
a) \(\frac{6^{10}.27^5}{4^5.81^6}=\frac{\left(2.3\right)^{10}.\left(3^3\right)^5}{\left(2^2\right)^5.\left(3^4\right)^6}=\frac{2^{10}.3^{10}.3^{15}}{2^{10}.3^{24}}=\frac{2^{10}.3^{25}}{2^{10}.3^{24}}=\frac{3^{25}}{3^{24}}=3\)
b) \(\frac{72^3.54^2}{108^4}=\frac{\left(2^3.3^2\right)^3.\left(2.3^3\right)^2}{\left(2^2.3^3\right)^4}=\frac{2^9.3^6.2^2.3^6}{2^8.3^{12}}=\frac{2^{11}.3^{12}}{2^8.3^{12}}=\frac{2^{11}}{2^8}=2^3=8\)
c) \(\frac{27^4.2^3-3^{10}.4^3}{6^4.9^3.4}=\frac{\left(3^3\right)^4.2^3-3^{10}.\left(2^2\right)^3}{\left(2.3\right)^4.\left(3^2\right)^3.2^2}=\frac{3^{12}.2^3-3^{10}.2^6}{2^4.3^4.3^6.2^2}\)
= \(\frac{3^{12}.2^3-3^{10}.2^6}{2^6.3^{10}}=\frac{3^{12}.2^3}{2^6.3^{10}}-\frac{3^{10}.2^6}{2^6.3^{10}}=\frac{3^2}{2^3}-1=\frac{9}{8}-1=\frac{1}{8}\)