\(\dfrac{8}{27}\)=\(\dfrac{2^n}{3\times3^m}\)
Số?
a) \(\dfrac{2}{5}=\dfrac{2\times3}{5\times3}=\dfrac{?}{?}\) \(\dfrac{4}{7}=\dfrac{4\times2}{7\times2}=\dfrac{?}{?}\) \(\dfrac{13}{54}=\dfrac{13\times3}{54\times3}=\dfrac{?}{?}\)
b) \(\dfrac{8}{20}=\dfrac{8:4}{20:4}=\dfrac{?}{?}\) \(\dfrac{10}{16}=\dfrac{10:2}{16:2}=\dfrac{?}{?}\) \(\dfrac{25}{65}=\dfrac{25:5}{65:5}=\dfrac{?}{?}\)
a) \(\dfrac{2}{5}=\dfrac{2\times3}{5\times3}=\dfrac{6}{15}=\dfrac{2}{5}\)
\(\dfrac{4}{7}=\dfrac{4\times2}{7\times2}=\dfrac{8}{14}=\dfrac{4}{7}\)
\(\dfrac{13}{54}=\dfrac{13\times3}{54\times3}=\dfrac{39}{162}=\dfrac{13}{54}\)
b) \(\dfrac{8}{20}=\dfrac{8:4}{20:4}=\dfrac{2}{5}\)
\(\dfrac{10}{16}=\dfrac{10:2}{16:2}=\dfrac{5}{8}\)
\(\dfrac{25}{65}=\dfrac{25:5}{65:5}=\dfrac{5}{13}\)
Quy đồng mẫu số các phân số (theo mẫu).
Mẫu: \(\dfrac{2}{3};\dfrac{3}{4}\) và \(\dfrac{7}{12}\) \(\dfrac{2}{3}=\dfrac{2\times4}{3\times4}=\dfrac{8}{12};\dfrac{3}{4}=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}\) |
a) \(\dfrac{3}{5};\dfrac{4}{7}\) và \(\dfrac{9}{35}\)
b) \(\dfrac{5}{6};\dfrac{7}{9}\) và \(\dfrac{19}{54}\)
a) \(\dfrac{3}{5}=\dfrac{21}{35};\dfrac{4}{7}=\dfrac{20}{35}\)
b) \(\dfrac{5}{6}=\dfrac{45}{54};\dfrac{7}{9}=\dfrac{42}{54}\)
\(C=\dfrac{5\times4^6\times9^4-3^9\times\left(-8\right)^4}{4\times2^{13}\times3^8+2\times8^4\times\left(-27\right)^3}\)
\(C=\dfrac{5\times2^{12}\times3^8-3^9\times2^{12}}{2^2\times2^{13}\times3^8+2\times2^{12}\times\left(-3^9\right)}=\dfrac{3^8\times2^{12}\times\left(5-3\right)}{2^{15}\times3^8+2^{13}\times\left(-3\right)^9}\)
\(=\dfrac{3^8\times2^{12}\times2}{2^{13}\times3^8\times\left(4-3\right)}=\dfrac{1}{1}=1\)
\(#PaooNqoccc\)
tính nhanh cho mn
\(3+\dfrac{4}{9}\times\dfrac{7}{25}\times\dfrac{27}{12}\times3\dfrac{4}{7}-\dfrac{7}{25}\)
\(=3+\dfrac{4}{9}\cdot\dfrac{7}{25}\cdot\dfrac{9}{4}\cdot\dfrac{25}{7}=3+\dfrac{4.7.9.25}{9.25.4.7}=3+1=4\)
\(=3+\dfrac{4}{9}\times\dfrac{7}{25}\times\dfrac{9}{4}\times\dfrac{25}{7}-\dfrac{7}{25}\)
\(=3+\dfrac{4.7.9.45}{9.25.4.7}-\dfrac{7}{25}\)
\(=3+1-\dfrac{7}{25}\)
\(=4-\dfrac{7}{25}\)
\(=\dfrac{93}{25}\)
Bài 1:
\(a,\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+...............+\dfrac{1}{1280}\)
\(b,\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+..............+\dfrac{1}{59049}\)
\(c,\dfrac{1}{2}\times3+\dfrac{1}{3}\times4+\dfrac{1}{4}\times5+\dfrac{1}{5}\times6\)
Tính (theo mẫu).
Mẫu: \(\dfrac{2}{5}\times3=\dfrac{2}{5}\times\dfrac{3}{1}=\dfrac{2\times3}{5\times1}=\dfrac{6}{5}\) Ta có thể viết gọn như sau: \(\dfrac{2}{5}\times3=\dfrac{2\times3}{5}=\dfrac{6}{5}\) |

a) \(\dfrac{9}{11}\times8\) b) \(\dfrac{4}{5}\times1\) c) \(\dfrac{15}{8}\times0\)
a) \(\dfrac{9}{11}\times8=\dfrac{9\times8}{11}=\dfrac{72}{11}\)
b) \(\dfrac{4}{5}\times1=\dfrac{4\times1}{5}=\dfrac{4}{5}\)
c) \(\dfrac{15}{8}\times0=\dfrac{15\times0}{8}=\dfrac{0}{8}=0\)
a: 9/11*8=(9*8)/11=72/11
b: 4/5*1=(4*1)/5=4/5
c: 15/8*0=(15*0)/8=0/8=0
Thực hiện phép tính
a, A = \(\left(\dfrac{1}{4\times9}+\dfrac{1}{9\times14}+\dfrac{1}{14\times19}+....+\dfrac{1}{44\times49}\right)\times\dfrac{1-3-5-7-....-49}{89}\)
b, B = \(\dfrac{2^{12}\times3^5-4^6\times9^2}{\left(2^2\times3\right)^6+8^4\times3^5}-\dfrac{5^{10}\times7^3-25^5\times49^2}{\left(125\times7\right)^3-5^9\times14^3}\)
\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{48\times49\times50}\)
\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{48\times49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{48\times49}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{2}-\dfrac{1}{2450}\)
\(=\dfrac{612}{1225}\)
\(\text{#}Toru\)
Tính giá trị của mỗi phân số sau:
\(E=\dfrac{11\times3^{29}-\left(3^2\right)^{15}}{2\times3^{14}\times2\times3^{14}}\)
\(G=\dfrac{5\times3^{11}+4\times3^{12}}{3^9\times5^2-3^0\times2^3}\)
\(H=\dfrac{\left(3\times4\times2^{16}\right)^2}{11\times2^{13}\times4^{11}-16^9}\)
\(E=\dfrac{11.3^{29}-3^{2^{15}}}{2.3^{14}.2.3^{14}}\)
\(=\dfrac{11.3-3^{30}}{2^2}=\dfrac{33-3^{30}}{4}\)