\(\dfrac{2^7.9^4}{4^3.3^9}\)=......
\(\dfrac{2^7.9^4}{4^4.3^9}\)
\(\dfrac{2^7.9^4}{4^4.3^9}\)
\(\dfrac{2^7.9^4}{4^4.3^9}\)
\(=\dfrac{2^7.\left(3^2\right)^4}{\left(2^2\right)^4.3^9}\)
\(=\dfrac{2^7.3^8}{2^8.3^9}\)
\(=\dfrac{1}{2.3}=\dfrac{1}{6}\)
\(#WendyDang\)
tìm B biết B=2^2/3.3^3/8^2.4^4/15^3.5^5/24^4.6^6/35^5.7^7/48^6.8^8/63^7.9^9/80^8
3.3+6.6+9.9+....+90.90
4/3.5+4/5.7+4/7.9+.....4/37.39
a/
\(A=3^2+3^2.2^2+3^2.3^2+3^2.4^2+...+3^2.30^2=\)
\(=3^2\left(1^2+2^2+3^2+...+30^2\right)\)
Đăt biểu thức trong dấu ngoặc là B
\(B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+30\left(31-1\right)=\)
\(=1.2+2.3+3.4+30.31-\left(1+2+3+...+30\right)=\)
\(C=1+2+3+...+30=\dfrac{30\left(1+30\right)}{2}=465\)
\(D=1.2+2.3+3.4+...+30.31\)
\(3D=1.2.3+2.3.3+3.4.3+...+30.31.3=\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+30.31.\left(32-29\right)=\)
\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-29.30.31+30.31.32=\)
\(=30.31.32\Rightarrow D=\dfrac{30.31.32}{3}=10.31.32\)
\(\Rightarrow A=3^2\left(C-D\right)=3^2\left(10.31.32-465\right)\)
b/
Đặt biểu thức là A
\(\dfrac{A}{2}=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{39-37}{37.39}=\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{37}-\dfrac{1}{39}=\)
\(=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{12}{39}\Rightarrow A=\dfrac{2.12}{39}=\dfrac{24}{39}=\dfrac{8}{13}\)
\(A=3.3+6.6+9.9+...+90.90\)
\(A=3^2\left(1+2^2+3^2+...+10^2\right)\)
\(A=9.\dfrac{10.\left(10+1\right)\left(2.10+1\right)}{6}\)
\(A=3.\dfrac{10.11.21}{2}\)
\(A=3465\)
Tính:
a,A=\(\dfrac{12^{15}.3^4-4^5.3^9}{27^3.2^{10}-32^3.3^9}\)
b. B= \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^3.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{99}{49^2.50^2}\)
\(A=\dfrac{12^{15}\cdot3^4-4^5\cdot3^9}{27^3\cdot2^{10}-32^3\cdot3^9}\\ =\dfrac{\left(2^2\cdot3\right)^{15}\cdot3^4-\left(2^2\right)^5\cdot3^9}{\left(3^3\right)^3\cdot2^{10}-\left(2^5\right)^3\cdot3^9}\\ =\dfrac{2^{30}\cdot3^{15}\cdot3^4-2^{10}\cdot3^9}{3^9\cdot2^{10}-2^{15}\cdot3^9}\\ =\dfrac{3^9\cdot2^{10}\left(2^{20}\cdot3^{10}\right)}{3^9\cdot2^{10}\left(1-2^5\right)}\\ =\dfrac{\left(2^2\right)^{10}\cdot3^{10}}{1-32}\\ =\dfrac{\left(2^2\cdot3\right)^{10}}{-31}\\ =\dfrac{-12^{10}}{31}\)
\(B=\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{99}{49^2\cdot50^2}\\ =\dfrac{2^2-1^2}{1^2\cdot2^2}+\dfrac{3^2-2^2}{2^2\cdot3^2}+...+\dfrac{50^2-49^2}{49^2\cdot50^2}\\ =\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{49^2}-\dfrac{1}{50^2}\\ =1-\dfrac{1}{2500}\\ =\dfrac{2499}{2500}\)
Bài 1: rút gọn
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}\)
\(\left(\dfrac{1}{5}\right)^{15}\).\(\left(\dfrac{1}{4}\right)^{20}\)
\(\dfrac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)=-\dfrac{1}{6}\)
\(\left(\dfrac{1}{5}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\dfrac{1}{5^{12}}.\dfrac{1}{4^{20}}=5^{-12}.4^{-20}=125^{-4}.1024^{-4}=\left(125.1024\right)^{-4}=128000^{-4}\)
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.2^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\dfrac{2.6}{3.7}=\dfrac{4}{7}\)
a,3/2-5/6:(1/2)2+\(\sqrt{0,25-\sqrt{\dfrac{1}{4}}}\)
b,-4/3:2/9+13/12:-13/8
c,(-1/2)2-[-1/6:|-1+5|-\(\sqrt{64}.\left(\dfrac{1}{4}\right)^2\)
d,15^11.5^7.9^2/5^18.27^6
\(a,=\dfrac{3}{2}-\dfrac{5}{6}:\dfrac{1}{4}+\sqrt{\dfrac{1}{4}-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{10}{3}+\sqrt{\dfrac{1}{2}}=-\dfrac{11}{6}+\dfrac{\sqrt{2}}{2}=\dfrac{-33+3\sqrt{2}}{6}\)
\(b,=-\dfrac{4}{3}\cdot\dfrac{9}{2}+\dfrac{13}{12}\cdot\left(-\dfrac{8}{13}\right)=6-\dfrac{2}{3}=\dfrac{16}{3}\\ c,=\dfrac{1}{4}-\left(-\dfrac{1}{6}:4-8\cdot\dfrac{1}{16}\right)=\dfrac{1}{4}-\left(-\dfrac{1}{24}-\dfrac{1}{2}\right)\\ =\dfrac{1}{4}-\dfrac{13}{24}=-\dfrac{7}{24}\\ d,=\dfrac{3^{11}\cdot5^{11}\cdot5^7\cdot3^4}{5^{18}\cdot3^{18}}=\dfrac{1}{3^3}=\dfrac{1}{27}\)
So sánh:
A = \(\dfrac{3^2}{2.5}+\dfrac{3^2}{5.8}+\dfrac{3^2}{8.11}\) và B = \(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
So sánh: A =\(\dfrac{3^2}{2.5}+\dfrac{3^2}{5.8}+\dfrac{3^2}{8.11}\) và B \(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
\(\dfrac{2^4.5^2.7}{2^3.5.7^2.11}\);\(\dfrac{2^3.3^3.5^3.7.8}{3.2^4.5^3.14}\)
\(\dfrac{4^2.5.11}{44.20}\);\(\dfrac{13.15.16}{18.65.7};\dfrac{7.2.8.5^2}{14.2.5}\)
\(\dfrac{2^3.3^3.5}{3.2^3.5^3}\)
a) \(\dfrac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^2\cdot11}=\dfrac{2^3\cdot5\cdot10\cdot7}{2^3\cdot5\cdot7\cdot77}=\dfrac{10}{77}\)
\(\dfrac{2^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot2^4\cdot5^3\cdot14}=\dfrac{2^3\cdot3\cdot5^3\cdot7\cdot3^2\cdot8}{3\cdot2^3\cdot2\cdot5^3\cdot14}=\dfrac{7\cdot3^2\cdot8}{2\cdot14}=\dfrac{63\cdot8}{2\cdot14}=18=\dfrac{1386}{77}\)
Tính
a, A = \(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
b, B = \(\left(\dfrac{3}{7}.\dfrac{4}{15}+\dfrac{1}{3}.9^{15}\right).\dfrac{1}{3}.\dfrac{6^8}{12^4}\)
c, C = \(\dfrac{10^4.81-16.15^2}{4^4.675}\)
a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)