1. thực hiện phép tính
a. A=[\(6.\left(\dfrac{-1}{3}\right)-3.\left(\dfrac{1}{3}\right)+1\)]:\(\left(-\dfrac{1}{3}-1\right)\)
b. B=\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
1. Thực hiện phép tính :
a. \(A=\left[6.\left(\dfrac{-1}{3}\right)-3.\left(\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
\(A=\left[-2-1+1\right]:\left(-\dfrac{4}{3}\right)\)
\(A=-2:\left(-\dfrac{4}{3}\right)\)
Vậy : \(A=\dfrac{8}{3}\)
Tính giá trị biểu thức
P=\(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^3.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
tính giá trị biểu thức sau
a) \(A=\dfrac{25^6}{5^3}\)
b) \(B=32.\left(\dfrac{3}{2}\right)^5\)
c) \(C=\left(\dfrac{1}{3}\right)^4.3^{-3}\)
d) \(D=4^{-2}.\left(\dfrac{2}{5}\right)^5.5^4\)
e) \(E=9^{-5}:\left(\dfrac{5}{3}\right)^4.25^2\)
f) \(F=\left(\dfrac{5}{8}\right)^{-2}:4^2\)
g) \(G=\left(\dfrac{5}{3}\right)^3.\left(\dfrac{9}{2}\right)^2:\left(\sqrt{3}\right)^4\)
a: \(A=\dfrac{25^6}{5^3}=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)
b: \(B=32\cdot\left(\dfrac{3}{2}\right)^5=32\cdot\dfrac{3^5}{2^5}=32\cdot\dfrac{243}{32}=243\)
c: \(C=\left(\dfrac{1}{3}\right)^4\cdot3^{-3}=3^{-4}\cdot3^{-3}=3^{-4-3}=3^{-7}\)
d: \(D=4^{-2}\cdot\left(\dfrac{2}{5}\right)^5\cdot5^4\)
\(=\dfrac{1}{4^2}\cdot\dfrac{2^5}{5^5}\cdot5^4\)
\(=\dfrac{1}{16}\cdot\dfrac{32}{5}=\dfrac{2}{5}\)
e: \(E=9^{-5}:\left(\dfrac{5}{3}\right)^4\cdot25^2\)
\(=\dfrac{1}{9^5}:\dfrac{5^4}{3^4}\cdot\left(5^2\right)^2\)
\(=\dfrac{1}{3^{10}}\cdot\dfrac{3^4}{5^4}\cdot5^4=\dfrac{1}{3^6}\)
f: \(F=\left(\dfrac{5}{8}\right)^{-2}:4^2\)
\(=\left(1:\dfrac{5}{8}\right)^2:4^2\)
\(=\left(\dfrac{8}{5}\right)^2\cdot\dfrac{1}{16}=\dfrac{64}{25}\cdot\dfrac{1}{16}=\dfrac{4}{25}\)
g: \(G=\left(\dfrac{5}{3}\right)^3\cdot\left(\dfrac{9}{2}\right)^2:\left(\sqrt{3}\right)^4\)
\(=\dfrac{5^3}{3^3}\cdot\dfrac{9^2}{2^2}:9\)
\(=\dfrac{5^3\cdot3^4}{3^3\cdot2^2}\cdot\dfrac{1}{3^2}\)
\(=\dfrac{125}{2^2\cdot3}=\dfrac{125}{3\cdot4}=\dfrac{125}{12}\)
\(A=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)
\(B=32.\left(\dfrac{3}{2}\right)^5=\dfrac{2^5.3^5}{2^5}=2^5\)
\(C=\left(\dfrac{1}{3}\right)^4.3^{-3}=\dfrac{1}{3^4.3^3}=\dfrac{1}{3^7}\)
\(D=4^{-2}.\left(\dfrac{2}{5}\right)^5.5^4=\dfrac{1}{\left(2^2\right)^2}.\dfrac{2^5}{5^5}.5^4=\dfrac{2}{5}\)
\(E=\dfrac{1}{9^5}.\dfrac{3^4}{5^4}.\left(5^2\right)^2=\dfrac{1}{3^{10}}.\dfrac{3^4}{5^4}.5^4=\dfrac{1}{3^6}\)
\(F=\dfrac{8^2}{5^2}:\left(2^2\right)^2=\dfrac{\left(2^3\right)^2}{5^2.2^4}=\dfrac{2^6}{5^2.2^4}=\dfrac{2^2}{5^2}\)
\(G=\dfrac{5^3}{3^3}.\dfrac{\left(3^2\right)^2}{2^2}:3^2=\dfrac{5^3}{3^3}.\dfrac{3^4}{2^2}.\dfrac{1}{3^2}=\dfrac{5^3}{3.2^2}\)
Tính
a)\(\left(5^{-5}\right)^{-1}.\left(\dfrac{1}{2}\right)^{-2}.\dfrac{1}{10^5}\)
b) \(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
c) \(\dfrac{4^2.25^2+32.125}{2^3.5^2}\)
a, (5-5)-1 . \(\left(\dfrac{1}{2}\right)^{-2}\) . \(\dfrac{1}{10^5}\)
= 55 . \(\dfrac{1^{-2}}{2^{-2}}\) . \(\dfrac{1}{10^5}\)
= (55 . \(\dfrac{1}{10^5}\)) . \(\dfrac{1}{4}\)
= \(\dfrac{5^5}{10^5}\) . \(\dfrac{1}{4}\) = \(\left(\dfrac{1}{2}\right)^5\). \(\dfrac{1}{4}\)
= \(\dfrac{1}{32}.\text{}\dfrac{1}{4}\)= \(\dfrac{1}{128}\)
b: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\dfrac{4}{5}\)
c: \(=\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}=\dfrac{2^4\cdot5^3\left(5+2\right)}{2^3\cdot5^2}=10\cdot7=70\)
Tính nhanh giá trị của các biểu thức sau :
\(A=\dfrac{-5}{11}+\left(\dfrac{-6}{11}+1\right)\)
\(B=\dfrac{2}{3}+\left(\dfrac{5}{7}+-\dfrac{2}{3}\right)\)
\(C=\left(\dfrac{-1}{4}+\dfrac{5}{8}\right)+\dfrac{-3}{8}\)
Tính giá trị của biểu thức sau:
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
Câu 1: Tính giá trị biểu thức:
a.A=\(\left(\dfrac{136}{15}-\dfrac{28}{5}+\dfrac{62}{10}\right)\).\(\dfrac{21}{24}\)
b.B=\(\dfrac{5}{6}\)+6\(\dfrac{5}{6}\)\(\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right)\):8\(\dfrac{1}{3}\)
c.C=1+3+6+10+15+...+1225.
Tính bằng cách hợp lí giá trị của biểu thức.
A = \(\left(3-\dfrac{1}{4} +\dfrac{3}{2}\right)\)- \(\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)\)-\(\left(6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)
B =\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)
\(a.19\dfrac{5}{8}:\dfrac{7}{12}-15\dfrac{1}{4}:\dfrac{7}{12} b.\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)
c.\(\left(3\dfrac{1}{3}+2,5\right):\left(3\dfrac{1}{6}-\left(4\dfrac{1}{5}\right)\right)-\dfrac{11}{31}\)
1.Tính giái trị biểu thức
a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)
\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)
b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)
\(=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)
\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)
