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\(a,A=\left(3\dfrac{5}{6}-1\dfrac{1}{3}\right)\left(3\dfrac{4}{15}-2\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left(3+\dfrac{5}{6}-1+\dfrac{1}{3}\right)\left(3+\dfrac{4}{15}-2+\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left[\left(3-1\right)+\left(\dfrac{5}{6}+\dfrac{1}{3}\right)\right]+\left[\left(3-2\right)+\left(\dfrac{4}{15}+\dfrac{3}{5}\right)\right]\)
\(\Leftrightarrow A=\left[2+\left(\dfrac{5}{6}+\dfrac{2}{6}\right)\right]+\left[1+\left(\dfrac{4}{15}+\dfrac{9}{15}\right)\right]\)
\(\Leftrightarrow A=\left(2+\dfrac{7}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=\left(2+1+\dfrac{1}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=3\dfrac{1}{6}+1\dfrac{13}{15}\)
Vậy...
\(b,B=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^{10}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(\Leftrightarrow B=\dfrac{6}{\left(2.3\right).5}\)
\(\Leftrightarrow B=\dfrac{6}{6.5}\)
\(\Leftrightarrow B=\dfrac{1}{5}\)
Vậy....
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)
\(B=\frac{300}{343}:\frac{1347}{343}\)
\(B=\frac{100}{449}\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)
\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)
\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)
\(A=\frac{-1}{2}+\frac{1710}{9}\)
\(A=\frac{-1}{2}+190\)
\(A=\frac{-1}{2}+\frac{380}{2}\)
\(A=\frac{379}{2}\)
\(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\dfrac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(=\dfrac{2^{12}.3^4.2}{2^{12}.3^5.4}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)
\(=\dfrac{1}{6}-\dfrac{-10}{3}\)
\(=\dfrac{7}{2}\)
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: \(=\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}=\dfrac{2^4\cdot5^3\left(5+2\right)}{2^3\cdot5^2}=10\cdot7=70\)
\(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\dfrac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(2.7\right)^3}\)
\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
= \(\dfrac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\dfrac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)
= \(\dfrac{2}{3.4}-\dfrac{5\left(-6\right)}{9}\)
= \(\dfrac{7}{2}\)
\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)
\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)
a: \(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\cdot\dfrac{18}{5}-\dfrac{6}{5}:\dfrac{-9}{5}+4\)
\(=\dfrac{18}{5}-\dfrac{6}{5}\cdot\dfrac{-5}{9}+4\)
\(=\dfrac{18}{5}+\dfrac{2}{3}+4\)
\(=\dfrac{124}{15}\)
b: \(=\dfrac{9}{25}\cdot\left(\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{1}{2}\right)-\dfrac{3}{8}:\dfrac{9}{8}\)
\(=\dfrac{9}{25}\cdot\dfrac{4}{10}-\dfrac{1}{3}\)
\(=-\dfrac{71}{375}\)
c: \(=\dfrac{7}{10}:\dfrac{4}{5}+\dfrac{2}{9}:\dfrac{5}{9}+\dfrac{1}{8}\)
\(=\dfrac{7}{10}\cdot\dfrac{5}{4}+\dfrac{2}{5}+\dfrac{1}{8}\)
=1+2/5
=7/5
d: \(=\dfrac{3}{7}\left(19+\dfrac{1}{3}-33-\dfrac{1}{3}\right)-\dfrac{2}{7}=\dfrac{3}{7}\cdot\left(-14\right)-\dfrac{2}{7}=-6-\dfrac{2}{7}=\dfrac{-44}{7}\)
e: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{11}\cdot3^{11}-2^{12}\cdot3^{12}}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{-2^{11}\cdot3^{11}\left(1+2\cdot3\right)}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)
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 : \(A=\dfrac{8}{3}\)