RÚT GỌN BIỂU THỨC SAU:
\(P=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)\)
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T
18 tháng 6 2019
#)Giải :
a) \(A=\frac{4^5.9^4-2^6.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=-\frac{1}{3}\)
18 tháng 6 2019
\(a,A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-1}{3}\)
Học tốt!!!!!!!!!!!!!
6 tháng 11 2017
a, \(\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
=\(\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot\left(3^2\right)^4}\)
=\(\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
=\(\dfrac{2^{11}\cdot3^6\cdot\left(2^2+3^3\right)}{2^{10}\cdot3^7\cdot\left(2^4+5\cdot3\right)}\)
=\(\dfrac{2^{11}\cdot3^6\cdot31}{2^{10}\cdot3^7\cdot31}\)
=\(\dfrac{2}{3}\)
6 tháng 11 2017
b, \(\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{4}{25}\cdot\dfrac{-125}{1728}}\)
=\(\dfrac{\dfrac{8\cdot9\cdot\left(-1\right)}{27\cdot16}}{\dfrac{4\cdot\left(-125\right)}{25\cdot1728}}\)
=\(\dfrac{\dfrac{-1}{6}}{\dfrac{-5}{432}}\)
=\(\dfrac{-1}{6}\cdot\dfrac{-432}{5}\)
=\(\dfrac{72}{5}\)
14 tháng 1 2017
\(A=\frac{\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right)}{\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)}\)
\(A=\frac{\left(\frac{15}{10}-\frac{4}{10}+\frac{1}{10}\right)}{\left(\frac{18}{12}-\frac{8}{12}+\frac{1}{12}\right)}\)
\(A=\frac{\frac{6}{5}}{\frac{11}{12}}=\frac{6}{5}:\frac{11}{12}=\frac{6}{5}\times\frac{12}{11}\)
\(A=\frac{72}{55}\)
31 tháng 3 2020
\(B=\left(\frac{3}{5}\right)^2\cdot5^2-\left(2\frac{1}{4}\right)^3:\left(\frac{3}{4}\right)^3+\frac{1}{2}\)
\(B=\left(\frac{3}{5}\cdot5\right)^2-\left(\frac{9}{4}:\frac{3}{4}\right)^3+\frac{1}{2}\)
\(B=3^2-\left(\frac{9}{4}\cdot\frac{4}{3}\right)^3+\frac{1}{2}\)
\(B=3^2-3^3+\frac{1}{2}=-18+\frac{1}{2}=-\frac{35}{2}\)
A
25 tháng 12 2018
\(\left(\frac{1}{x}+1-\frac{3}{x^3+1}-\frac{3}{x^2-x+1}\right)\cdot\frac{3x^2-3x+3}{\left(x+1\right).\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(=\left(\frac{x+1}{x}-\frac{3}{\left(x+1\right).\left(x^2-x+1\right)}+\frac{3.\left(x+1\right)}{\left(x+1\right).\left(x^2-x+1\right)}\right)\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{\left(x+1\right)^2.\left(x^2-x+1\right)-3x+3x^2+3x}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{x^4+x^3+x+1+3x^2}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2x^3+2x^2-2x-2}{x.\left(x+1\right)^2.\left(x+2\right)}\)
\(=\frac{3x^4+x^3+7x^2+5x+5}{x.\left(x+1\right)^2.\left(x+2\right)}\)
\(P=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)\)
\(P=\left(\frac{15}{10}-\frac{4}{10}+\frac{1}{10}\right):\left(\frac{18}{12}-\frac{8}{12}+\frac{1}{12}\right)\)
\(P=\frac{12}{10}:\frac{11}{12}\)
\(P=\frac{6}{5}\times\frac{12}{11}\)
\(P=\frac{72}{55}\)
\(P=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)\)
\(\Rightarrow P=\frac{\frac{3}{2}-\frac{2}{5}+\frac{1}{10}}{\frac{3}{2}-\frac{2}{3}+\frac{1}{12}}\)\(\Rightarrow P=\frac{\frac{15}{10}-\frac{4}{10}+\frac{1}{10}}{\frac{18}{12}-\frac{8}{12}+\frac{1}{12}}\)
\(\Rightarrow P=\frac{1}{\frac{3}{4}}=\frac{4}{3}\)