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AD
22 tháng 7 2023
\(1,\left[\left(-\dfrac{2}{5}\right)+\dfrac{1}{3}\right]-\left(\dfrac{3}{5}-\dfrac{1}{3}\right)=-\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{1}{3}\\ =\left(-\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{1}{3}+\dfrac{1}{3}\right)\\ =\dfrac{-2-3}{5}+\dfrac{1+1}{3}\\ =-\dfrac{5}{5}+\dfrac{2}{3}\\ =-1+\dfrac{2}{3}\\ =\dfrac{-3+2}{3}=-\dfrac{1}{3}\\ b,\left(\dfrac{3}{2}-\dfrac{3}{4}\right)-\left(0,25+\dfrac{1}{2}\right)\\ =\left(\dfrac{3}{2}-\dfrac{3}{4}\right)-\left(\dfrac{1}{4}+\dfrac{1}{2}\right)\\ =\dfrac{3}{2}-\dfrac{3}{4}-\dfrac{1}{4}-\dfrac{1}{2}\\ =\left(\dfrac{3}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{3}{4}-\dfrac{1}{4}\right)\\ =\dfrac{3-1}{2}+\dfrac{-3-1}{4}\\ =\dfrac{2}{2}-\dfrac{4}{4}=1-1=0\)
22 tháng 7 2023
1: =-2/5+1/3-3/5+1/3
=-1+2/3=-1/3
2: =3/2-3/4-1/4-1/2
=1-1=0
1 tháng 4 2018
\(A=\left(0,25\right)^{-1}.\left(\frac{1}{4}\right)^{-2}.\left(\frac{4}{3}\right)^2.\left(\frac{5}{4}\right)^{-1}.\left(\frac{2}{3}\right)^{-3}\)
\(\Rightarrow A=4^1.4^2.\frac{16}{9}.\frac{4}{5}\frac{27}{8}\)
\(\Rightarrow A=\frac{64}{1}.\frac{16}{9}.\frac{4}{5}.\frac{27}{8}\)
\(\Rightarrow A=\frac{1536}{5}\)
Vậy \(A=\frac{1536}{5}\)
TN
6 tháng 8 2016
1)
a. \(\left(3x^2-50\right)^2=5^4\)
\(\Leftrightarrow3x^4-50=625\)
\(\Leftrightarrow3x^4=675\)
\(\Leftrightarrow x^4=225\)
\(\Leftrightarrow x=\sqrt{15}\)
2)
a. \(\frac{\left(3^4-3^3\right)^4}{27^3}=\frac{3^{16}-3^{12}}{\left(3^3\right)^3}=\frac{3^{12}.3^4-3^{12}}{3^9}=\frac{3^{12}\left(3^4-1\right)}{3^9}\)
\(=\frac{3^{12}.80}{3^9}=3^3.80=27.80=2160\)
b. \(\frac{25^3}{\left(5^5-5^3\right)^2}=\frac{\left(5^2\right)^3}{5^{10}-5^6}=\frac{5^6}{5^6.5^4-5^6}=\frac{5^6}{5^6\left(5^4-1\right)}\)
\(=\frac{5^6}{5^6.624}=\frac{1}{624}\)
1 tháng 10 2021
a: \(0.4\cdot\sqrt{0.25-\sqrt{\dfrac{1}{4}}}=0.4\cdot\sqrt{0.25-0.5}\)(đề này sai rồi bạn)
b: \(\dfrac{3}{2}+2\left(x-1\right)=-5\dfrac{1}{2}\)
\(\Leftrightarrow2\left(x-1\right)=\dfrac{-11}{2}-\dfrac{3}{2}=-7\)
\(\Leftrightarrow x-1=\dfrac{-7}{2}\)
hay \(x=-\dfrac{5}{2}\)
Trước hết ta có \(\left(\dfrac{x}{y}\right)^{-z}=\dfrac{1}{\left(\dfrac{x}{y}\right)^z}=\dfrac{1}{\dfrac{x^z}{y^z}}=\dfrac{y^z}{x^z}\)
Suy ra:
\(A=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)
\(A=\left(\dfrac{1}{4}\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)
\(A=4\cdot4^2\cdot\dfrac{3^2}{4^2}\cdot\dfrac{4}{5}\cdot\dfrac{3^3}{2^3}=4^2\cdot3^5\text{}\div5\div2^3\)
\(A=2^4\div2^3\cdot3^5\div5=2\cdot3^5\div5=2\cdot243\div5=\dfrac{486}{5}\)