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\(a,2010:\left(-5\right)+400-1\\ =-402+400-1\\ =-3\\ b,\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)\\ =\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c,\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\\ =\dfrac{1}{12}.\left(\dfrac{1}{20}\right)^2\\ =\dfrac{1}{12}.\dfrac{1}{400}\\ =\dfrac{1}{4800}\)
a) \(2010:\left(-5\right)+400-1=-400+400-1=-1\)
b) \(\dfrac{2}{3}+\dfrac{3}{4}\cdot\dfrac{-4}{9}=\dfrac{2}{3}+\dfrac{-1}{3}=\dfrac{1}{3}\)
c) \(\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\dfrac{1}{12}\cdot\dfrac{1}{400}=\dfrac{1}{4800}\)
a)\(\frac{{ - 2}}{5} + \frac{3}{7} = \frac{{ - 14}}{{35}} + \frac{{15}}{{35}} = \frac{1}{{35}}\)
b)\(0,123 - 0,234 = - \left( {0,234 - 0,123} \right) = - 0,111.\)
1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)
\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)
\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)
\(=5-\dfrac{4}{2}\)
\(=5-2\)
\(=3\)
b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)
\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)
\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)
\(=0+0+0+2022\)
\(=2022\)
2) \(0,7^2\cdot x=0,49^2\)
\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)
\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)
\(\Rightarrow x=\left(0,7\right)^2\)
\(\Rightarrow x=0,49\)
b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)
\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)
\(\Rightarrow x=\left(-0,5\right)^5\)
\(\Rightarrow x=-\dfrac{1}{32}\)
2:
a: =>x*0,49=0,49^2
=>x=0,49
b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5
a)\(\frac{1}{8}.\frac{3}{5} = \frac{{1.3}}{{8.5}} = \frac{3}{{40}}\)
b)\(\frac{{ - 6}}{7}:\left( { - \frac{5}{3}} \right) = \frac{{ - 6}}{7}.\frac{{ - 3}}{5} = \frac{{18}}{{35}}\)
c)\(0,6.\left( { - 0,15} \right) = \frac{6}{{10}}.\frac{{ - 15}}{{100}} = \frac{{ - 90}}{{1000}} = \frac{{ - 9}}{{100}}\).
a)
\(-7x^2\left(3x-4y\right)\)
\(=-21x^3+28x^2y\)
b)
\(\left(x-3\right)\left(5x-4\right)\)
\(=x\left(5x-4\right)-3\left(5x-4\right)\)
\(=5x^2-4x-15x+12\)
\(=5x^2-\left(4x+15x\right)+12\)
\(=5x^2-19x+12\)
a: =-7x^2*3x+7x^2*4y
=-21x^3+28x^2y
b: =5x^2-4x-15x+12
=5x^2-19x+12
a) \({x^2} + \dfrac{1}{4}{x^2} - 5{x^2} = (1 + \dfrac{1}{4} - 5){x^2} = - \dfrac{{15}}{4}{x^2}\);
b) \({y^4} + 6{y^4} - \dfrac{2}{5}{y^4} = (1 + 6 - \dfrac{2}{5}){y^4} = \dfrac{{33}}{5}{y^4}\).
a) \(\dfrac{4}{9}x + \dfrac{2}{3}x = (\dfrac{4}{9} + \dfrac{2}{3})x = (\dfrac{4}{9} + \dfrac{6}{9})x = \dfrac{{10}}{9}x\);
b) \( - 12{y^2} + 0,7{y^2} = ( - 12 + 0,7){y^2} = - 11,3{y^2}\);
c) \( - 21{t^3} - 25{t^3} = ( - 21 - 25){t^3} = - 46{t^3}\).
\(=\left(3.5-0.5\right)\cdot\dfrac{1}{27}+\dfrac{4}{3}=\dfrac{1}{9}+\dfrac{4}{3}=\dfrac{1}{9}+\dfrac{12}{9}=\dfrac{13}{9}\)