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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
\(a,7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=\frac{68}{9}-\frac{11}{4}-\frac{32}{9}\)
\(=4-\frac{11}{4}\)
\(=\frac{5}{4}\)
\(b,\left[\frac{3}{5}.\left(\frac{-7}{10}\right)+\frac{2}{10}.\frac{3}{5}\right]:\left(-3\right)\)
\(=\left(\frac{-21}{50}+\frac{3}{25}\right):\left(-3\right)\)
\(=\frac{-3}{10}:\left(-3\right)\)
\(=\frac{1}{10}\)
Ta sẽ dùng quy tắc đổi dấu :
-( 3/5 + 3/4 ) - ( -3/4 + 2/5 )
= -3/5 - 3/4 + 3/4 - 2/5
= -3/5 - ( -2/5 )
= -1/5
-3/5-3/4+3/4-2/5
=(-3/4+3/4)-3/5+2/5
=0-3/5+2/5
=1/5
nhớ tk nhe mn
a)(-7 / 5 + 3 / 8 ) : 2009 / 2010 + (-3 / 5+5 / 8) : 2009 / 2010
=[ (-7 / 5 + 3 / 8 ) + (-3 / 5+5 / 8) ] : 2009/2010
=[ -7/5 +3/8 + (-3)/5+5/8 ] : 2009/2010
=[ (-7/5 + (-3)/5) + (3/8 + 5/8) ] :2009/2010
=[-2+1] : 2009/2010
=-1 :2009/2010
=-2009/2010
b)\(\frac{9^8\cdot4^3}{27^4\cdot6^5}=\frac{\left(3^2\right)^8\cdot\left(2^2\right)^3}{\left(3^3\right)^4\cdot\left(2\cdot3\right)^5}=\frac{3^{16}\cdot2^6}{3^{12}\cdot2^5\cdot3^5}=\frac{3^{16}\cdot2^5\cdot2}{3^{16}\cdot3^1\cdot2^5}=\frac{2}{3^1}=\frac{2}{3}\)
\(a,2010:\left(-5\right)+400-1\\ =-402+400-1\\ =-3\\ b,\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)\\ =\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c,\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\\ =\dfrac{1}{12}.\left(\dfrac{1}{20}\right)^2\\ =\dfrac{1}{12}.\dfrac{1}{400}\\ =\dfrac{1}{4800}\)
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) \(\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\dfrac{1}{12}\cdot\dfrac{1}{400}=\dfrac{1}{4800}\)
\(a,8\times5+(-3)\times9=40+(-27)=13\)
\(b,\frac{2}{5}+\frac{1}{5}\times\left[\frac{-3}{4}\right]\)
\(=\frac{2}{5}+\frac{-3}{20}\)
\(=\frac{8}{20}+\frac{-3}{20}=\frac{5}{20}=\frac{1}{4}\)
\(c,2:\left[\frac{1}{2}-\frac{2}{3}\right]=2:\left[\frac{3}{6}-\frac{4}{6}\right]=2:\frac{-1}{6}=2\times\frac{6}{-1}=-12\)