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\(\frac{72^2}{24^2}=\frac{3^2.24^2}{24^2}=3^2=9\)
\(\frac{\left(-7,5\right)^2}{2,5^2}=\frac{\left(7,5\right)^2}{\left(2,5\right)^2}=\frac{\left(2,5\right)^2.3^2}{\left(2,5\right)^2}=9\)
\(\frac{15^2}{27}=\frac{3^2.5^2}{27}=\frac{9.25}{27}=\frac{25}{3}\)
\(\frac{72^2}{24^2}=\left(\frac{72}{24}\right)^2=3^2=9\)
\(\frac{\left(-7,5\right)^3}{\left(2,5\right)^3}=\left(\frac{-7,5}{2,5}\right)^3=\left(-3\right)^3=-27\)
\(\frac{15^3}{27}=\frac{15^3}{3^3}=\left(\frac{15}{3}\right)^3=5^3=125\)
Chúc bạn hok tốt
a) \(0,2 + 2,5:\frac{7}{2} = \frac{2}{{10}} + \frac{25}{10}:\frac{7}{2} = \frac{1}{5} + \frac{25}{10}.\frac{2}{7} \\= \frac{1}{5} + \frac{5}{7} = \frac{7}{{35}} + \frac{{25}}{{35}} = \frac{{32}}{{35}}\)
b)
\(\begin{array}{l}9.{\left( {\frac{{ - 1}}{3}} \right)^2} - {\left( { - 0,1} \right)^3}:\frac{2}{{15}}\\ = 9.\frac{1}{9} - {\left( {\frac{{ - 1}}{{10}}} \right)^3}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}.\frac{{15}}{2}\\ = 1 + \frac{3}{{400}}\\=\frac{400}{400}+\frac{3}{400}\\ = \frac{{403}}{{400}}\end{array}\)
\(\frac{\left(-7,5\right)^3}{2,5^3}=\left(\frac{-7,5}{2,5}\right)^3\)
\(=\left(-3\right)^3\)
\(=-27\)
\(\frac{\left(-7,5\right)^3}{\left(2,5\right)^3}=\frac{-7,5\cdot\left(-7,5\right)\cdot\left(-7,5\right)}{2,5\cdot2,5\cdot2,5}=-3\cdot\left(-3\right)\cdot\left(-3\right)=\left(-3\right)^3=-27\)
a) ( -2.5 ) . ( 7,5) .( -4 )
= [(-2,5).(-4)].(7,5)
= 10 . 7,5
= 75
b) \(1\frac{4}{23}+\frac{8}{21}-\frac{4}{23}+0,6+\frac{13}{21}\)
=\(1\frac{4}{23}-\frac{4}{23}+\frac{8}{21}+\frac{13}{21}-0,6\)
\(=1+1-0,6\)
\(=2-0,6\)
= 1,4
c) \(\frac{2}{7}.15\frac{1}{3}-\frac{2}{7}.20.\frac{1}{3}+4\frac{1}{3}\)
\(=\frac{2}{7}.5-\frac{1}{3}.\frac{40}{7}+4\frac{1}{3}\)
= \(=\frac{10}{7}-\frac{17}{7}\)
= -1
d) \(2\frac{1}{4}:\left(\frac{-3}{5}\right)-1\frac{1}{4}:\left(\frac{-3}{5}\right)\)
\(=\frac{9}{4}.\left( \frac{-5}{3}\right)-\frac{5}{4}.\left(\frac{-5}{3}\right)\)
=\(\left(\frac{-5}{3}\right).\left(\frac{9}{4}-\frac{5}{4}\right)\)
\(=\frac{-5}{3}.1\)
\(=\frac{-5}{3}\)
\(A=\left|4x-3\right|+\left|5y+7,5\right|+10\)
Mà \(\left|4x-3\right|\ge0\)với mọi x
\(\left|5y+7,5\right|\ge0\)với mọi y
\(\Rightarrow A\)có GTNN là 10
Để A có GTNN thì :
\(4x-3=0\) \(5y+7,5=0\)
\(4x=3\) \(5y=-7,5\)
\(x=\frac{3}{4}\) \(y=-1,5\)
\(B=\frac{5,8}{\left|2,5-x\right|+5,8}\)
Mà \(\left|2,5-x\right|\ge0\)
\(\Rightarrow\)GTNN \(\left|2,5-x\right|+5,8=5,8\)
Để B có GTLN \(\Rightarrow2,5-x=0\)
\(\Rightarrow x=2,5\)
722/24 = 216
(-7,5)3/(2,5)3 = 27
153/27 = 125