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a) \(\left(\dfrac{1}{5}\right)^5.5^5=1\)
b) \(\left(0,125\right)^3.512=1\)
c) \(\left(0,25\right)^4.1024=4\)
a) (1/5)^5 . 5^5 = (1/5. 5)^5 = 1^5= 1
b) (0,125)^3. 512= (0,125)^3 . 8^3 = (0,125. 8)^3 = 1^3= 1
c) (0,25)^4. 1024= [(0,25)^2]^2. 32^2= (1/6)^2. 32^2=(1/6.32)^2= (32/6)^2 =2^2= 4
a/ \(\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)
b/ \(\left(0,125\right)^3.512=\left(0,125\right).8^3=\left(0,125.8\right)^3=1^3=1\)
c/ \(\left(0,25\right)^4.1024=\left(0,25^2\right)^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=16^2\)
\(a,\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)
\(b,\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)
\(c,\left(0,25\right)^4.1024=\left(0,25\right)^4.4^4.4=\left(0,25.4\right)^4.4=1^4.4=1.4=4\)
a) \(\left(\dfrac{1}{5}\right)^5.5^5\)
\(=\dfrac{1}{3125}.3125\)
= 1
b) \(\left(0,125\right)^3.512\)
\(=\dfrac{1}{512}.512\)
= 1
c) \(\left(0,25\right)^4.1024\)
= \(\dfrac{1}{256}.1024\)
= 4
\(a,\left(\dfrac{1}{7}\right)^7.7^7=\left(\dfrac{1}{7}.7\right)^7=1\)
\(b,\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)\(c,\left(0,25\right)^4.1024=\left(\left(0,25\right)^2\right)^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=2^2=4\)
\(d,\dfrac{90^3}{15^3}=\left(\dfrac{90}{15}\right)^3=6^3=216\)
a) \({\left( {\frac{8}{9}} \right)^3} \cdot \frac{4}{3} \cdot \frac{2}{3} = {\left( {\frac{8}{9}} \right)^3}.\frac{8}{9} = {\left( {\frac{8}{9}} \right)^{3+1}}={\left( {\frac{8}{9}} \right)^4}\)
b) \({\left( {\frac{1}{4}} \right)^7} \cdot 0,25 = {\left( {0,25} \right)^7}.0,25 ={\left( {0,25} \right)^{7+1}}= {\left( {0,25} \right)^8}\)
c) \({( - 0,125)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^{6-1}}= {\left( {\frac{{ - 1}}{8}} \right)^5}\)
d) \({\left[ {{{\left( {\frac{{ - 3}}{2}} \right)}^3}} \right]^2} = {\left( {\frac{{ - 3}}{2}} \right)^{3.2}} = {\left( {\frac{{ - 3}}{2}} \right)^6}\)
a) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}^3\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{18}\)
\(=\left(\frac{3}{7}\right)^{21-18}\)
\(=\left(\frac{3}{7}\right)^3\)
\(=\frac{27}{343}\)
k nha
\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)
\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)