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a)\(\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.5\right)^{10}=1^{10}=1\)
b)\(5^2.3^5.\left(\frac{3}{5}\right)^2=\left(\frac{3}{5}.5\right)^2.3^5=3^2.3^5=3^7\)
c)\(\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{2.3}:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{6+2}=\left(\frac{1}{8}\right)^8\)
\(a.\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.\left(5^2\right)^{10}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.25\right)^{10}=5^{10}.\)
\(b.5^2.3^5.\left(\frac{3}{5}\right)^2=\left[5^2.\left(\frac{3}{5}\right)^2\right].3^5=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)\(c.\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left[\left(\frac{1}{4}\right)^2\right]^3:\left[\left(\frac{1}{2}\right)^3\right]^2=\left(\frac{1}{4}\right)^6:\left(\frac{1}{2}\right)^6=\left(\frac{1}{4}:\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^6\)
b) (5/2-3x)=25/9
3x = 5/2-25/9
3x =-5/18
x =-5/18:3
x=-5/54
\(e.\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(x-1=-2\)
\(x\) \(=-2+1\)
\(x\) \(=-1\)
Vậy \(x=-1\)
Bài 8:
a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)
b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)
a,\(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2\)
\(=\left(\dfrac{13}{14}\right)^2\)
\(=\dfrac{169}{196}\)
b,\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=\left(\dfrac{-1}{12}\right)^2\)
\(=\dfrac{1}{144}\)
c,\(\dfrac{5^4.20^4}{25^5.4^5}\)
\(=\dfrac{100^4}{100^5}\)
\(=\dfrac{1}{100}\)
d,\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4\)
\(=\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-6}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=\left(\dfrac{\left(-10\right)}{3}.\dfrac{\left(-6\right)}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=4^4.\left(\dfrac{-10}{3}\right)\)
\(=256.\left(\dfrac{-10}{3}\right)\)
\(=\dfrac{-2560}{3}\)
\(\left(-0,75\right)-\left(-1+\dfrac{2}{3}\right):0,5+\left(-\dfrac{1}{4}\right)\)
\(=\left(-0,75\right)-\left(-1-\dfrac{2}{3}\right)\cdot\dfrac{1}{2}-0,25\)
\(=\left(-0,75-0,25\right)+\dfrac{5}{6}\)
\(=-1+\dfrac{5}{6}\)
\(=-\dfrac{11}{6}\)
_________________
\(\left[\left(-\dfrac{3}{2}\right)+\dfrac{2}{3}\right]^2\cdot\dfrac{24}{25}-\dfrac{1}{5}\)
\(=\left(-\dfrac{9}{6}+\dfrac{4}{6}\right)^2\cdot\dfrac{24}{25}-\dfrac{1}{5}\)
\(=\left(\dfrac{-5}{6}\right)^2\cdot\dfrac{24}{25}-\dfrac{1}{5}\)
\(=\dfrac{25}{36}\cdot\dfrac{24}{25}-\dfrac{1}{5}\)
\(=\dfrac{2}{3}-\dfrac{1}{5}\)
\(=\dfrac{7}{15}\)
\(a,\left(-0,75\right)-\left(-1+\dfrac{2}{3}\right):0,5-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(\dfrac{-3+2}{3}\right):\dfrac{1}{2}-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(-\dfrac{1}{3}\right):\dfrac{1}{2}-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(-\dfrac{1}{3}\right)\times2-\dfrac{1}{4}\\ =-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{4}\\ =\left(-\dfrac{3}{4}-\dfrac{1}{4}\right)+\dfrac{2}{3}\\ =-\dfrac{4}{4}+\dfrac{2}{3}\\ =-1+\dfrac{2}{3}\\ =\dfrac{-3+2}{3}=-\dfrac{1}{3}\)
\(b,\left[\left(-\dfrac{3}{2}\right)+\dfrac{2}{3}\right]^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\left(\dfrac{-3\times3+2\times2}{6}\right)^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\left(-\dfrac{5}{6}\right)^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\dfrac{25}{36}\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\dfrac{2}{3}-\dfrac{1}{5}\\ =\dfrac{2\times5-3}{15}=\dfrac{7}{15}\)