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\(\frac{1}{20}\left(x-\frac{8}{15}\right)=-\frac{1}{30}\) \(\left(28+\frac{1}{5}\right).\left(\frac{3}{5}.x+\frac{4}{7}\right)=0\)
\(x-\frac{8}{15}=-\frac{1}{30}:\frac{1}{20}\) \(\frac{141}{5}.\left(\frac{3}{5}.x+\frac{4}{7}\right)=0\)
\(x-\frac{8}{15}=-\frac{2}{3}\) \(\frac{3}{5}.x+\frac{4}{7}=0\)
\(x=-\frac{2}{3}+\frac{8}{15}\) \(\frac{3}{5}.x=-\frac{4}{7}\)
\(x=-\frac{2}{15}\) \(x=-\frac{20}{21}\)
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
a) 12. \(\frac{4}{9}\)+\(\frac{4}{3}\)=\(\frac{16}{3}\)+\(\frac{4}{3}\)=\(\frac{20}{3}\)
b) (\(\frac{-5}{7}\)) . (12,5+1,5)= (\(\frac{-5}{7}\)).14=-10
a) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}=12.\frac{4}{9}+\frac{4}{3}=\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)
b) \(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)=-\frac{5}{7}.\left(12,5+1,5\right)=-\frac{5}{7}.14=-10\)
c) \(1:\left(\frac{2}{3}-\frac{3}{4}\right)^2=1:\left(-\frac{1}{12}\right)^2=1:\frac{1}{144}=1.144=144\)
d) \(15.\left(-\frac{2}{3}\right)^2-\frac{7}{3}=15.\frac{4}{9}-\frac{7}{3}=\frac{20}{3}-\frac{7}{3}=\frac{13}{3}\)
e) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(-1\right)^{2007}=\frac{1}{2}.8-\frac{2}{5}+\left(-1\right)=4-\frac{2}{5}-1=\frac{13}{5}\)
#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
a) \(\left(-\frac{5}{2}\right)^2:\left(-15\right)-\left(-0,45+\frac{3}{4}\right).\left(-1\frac{5}{9}\right)\)
= \(-\frac{25}{4}:\left(-15\right)-\left(\frac{9}{20}+\frac{15}{20}\right).\left(-\frac{14}{9}\right)\)
=\(-\frac{25}{4}.\frac{1}{-15}-\frac{6}{5}.\left(-\frac{14}{9}\right)\)
= \(\frac{-5}{12}-\frac{8}{5}\)
= \(\frac{\left(-25\right)-96}{60}\)
= \(\frac{\left(-25\right)+\left(-96\right)}{60}\)
=\(\frac{121}{60}\)
b) \(\left(\frac{-1}{3}\right)-\left(\frac{-3}{5}\right)^0+\left(1-\frac{1}{2}\right)^2:2\)
= \(\left(\frac{-1}{3}\right)-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}\)
=\(\left(\frac{-1}{3}\right)-\frac{3}{3}+\frac{1}{4}.\frac{1}{2}\)
= \(\frac{-4}{3}+\frac{1}{8}\)=\(\frac{-32+3}{24}\)
=\(\frac{-29}{24}\)
c) E=\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{2^{10}.3^8-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{3}{5}\)
d)\(\frac{5^4.20^4}{25^5.4^5}\)
=\(\frac{\left(5.20\right)^4}{\left(25.4\right)^5}\)
=\(\frac{100^4}{100^5}\)
=\(\frac{1}{100}\)
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
\(\left(\frac{1}{2}\right)^{15}\cdot\left(\frac{1}{4}\right)^{20}=\left(\frac{1}{2}\right)^{15}\cdot\left[\left(\frac{1}{2}\right)^2\right]^{20}=\left(\frac{1}{2}\right)^{15}\cdot\left(\frac{1}{2}\right)^{40}=\left(\frac{1}{2}\right)^{15+40}=\left(\frac{1}{2}\right)^{55}\)
\(\left(\frac{1}{9}\right)^{25}:\left(\frac{1}{3}\right)^{30}=\left[\left(\frac{1}{3}\right)^2\right]^{25}:\left(\frac{1}{3}\right)^{30}=\left(\frac{1}{3}\right)^{50}:\left(\frac{1}{3}\right)^{30}=\left(\frac{1}{3}\right)^{50-30}=\left(\frac{1}{3}\right)^{20}\)
NẾU THẤY ĐÚNG THÌ NHỚ K CHO MÌNH VỚI ĐÓ !!! :33