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\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
Ta có: F= (100-12) (100-22)...(100-252)
=> F= (100-12)...(100-102)...(100-252)
=> F= (100-12)...0...(100-252)
=> F= 0
Vậy F= 0
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
Bài 1:
a) \(\left(\frac{1}{2}\right)^2\) và \(\left(\frac{1}{2}\right)^5\)
Ta có: \(\left(\frac{1}{2}\right)^2=\frac{1}{4}.\)
\(\left(\frac{1}{2}\right)^5=\frac{1}{32}.\)
Vì \(\frac{1}{4}< \frac{1}{32}.\)
=> \(\left(\frac{1}{2}\right)^2< \left(\frac{1}{2}\right)^5.\)
b) \(\left(2,4\right)^3\) và \(\left(2,4\right)^2\)
Ta có: \(\left(2,4\right)^3=13,824.\)
\(\left(2,4\right)^2=5,76.\)
Vì \(13,284>5,76.\)
=> \(\left(2,4\right)^3>\left(2,4\right)^2.\)
c) \(\left(-1\frac{1}{2}\right)^2\) và \(\left(-1\frac{1}{2}\right)^3\)
Ta có: \(\left(-1\frac{1}{2}\right)^2=\left(-\frac{3}{2}\right)^2=\frac{9}{4}.\)
\(\left(-1\frac{1}{2}\right)^3=\left(-\frac{3}{2}\right)^3=-\frac{27}{8}.\)
Vì số dương luôn lớn hơn số âm nên \(\frac{9}{4}>-\frac{27}{8}.\)
=> \(\left(-1\frac{1}{2}\right)^2>\left(-1\frac{1}{2}\right)^3.\)
Chúc bạn học tốt!