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Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
d) \(\left(-45,7\right)+\left[\left(+5,7\right)+\left(+5,75\right)+\left(-0,75\right)\right]\)
\(=\left(-45,7\right)+\left[5,7+5,75-0,75\right]\)
\(=\left(-45,7\right)+5,7+5,75-0,75\)
\(=\left[\left(-45,7+5,7\right)\right]+\left[5,75-0,75\right]\)
\(=-40+5=-35\)
e) \(11,26-5,13:\left(5\frac{5}{18}-1\frac{8}{9}\cdot1,25+1\frac{16}{63}\right)\)
\(=11,26-5,13:\left(\frac{95}{18}-\frac{17}{9}\cdot\frac{5}{4}+\frac{79}{63}\right)\)
\(=11,26-5,13:\left(\frac{95}{18}-\frac{85}{36}+\frac{79}{63}\right)\)
\(=\frac{563}{50}-\frac{513}{100}:\frac{1051}{252}\)
\(=\frac{563}{50}-\frac{513}{100}\cdot\frac{252}{1051}\)
\(=\frac{563}{50}-\frac{129276}{105100}=\frac{21083}{2102}\)
Số lớn quá!
j) \(\sqrt{8^2+6^2}\cdot\sqrt{16}+\frac{1}{2}\cdot\sqrt{\frac{4}{5}}\)
\(=\sqrt{64+36}\cdot\sqrt{16}+\frac{1}{2}\cdot\sqrt{\frac{4}{5}}\)
\(=\sqrt{100}\cdot4+\frac{1}{2}\cdot\frac{2\sqrt{5}}{5}\)
\(=10\cdot4+\frac{\sqrt{5}}{5}=40+\frac{\sqrt{5}}{5}=\frac{200+\sqrt{5}}{5}\)
h) Cái đây mình có làm rồi
\(\frac{3}{4}+\frac{1}{4}:\left(-\frac{2}{3}\right)-\left(-5\right)\)
\(=\frac{3}{4}+\frac{1}{4}.\left(-\frac{3}{2}\right)+5\)
\(=\frac{3}{4}-\frac{3}{8}+5\)
\(=\frac{3}{8}+5=\frac{43}{8}\)
\(12.\left(\frac{2}{5}-\frac{5}{6}\right)^2=12.\left(-\frac{13}{30}\right)^2=12.\frac{169}{900}=\frac{169}{75}\)
\(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}=4+6-3+5=12\)
\(\left(9\frac{3}{4}:3.4.2\frac{7}{34}\right):\left(-1\frac{9}{16}\right)=\left(\frac{39}{4}:3.4.\frac{75}{34}\right):\left(-\frac{25}{16}\right)=\frac{975}{34}.\left(-\frac{16}{25}\right)=-\frac{312}{17}\)
\(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}=\frac{3+39}{91-7}=\frac{42}{84}=\frac{1}{2}\)
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
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) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(-1\right)^{2007}=\frac{1}{2}.8-\frac{2}{5}+\left(-1\right)=4-\frac{2}{5}-1=\frac{13}{5}\)
d: \(D=-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-8\cdot\dfrac{1}{2}:\dfrac{27-14}{12}\)
\(=-4:\dfrac{13}{12}\)
\(=-4\cdot\dfrac{12}{13}=-\dfrac{48}{13}\)
e: \(E=5\cdot4-4\cdot3+5-0.3\cdot20\)
=20-12+5-6
=8+5-6
=13-6=7
f: \(F=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{2}:6\)
\(=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{12}\)
\(=\dfrac{27}{12}+\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{34}{12}=\dfrac{17}{6}\)
\(A=\left(0,3.5-0,5:\frac{1}{3}\right)\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)
\(A=\left(0,3.5-0,5.3\right)\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)
\(A=\left(1,5-1,5\right)\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)
\(A=0.\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)
\(A=0\)
VẬY \(A=0\)
(0,3.5-0,5:1/3).(1/2006^2+1/2008^2)
(1,5-1,5).(1/200^2+1/2008^2)
0.(1/2006^2+1/2008^2)
0
\(\left(\sqrt{0,36}.5-\sqrt{\left(0,5\right)^2}\div\frac{1}{6}\right).\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)
= \(\left(0,6.5-0,5.6\right).\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)
= \(\left(3-3\right).\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)
= \(0.\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)
= 0