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\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{1}{18}\)
\(b,2^8:2^5+3^3.2-12\)
\(=2^3+9.2-12\)
\(=8+18-12\)
\(=26-12\)
\(=14\)
Câu c,d em chưa học nên không biết làm ạ, mong mọi người thông cảm!!!
Sửa lại câu b
\(=2^3+27.2-12\)
\(=8+54-12\)
\(=62-12\)
\(=50\)
\(\sqrt{\frac{1}{9}+\frac{1}{16}}\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}\)
\(-\frac{5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)=-\frac{5}{9}\left(\frac{3}{10}-\frac{4}{10}\right)=-\frac{5}{9}.\frac{-1}{10}=\frac{1}{18}\)
\(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}=\frac{1}{2}.8-\frac{3}{5}+1=4+\frac{2}{5}=\frac{22}{5}\)
\(2^8:2^5+3^2.2-12=2^3+9.2-12=8+18-12=8+6=14\)
\(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+\frac{9}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+3=9\frac{4}{9}\)
\(3^x+\frac{4}{9}=9+\frac{4}{9}\)
\(\Rightarrow3^x=9+\frac{4}{9}-\frac{4}{9}\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
a) \(\frac{15^{15}.5^{10}}{9^7.25^{13}}=\frac{3^{15}.5^{15}.5^{10}}{3^{14}.5^{26}}=\frac{3.5^{25}}{5^{26}}=\frac{3}{5}\)
b) \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\sqrt{\frac{2^2}{9^2}}:\sqrt{\frac{5^2}{9^2}}-\frac{7}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{9}.\frac{9}{5}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}\)
\(=\frac{-5}{5}=-1\)
c) \(\left(3^2\right)^2-625+64\)
\(=3^4-625+64\)
\(=81-625+64\)
\(=-480\)
d) \(\frac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{7^2}-\sqrt{91^2}}\)
\(=\frac{3-39}{7-91}\)
\(=\frac{-36}{-84}\)
\(=\frac{3}{7}\)
1. a) 3+2=5
b) 0,5-0,1=0,4
c) 4/5-1/9=31/45
d) 2-0,6=1,4
2. a) 8-4+3=7
b) 11+5-3=13
c) 3/2-4/6-7-37/6
d) 4+5-6=3
\(\frac{2}{3}\sqrt{81}-\left(\frac{-3}{4}\right):\sqrt{\frac{9}{64}}-\left(\frac{\sqrt{5}}{2011}\right)^0\)
\(=\frac{2}{3}\cdot9+\frac{3}{4}\cdot\frac{8}{3}-1\)
\(=6+2-1\)
\(=7\)
\(\frac{2}{3}\sqrt{81}-\left(-\frac{3}{4}\right):\sqrt{\frac{9}{64}}-\left(\frac{\sqrt{5}}{2011}\right)^0=\frac{2}{3}.9-\left(-2\right)-1=6+2-1=7\)