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a)
\(=-\frac{1}{2187}.2187\)
\(=-1\)
b)
\(=\frac{1}{512}.512\)
\(=1\)
c)
\(=\frac{8100}{225}=36\)
d) \(=10000\)
a) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{47}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^{21-12}=\left(\frac{3}{7}\right)^9\)
b) \(\frac{390}{130}^4=3^4=81\)
c) \(\left(0,125\right)^3.512=\frac{1}{512}.512=1\)
Bài làm :
\(\text{a)}=2,5-1,65.\frac{10}{11}=2,5-1.5=1\)
\(b\text{)}=\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{2015-2012}{2012.2015}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-...+\frac{1}{2012}-\frac{1}{2015}\)
\(=\frac{1}{5}-\frac{1}{2015}\)
\(=\frac{402}{2015}\)
\(a,\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right).\frac{10}{11}\)
\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right).\frac{10}{11}\)
\(=\frac{5}{2}-\frac{33}{20}.\frac{10}{11}\)
\(=\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1\)
\(\left(\dfrac{1}{7}\right)^7\cdot7^7=\left(\dfrac{1}{7}\cdot7\right)^7=1^7=1\\ \left(0,125\right)^3\cdot512=\left(0,125\right)^3\cdot8^3=\left(0,125\cdot8\right)^3=1^3=1\\ \left(0,25\right)^4\cdot1024=\left(0,25\right)^4\cdot256\cdot4=\left(0,25\right)^4\cdot4^4\cdot4=\left(0,25\cdot4\right)^4\cdot4=1^4\cdot4=4\)
a) \(\left(\dfrac{1}{7}\right)^7.7^7=\left(\dfrac{1}{7}.7\right)^7=1^7=1\)
b) \(\left(0.125\right)^3.512=\left(0.125\right)^3.8^3=\left(0.125\cdot8\right)^3=1^3=1\)
c) \(\left(0.25\right)^4.1024=\left(0.25\right)^4.4^5=\left(0.25\right)^4.4^4.4=\left(0.25.4\right)^4.4=1^4.4=1.4=4\)
A) (1/5)^5 . 5^5 = 1/5^5 . 5^5 = 5^5 / 5^5 = 1.
B)(0,125)^3 . 512 = (1/8)^3 . 512 = 1/8^3 . 512 = 1/512 . 512 = 1.
C) (0,25)^4 . 1024 = (1/4)^4 . 1024 = 1/4^4 . 1024 = 1/256 . 1024 = 4.
a) (-0,125)3 . (-8) = (-0,125). (-0,125)2 . (-8) = [(-0,125) . (-8)] . (-0,125)2 = 1.1/64 = 1/64
b) \(\frac{27^{15}}{9^{21}}=\frac{\left(3^3\right)^{15}}{\left(3^2\right)^{21}}=\frac{3^{45}}{3^{42}}=3^3=27\)
c) \(\left(-2,5\right)^3+\frac{2496^5}{\left(-832\right)^5}-\frac{98^{17}}{98^{16}}=-\frac{125}{8}+\left(-243\right)-98=-\frac{2853}{8}\)
d) \(\left(1-\frac{1}{3}-\frac{1}{6}\right)^2\cdot\left(16^{37}:2^{145}-1963^0\right)=\left(\frac{6}{6}-\frac{2}{6}-\frac{1}{6}\right)^2\left(16^{37}:2^{145}-1963^0\right)\)
\(\left(\frac{1}{2}\right)^2\cdot7=\frac{1}{4}\cdot7=\frac{7}{4}\)
a) \(\left(-0,125\right)^3.\left(-8\right)=\left(\frac{-1}{8}\right)^3.\left(-8\right)=\left(\frac{-1}{8}\right)^3\div\left(\frac{-1}{8}\right)=\left(\frac{-1}{8}\right)^2=\frac{1}{64}\)
b) \(27^{15}\div9^{21}=\left(3^3\right)^{15}\div\left(3^2\right)^{21}=3^{45}\div3^{42}=3^3=27\)
c) \(\left(-2,5\right)^3+2496^5\div\left(-832\right)^5-98^{17}\div98^{16}=\left(\frac{-5}{2}\right)^3-3^5-98\)
\(=\left(\frac{-125}{8}\right)-243-98=\frac{-2853}{8}\)
d) \(\left(1-\frac{1}{3}-\frac{1}{6}\right).\left(16^{37}\div2^{125}-1963^0\right)=\frac{1}{2}.\left[\left(2^4\right)^{37}\div2^{125}-1\right]\)
\(=\frac{1}{2}.\left[2^{148}\div2^{125}-1\right]=\frac{1}{2}.\left[2^{23}-1\right]=\frac{2^{23}-1}{2}\)
512-\(\frac{512}{2}\)-\(\frac{512}{2^2}\)-\(\frac{512}{2^3}\)-....-\(\frac{512}{2^{10}}\)
=512-256-\(\frac{2^9}{2^2}\)-\(\frac{2^9}{2^3}\)-\(\frac{2^9}{2^4}\)-\(\frac{2^9}{2^5}\)-\(\frac{2^9}{2^6}\)-\(\frac{2^9}{2^7}\)-\(\frac{2^9}{2^8}\)-\(\frac{2^9}{2^9}\)-\(\frac{2^9}{2^{10}}\)
=512-256-128-64-32-16-8-4-2-\(\frac{1}{2}\)
=\(\frac{3}{2}\)
Đặt \(Q=512-\frac{512}{2}-\frac{512}{2^2}-...-\frac{512}{2^{10}}\)
\(=512-512\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
Đặt A là tên biểu thức trong ngoặc ta cs:
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)
Thay A vào Q ta được:
\(Q=512-512\left(1-\frac{1}{2^{10}}\right)=512-512+\frac{512}{2^{10}}=\frac{2^9}{2^{10}}=\frac{1}{2}\)
\(16:\left(0,125\right)^3.512\)
\(=16:\dfrac{1}{512}.512\)
\(=16:\left(\dfrac{1}{512}.512\right)\)
\(=16:1\)
\(=16\)