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làm bài 3 BĐT
theo bảng xét dấu
còn bài 1,2 ở trên là 1.1 và 1.2 đều trg bài 1.2
bài 1.2 (tức bài 2 ở trên )làm a,b,c,d
\còn bài 2( tức bài 2 ở trên) làm hết
a) \(\left(\dfrac{1}{5}\right)^5.5^5=1\)
b) \(\left(0,125\right)^3.512=1\)
c) \(\left(0,25\right)^4.1024=4\)
a) (1/5)^5 . 5^5 = (1/5. 5)^5 = 1^5= 1
b) (0,125)^3. 512= (0,125)^3 . 8^3 = (0,125. 8)^3 = 1^3= 1
c) (0,25)^4. 1024= [(0,25)^2]^2. 32^2= (1/6)^2. 32^2=(1/6.32)^2= (32/6)^2 =2^2= 4
a/ \(\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)
b/ \(\left(0,125\right)^3.512=\left(0,125\right).8^3=\left(0,125.8\right)^3=1^3=1\)
c/ \(\left(0,25\right)^4.1024=\left(0,25^2\right)^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=16^2\)
\(a,\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)
\(b,\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)
\(c,\left(0,25\right)^4.1024=\left(0,25\right)^4.4^4.4=\left(0,25.4\right)^4.4=1^4.4=1.4=4\)
Giải:
a) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=30^3=2700\)
b) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=30^4=810000\)
c) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^2=8^2=64\)
Đáp số: a) 2700; b) 810000; c) 64.
Chúc bạn học tốt!!!
a) \(\left(\dfrac{1}{5}\right)^5.5^5\)
\(=\dfrac{1}{3125}.3125\)
= 1
b) \(\left(0,125\right)^3.512\)
\(=\dfrac{1}{512}.512\)
= 1
c) \(\left(0,25\right)^4.1024\)
= \(\dfrac{1}{256}.1024\)
= 4
a) $\dfrac{120^3}{40^3}=(\dfrac{120}{40})^3=3^3=27$
b) $\dfrac{390^4}{130^4}=(\dfrac{390}{130})^4=3^4=81$
c) $\dfrac{3^2}{(0,375)^2}=(3:0,375)^2=(3:\dfrac{3}{8})^2=8^2=64$
\(a,\left(\dfrac{1}{7}\right)^7.7^7=\left(\dfrac{1}{7}.7\right)^7=1\)
\(b,\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)\(c,\left(0,25\right)^4.1024=\left(\left(0,25\right)^2\right)^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=2^2=4\)
\(d,\dfrac{90^3}{15^3}=\left(\dfrac{90}{15}\right)^3=6^3=216\)
a.\(12,5.\left(-\dfrac{5}{7}\right)+1,5.\left(-\dfrac{5}{7}\right)\)
\(=\left(-\dfrac{5}{7}\right).\left(12,5+1,5\right)\)
\(=-10\)
b,\(\left(-\dfrac{2}{5}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{5}-\dfrac{3}{7}-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=-\dfrac{3}{5}:\dfrac{4}{5}\)
\(=-\dfrac{3}{4}\)
c,\(12.\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)
\(=12.\dfrac{4}{9}+\dfrac{4}{3}\)
\(=\dfrac{16}{3}+\dfrac{4}{3}\)
\(=\dfrac{20}{3}\)
d,\(1:\left(\dfrac{2}{3}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{1}{1}:\dfrac{1}{144}\)
\(=144\)
e,\(15.\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15.\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
a) = ( 12,5 +1,5 ). \(\left(-\dfrac{5}{7}\right)\)
= 14 . \(\left(-\dfrac{5}{7}\right)\)
= -10
b) = (\(-\dfrac{2}{5}+-\dfrac{1}{5}\)) + \(\left(\dfrac{3}{7}-\dfrac{3}{7}\right)\): \(\dfrac{4}{5}\)
= \(\left(-\dfrac{3}{5}+0\right)\): \(\dfrac{4}{5}\)
= \(\dfrac{3}{4}\)
c) = \(\left(12.-\dfrac{2}{9}\right)\) + \(\dfrac{4}{3}\)
= \(\dfrac{8}{3}\) + \(\dfrac{4}{3}\)
= \(-\dfrac{4}{3}\)
d) = 1: \(\dfrac{23}{48}\)
=\(\dfrac{48}{23}\)
e) =\(\left(15.-\dfrac{2}{9}\right)-\dfrac{7}{3}\)
= \(\left(-\dfrac{10}{3}\right)-\dfrac{7}{3}\)
=\(-\dfrac{17}{3}\)
f) = 10 485.76
a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)=\dfrac{7}{8}.\left(\dfrac{10}{60}+\dfrac{24}{60}\right)=\dfrac{7}{8}.\dfrac{17}{30}=\dfrac{114}{240}\)
b)\(\dfrac{3}{2}-\dfrac{5}{6}\left(\dfrac{1}{2}\right)^2+\sqrt{4}=\dfrac{3}{2}-\dfrac{5}{6}.\dfrac{1}{4}+2=\dfrac{3}{2}-\dfrac{5}{24}+2=\dfrac{36}{24}-\dfrac{5}{24}+\dfrac{48}{24}=\dfrac{79}{24}\)c)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-1\dfrac{15}{17}=1+\left(\dfrac{7}{21}+\dfrac{14}{21}\right)-\dfrac{32}{17}=1+1-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{34}{17}-\dfrac{32}{17}=\dfrac{2}{17}\)d)\(\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)=-8.\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-8.\dfrac{2}{4}:\left(\dfrac{54}{24}-\dfrac{28}{24}\right)=-8.\dfrac{2}{4}:\dfrac{13}{12}=-4.\dfrac{12}{13}=\dfrac{-48}{13}\)e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)=16\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)+28\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)=\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right).\left(-\dfrac{5}{3}\right)=\left(\dfrac{120}{7}+\dfrac{196}{7}\right).\left(-\dfrac{5}{3}\right)=\dfrac{316}{7}.\left(-\dfrac{5}{3}\right)=-\dfrac{1580}{21}\)
a) \(\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)
b) \(\left(0,125\right)^3.512=\left(0,512\right)^3.8^3=\left(0,512.8\right)^3=1^3=1\)
c) \(\left(0,25\right)^4.1024=\left[\left(0,25\right)^2\right]^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=2^2=4\)
d) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=3^3=64\)
e) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=3^4=81\)
g) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^3=8^3=512\)