So sánh:
1/ \(\left(2^2\right)^3\) và \(2^6\)
2/ \(\left(\frac{-1^2}{2}\right)^6\) và \(\left(\frac{-1}{2}\right)^{10}\)
3/ \(3^{4000}\) và \(9^{2000}\)
4/ \(\left(2.5\right)^2\) và \(2^2.5^2\)
Những câu hỏi liên quan
a) \(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
b) \(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)
\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)
\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)
\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
Mình làm tiếp câu b nha !
b, Bài giải
\(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)
\(\left(\frac{x-2018}{2}+1\right)+\left(\frac{x-2020}{4}+1\right)=\left(\frac{x-2040}{8}+1\right)+\left(\frac{x-2030}{14}+1\right)\)
\(\frac{x-2016}{2}+\frac{x-2016}{4}=\frac{x-2032}{8}+\frac{x-2016}{14}\)
\(\left(x-2016\right)\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{x-2016}{8}-2+\frac{x-2016}{14}\)
\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\left(\frac{1}{8}+\frac{1}{14}\right)-2\)
\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\cdot\frac{11}{56}-2\)
\(\left(x-2016\right)\cdot\frac{3}{4}-\left(x-2016\right)\cdot\frac{11}{56}=-2\)
\(\left(x-2016\right)\left(\frac{3}{4}-\frac{11}{56}\right)=-2\)
\(\left(x-2016\right)\cdot\frac{31}{56}=-2\)
\(x-2016=-2\text{ : }\frac{31}{56}\)
\(x-2016=-\frac{112}{31}\)
\(x=-\frac{112}{31}+2016\)
\(x=\frac{62384}{31}\)
tính và so sánh :
( 2.5 ) 2 và 22 . 52
\(\left(\frac{1}{2}\times\frac{3}{4}\right)^3\)và \(\left(\frac{1}{2}\right)^3\) \(\times\left(\frac{3}{4}\right)^3\)
(2.5)2 =22.52
\(\left(\frac{1}{2}.\frac{3}{4}\right)^3=\left(\frac{1}{2}\right)^3.\left(\frac{3}{4}\right)^3\)
luy thua cua mot tich= tich cac luy thua
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Tính và so sánh.
a)\({\left[ {{{\left( { - 2} \right)}^2}} \right]^3}\) và \({\left( { - 2} \right)^6}\) b) \({\left[ {{{\left( {\frac{1}{2}} \right)}^2}} \right]^2}\) và \({\left( {\frac{1}{2}} \right)^4}\).
a) \({\left[ {{{\left( { - 2} \right)}^2}} \right]^3} = {\left( { - 2} \right)^2}.{\left( { - 2} \right)^2}.{\left( { - 2} \right)^2} = {\left( { - 2} \right)^{2 + 2 + 2}} = {\left( { - 2} \right)^6}\)
Vậy \({\left[ {{{\left( { - 2} \right)}^2}} \right]^3}\) = \({\left( { - 2} \right)^6}\)
b) \({\left[ {{{\left( {\frac{1}{2}} \right)}^2}} \right]^2} = {\left( {\frac{1}{2}} \right)^2}.{\left( {\frac{1}{2}} \right)^2} = {\left( {\frac{1}{2}} \right)^4}\)
Vậy \({\left[ {{{\left( {\frac{1}{2}} \right)}^2}} \right]^2}\) = \({\left( {\frac{1}{2}} \right)^4}\).
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rút gona,frac{2.5^{22}-9.5^{21}}{25^{10}}b,frac{5left(3.7^{15}-19.7^{14}right)}{7^{16}+3.7^{15}}c,frac{left(left(-2right)^2right)^3.left(-4right)^2}{left(-2right)^3.left(-2right)^2} d,frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}e,2^3+3.left(frac{1}{8}right)^0-left(frac{1}{2^2}right).4+left[left(-2right)^2:frac{1}{2}right].8f,frac{left(frac{2}{5}right)^7.5^5+left(frac{9}{4}right)^3:left(frac{3}{16}right)^3}{2^7.5^2+512}
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rút gon
a,\(\frac{2.5^{22}-9.5^{21}}{25^{10}}\)
b,\(\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
c,\(\frac{\left(\left(-2\right)^2\right)^3.\left(-4\right)^2}{\left(-2\right)^3.\left(-2\right)^2}\)
d,\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
e,\(2^3+3.\left(\frac{1}{8}\right)^0-\left(\frac{1}{2^2}\right).4+\left[\left(-2\right)^2:\frac{1}{2}\right].8\)
f,\(\frac{\left(\frac{2}{5}\right)^7.5^5+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\)
Cũng khuya rồi , mình làm câu 1 thôi nhé !
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{2.5^{22}-9.5^{21}}{\left(5^2\right)^{10}}\)
\(\frac{5^{21}.\left(2.5-9\right)}{5^{20}}=5.\left(10-9\right)=5\)
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bài 1: tính
a) \(\left(\frac{3}{7}\right)^0=\frac{7}{9}:\left(\frac{2}{3}\right)^2-\left|-\frac{4}{5}\right|\)
b) \(\frac{10^3+2.5^3+5^3}{55}\)
bài 2: so sánh
\(^{3^{2009}}\) và \(9^{1005}\)
thanksss~~~
a.
\(\left(\frac{3}{7}\right)^0+\frac{7}{9}\div\left(\frac{2}{3}\right)^2-\left|-\frac{4}{5}\right|=0+\frac{7}{9}\div\frac{4}{9}-\frac{4}{5}=\frac{7}{9}\times\frac{9}{4}-\frac{4}{5}=\frac{7}{4}-\frac{4}{5}=\frac{35}{20}-\frac{16}{20}=\frac{19}{20}\)
b.
\(\frac{10^3+2\times5^3+5^3}{55}=\frac{\left(2\times5\right)^3+2\times5^3+5^3}{55}=\frac{2^3\times5^3+2\times5^3+5^3}{5\times11}=\frac{5^3\times\left(2^3+2+1\right)}{5\times11}=\frac{5^2\times11}{11}=5^2=25\)
c.
\(3^{2009}< 3^{2010}=\left(3^2\right)^{1005}=9^{1005}\)
Vậy 32009 < 91005
Chúc bạn học tốt ^^
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So sánh:
a) \({( - 2)^4} \cdot {( - 2)^5}\) và \({( - 2)^{12}}:{( - 2)^3}\);
b) \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6}\) và \({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2}\)
c) \({(0,3)^8}:{(0,3)^2}\) và \({\left[ {{{(0,3)}^2}} \right]^3}\);
d) \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3}\) và \({\left( {\frac{3}{2}} \right)^2}\).
a) \({( - 2)^4} \cdot {( - 2)^5} = {\left( { - 2} \right)^{4 + 5}} = {\left( { - 2} \right)^9}\)
\({( - 2)^{12}}:{( - 2)^3} = {\left( { - 2} \right)^{12 - 3}} = {\left( { - 2} \right)^9}\)
Vậy \({( - 2)^4} \cdot {( - 2)^5}\) = \({( - 2)^{12}}:{( - 2)^3}\);
b) \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6} = {\left( {\frac{1}{2}} \right)^{2 + 6}} = {\left( {\frac{1}{2}} \right)^8}\)
\({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2} = {\left( {\frac{1}{2}} \right)^{4.2}} = {\left( {\frac{1}{2}} \right)^8}\)
Vậy \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6}\) = \({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2}\)
c) \({(0,3)^8}:{(0,3)^2} = {\left( {0,3} \right)^{8 - 2}} = {\left( {0,3} \right)^6}\)
\({\left[ {{{(0,3)}^2}} \right]^3} = {\left( {0,3} \right)^{2.3}} = {\left( {0,3} \right)^6}\)
Vậy \({(0,3)^8}:{(0,3)^2}\)= \({\left[ {{{(0,3)}^2}} \right]^3}\).
d) \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3} = {\left( { - \frac{3}{2}} \right)^{5 - 3}} = {\left( { - \frac{3}{2}} \right)^2} = {\left( {\frac{3}{2}} \right)^2}\)
Vậy \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3}\) = \({\left( {\frac{3}{2}} \right)^2}\).
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(-2) ^4 . (-2) 65 và ( -2) ^ 12 : ( -2) ^3
=( -2) ^ 4+5 =(-2)^9 và (-2) ^12-3 = ( -2) ^9
vậy ( -2) ^9 = (-2) ^9
Nên (-2) ^4 .( -2) ^5 = ( -2) ^ 12 : ( -2) ^3
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Cho A=\(\frac{\left(2^4+\frac{4}{2^4}\right)\left(4^4+\frac{4}{2^4}\right)\left(6^4+\frac{4}{2^4}\right)...\left(32^4+\frac{^4}{2^4}\right)}{\left(1^4+\frac{4}{2^4}\right)\left(3^4+\frac{4}{2^4}\right)\left(5^4+\frac{4}{2^4}\right)...\left(31^4+\frac{4}{2^4}\right)}\) và B =2010. So sánh A và B
mk ko biết mk mới học lớp nhỏ thôi . Đó là lớp này nè bn...... tự vào trang của mk coi đi nhé
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Aduf Lớp 8? Mh mới lớp 2 thui!!!!!
1, tim x bết:,frac{left(-5^4right).left(-15^2right)-5^4.left(-3^2.5right)}{left(-3^4right).25^2-left(-15^2right).225.5}:frac{x}{5}frac{-1}{6}2, cho Afrac{20}{30}+frac{20}{70}+frac{20}{126}+...+frac{20}{798}Bleft(frac{41}{2}.frac{42}{2}.frac{43}{2}...frac{80}{2}right):left(1.3.5...79right)So sánh A và B.3, Tính nhanh.frac{0,875+frac{1}{2}-7%-frac{1}{58}}{frac{1}{25}-frac{1}{2}-frac{2}{7}+frac{2}{203}}-125%HELP ME PLEASE......
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1, tim x bết:
,\(\frac{\left(-5^4\right).\left(-15^2\right)-5^4.\left(-3^2.5\right)}{\left(-3^4\right).25^2-\left(-15^2\right).225.5}\)\(:\)\(\frac{x}{5}\)\(=\frac{-1}{6}\)
2, cho A=\(\frac{20}{30}+\frac{20}{70}+\frac{20}{126}+...+\frac{20}{798}\)
B=\(\left(\frac{41}{2}.\frac{42}{2}.\frac{43}{2}...\frac{80}{2}\right):\left(1.3.5...79\right)\)
So sánh A và B.
3, Tính nhanh.
\(\frac{0,875+\frac{1}{2}-7\%-\frac{1}{58}}{\frac{1}{25}-\frac{1}{2}-\frac{2}{7}+\frac{2}{203}}\)\(-125\%\)
HELP ME PLEASE......
Tínha,frac{2^{10}cdot55+2^{10}cdot26}{2^8cdot27}b,120:{300:[150-(2.53-23.25)]}c,left[left(frac{40}{130}-frac{12}{13}right)cdot40%+0,15right]:frac{-5}{52}d,frac{0,8:left(frac{4}{5}cdot1,25right)}{0,64-frac{1}{25}}+frac{left(1,08-frac{2}{25}right):frac{4}{7}}{left(6frac{5}{9}-3frac{1}{4}right)cdot2frac{2}{17}}+left(1,2+0,5right):frac{1}{5}
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Tính
a,\(\frac{2^{10}\cdot55+2^{10}\cdot26}{2^8\cdot27}\)
b,120:{300:[150-(2.53-23.25)]}
c,\(\left[\left(\frac{40}{130}-\frac{12}{13}\right)\cdot40\%+0,15\right]:\frac{-5}{52}\)
d,\(\frac{0,8:\left(\frac{4}{5}\cdot1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right)\cdot2\frac{2}{17}}+\left(1,2+0,5\right):\frac{1}{5}\)