\(\frac{\left(1,2\right)^6}{\left(0,3\right)^5.\left(0,2\right)^7}\)tính????
\(\frac{\left(1,2\right)^6}{\left(0,3\right)^5.\left(0,2\right)^7}\)tính
Tính: \(\frac{\left(0,6\right)^5.\left(0,3\right)^3}{\left(0,2\right)^6.\left(0,3\right)^7}\)
\(\frac{\left(0,6\right)^5.\left(0,3\right)^3}{\left(0,2\right)^6.\left(0,3\right)^7}\)
\(=\frac{\left(\frac{6}{10}\right)^5.\left(\frac{3}{10}\right)^3}{\left(\frac{2}{10}\right)^6.\left(\frac{3}{10}\right)^7}\)
\(=\frac{6^5.3^3.\frac{1}{10^8}}{2^6.3^7.\frac{1}{10^{13}}}\)
\(=\frac{2^5.3^5.3^3}{2^6.3^7.\frac{1}{10^5}}\)
\(=\frac{10^5.3}{2}\)
\(=150000\)
Tính
\(\frac{\left(\frac{2}{7}+0,2+\frac{2}{25}\right):\left(\frac{2}{7}+0,2+\frac{2}{25}\right)}{\left(0,5-\frac{1}{5}+0,25-0,2\right):\left(0,3-\frac{1}{5}\right)}\)
nhầm sửa lạ nha
tử bằng 1
mẫu bằng
(1/2-1/5+1/4-1/5):(3/10-1/5)=7/20:1/10=7/2
vậy 1:7/2=2/7
Tính:
a)\(\left\{\left[\left(6,2:0,31-\frac{5}{6}.0,9\right).0,2+0,15\right]:0,2\right\}:\left[\left(2+1\frac{4}{11}:0,1\right).\frac{1}{33}\right]\)
b)\(0,4\left(3\right)+0,6\left(2\right)-2\frac{1}{2}.\left[\left(\frac{1}{2}+\frac{1}{3}:0,5\left(8\right)\right)\right]:\frac{50}{53}\)
c)\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
Tính
a, \(\frac{\left(\frac{1}{6}\right)^2.6^2+\left(0,6\right)^5}{\left(0,2\right)^5}\)
b, \(\left(-3\right)^2+\left(\frac{-1}{2}\right)^3-\left(3,5\right)^0\)
c, \(2\frac{1}{4}+\frac{2}{5}.0,3-\left(\frac{12}{5}-3\right)^2\)
mình cần tính nhanh cơ hoặc tính ra đày đủ đó
Tính giá trị các biểu thức.
a)\(\frac{{{4^3}{{.9}^7}}}{{{{27}^5}{{.8}^2}}};\)
b)\(\frac{{{{\left( { - 2} \right)}^3}.{{\left( { - 2} \right)}^7}}}{{{{3.4}^6}}};\)
c)\(\frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,09} \right)}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}};\)
d)\(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}}.\)
a)
\(\frac{{{4^3}{{.9}^7}}}{{{{27}^5}{{.8}^2}}} = \frac{{{{\left( {{2^2}} \right)}^3}.{{\left( {{3^2}} \right)}^7}}}{{{{\left( {{3^3}} \right)}^5}.{{\left( {{2^3}} \right)}^2}}} =\frac{2^{2.3}.3^{2.7}}{3^{3.5}.2^{2.3}}= \frac{{{2^6}{{.3}^{14}}}}{{{3^{15}}{{.2}^6}}} = \frac{1}{3}\)
b)
\(\frac{{{{\left( { - 2} \right)}^3}.{{\left( { - 2} \right)}^7}}}{{{{3.4}^6}}} =\frac{(-2)^{3+7}}{3.(2^2)^6}= \frac{{{{\left( { - 2} \right)}^{10}}}}{{3.{{\left( {{2^{2.6}}} \right)}}}} = \frac{{{2^{10}}}}{{{{3.2}^{12}}}} = \frac{1}{{{{3.2}^2}}} = \frac{1}{{12}}\)
c)
\(\begin{array}{l}\frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,09} \right)}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left[ {{{\left( {0,3} \right)}^2}} \right]}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,3} \right)}^6}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}}\\ = \frac{{{{\left( {0,3} \right)}^2}}}{{{{\left( {0,2} \right)}^2}}} = \frac{{0,9}}{{0,4}} = \frac{9}{4}\end{array}\)
d)
Cách 1: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{8 + 16 + 32}}{{49}} = \frac{{56}}{{49}} = \frac{8}{7}\)
Cách 2: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{2^3.(1+2+2^2)}}{{7^2}} = \frac{{2^3.7}}{{7^2}} = \frac{8}{7}\)
Tính
a) \(0,76-\left|1,2-3,4\right|+\left(0,3\right)^2\)
b)\(\left(-7\right)^2+\left(\frac{1}{2}\right)^2+\left|5-17\right|\)
\(a.0,76-\left|1,2-3,4\right|+\left(0,3\right)^2=0,76-2,2+0,09=-1,35\)
\(b.\left(-7\right)^2+\left(\frac{1}{2}\right)^2+\left|5-17\right|=49+\frac{1}{4}+12=61,25\)
Đơn giản biểu thức sau:
a) \(\frac{2^3.2^4}{2^5}\)
b)\(\frac{\left(0,2\right)^5.\left(0,6\right)^4}{\left(0,2\right)^7.\left(0,3\right)^4}\)
c)\(\frac{3^3.12^4}{6^5.9^4}\)
d)\(\frac{2^3+2^4+2^5}{7^2}\)
e)\(\frac{2^{15}.9^4}{6^6.8^3}\)
\(a,\frac{2^3.2^4}{2^5}=\frac{2^7}{2^5}=2^2=4\)
\(b,\frac{\left(0,2\right)^5.\left(0,6\right)^4}{\left(0,2\right)^7.\left(0,3\right)^4}=\frac{\left(0,2\right)^5.\left(0,3\right)^4.2^4}{\left(0,2\right)^7.\left(0,3\right)^4}=\frac{2^4}{\left(0,2\right)^2}\)
\(c,\frac{3^3.12^4}{6^5.9^4}=\frac{3^3.6^4.2^4}{6^5.3^8}=\frac{2^4}{6.3^5}=\frac{2^4}{2.3.3^5}=\frac{2^3}{3^6}\)
\(d,\frac{2^3+2^4+2^5}{7^2}=\frac{8+16+32}{49}=\frac{56}{49}=\frac{8}{7}\)
\(e,\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
Viết kết quả của mỗi phép tính sau dưới dạng một lũy thừa của a:
a)\({\left[ {{{\left( { - \frac{1}{6}} \right)}^3}} \right]^4}\) với \(a = - \frac{1}{6}\).
b)\({\left[ {{{\left( { - 0,2} \right)}^4}} \right]^5}\) với \(a = - 0,2\).
a)\({\left[ {{{\left( { - \frac{1}{6}} \right)}^3}} \right]^4}\) (với \(a = - \frac{1}{6}\))
\(=(- \frac{1}{6})^{3. 4}=(- \frac{1}{6})^{12}\)
b)\({\left[ {{{\left( { - 0,2} \right)}^4}} \right]^5}\) (với \(a = - 0,2\))
\(=(-0,2)^{4.5}=(-0,2)^{20}\)