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\(A=\frac{25^3.5^5}{6.5^{10}}=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}=\frac{5^6.5^5}{6.5^{10}}=\frac{5^{11}}{6.5^{10}}=\frac{5}{6}\)
\(B=\frac{2^5.6^3}{8^2.9^2}=\frac{2^5.2^3.3^3}{\left(2^3\right)^2.\left(3^2\right)^2}=\frac{2^8.3^3}{2^6.3^4}=\frac{4}{3}\)
sorry mk chỉ giải đk câu a thui nhé
(2.5)^2= =100 ; (6.5)^3 =27000
k cho mk vs dù sao thì mk cũng giúp đk bn mà ^ ^
Tính nhanh \(A=\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}\)
A= \(\frac{25^3.5^3}{6.5^{10}}\)= \(\frac{\left(5^2\right)^3.5^3}{6.5^{10}}\)= \(\frac{5^6.5^3}{6.5^{10}}\)= \(\frac{5^9}{6.5^{10}}\)= \(\frac{5}{6}\)
B = \(\frac{2^5.6^3}{8^2.9^2}\)= \(\frac{2^5.\left(2.3\right)^3}{\left(2^3\right)^2.\left(3^2\right)^2}\)=\(\frac{2^5.2^3.3^3}{2^6.3^4}\)= \(\frac{2^8.3^3}{2^6.3^4}\)= \(\frac{2^2}{3}\)= \(\frac{4}{3}\)
C = \(\frac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)= \(\frac{5^3.3^3+5.5^2.3^2-5^3}{6^3.3^3+6.6^2.3^2-6^3}\)= \(\frac{5^3+5^3.3^2-5^3}{6^3.3^3+6^3.3^2-6^3}\)= \(\frac{5^3.\left(1+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)= \(\frac{5^3.9}{6^3.35}\)
=\(\frac{5^3.3^2}{2^3.3^3.7.5}\)
= \(\frac{25}{168}\)
D = \(\frac{\left(7^4-7^3\right)^2}{49^3}\)= \(\frac{[7^3\left(7-1\right)]^2}{\left(7^2\right)^3}\)= \(\frac{7^6.6^2}{7^6}\)= \(36\)
Mình mới lớp 6 nên chỉ làm được câu thôi.
\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}=\frac{3.\left\{\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right\}}{11.\left\{\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right\}}\)
\(=\frac{3}{11}\)
cho mình nha các bạn.
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
a . Ta có : \(A=\frac{25^3.5^5}{6.5^{10}}=\frac{5^{11}}{6.5^{10}}=\frac{5}{6}\)
b . Ta có : \(B=\frac{2^6.6^3}{8^2.9^2}=\frac{2^5.\left(2.3\right)^3}{\left(2^3\right)^2.\left(3^2\right)^2}=\frac{2^8.3^3}{2^6.3^4}=\frac{4}{3}\)
c . Ta có :
\(C=\frac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}=\frac{\left(3.5\right)^3+5.\left(3.5\right)^2-5^3}{\left(6.3\right)^3+6.\left(3.6\right)^2-6^3}=\frac{5^3.\left(3^3-3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}=\frac{5^3}{6^3}\)
d . Ta có :
\(D=\frac{\left(7^4-7^3\right)^2}{49^3}=\frac{7^4-7^3}{49^3}.\left(7^4-7^3\right)=\left(\frac{7^4}{49^3}-\frac{7^3}{49^3}\right).\left(7^4-7^3\right)\)
\(=\left(\frac{7^4}{7^6}-\frac{7^3}{7^6}\right).\left(7^4-7^3\right)=\left(\frac{1}{7^2}-\frac{1}{7^3}\right).\left(7^4-7^3\right)\)
\(=\frac{6}{7^3}.\left(7^4-7^3\right)=\frac{6}{7^3}.7^3.\left(7-1\right)=36\)
a) \(A=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}=\frac{5^6.5^5}{6.5^{10}}=\frac{5^{11}}{6.5^{10}}=\frac{5}{6}\)
b) \(B=\frac{2^5.6^3}{8^2.9^2}=\frac{2^5.\left(2.3\right)^3}{\left(2^3\right)^2.\left(3^2\right)^2}=\frac{2^5.2^3.3^3}{2^6.3^4}=\frac{2^8.3^3}{2^6.3^4}=\frac{2^2}{3}=\frac{4}{3}\)