Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(B=\frac{2^5.6^3}{8^2.9^2}=\frac{2^5.\left(2.3\right)^3}{\left(2^3\right)^2.\left(3^2\right)^2}=\frac{2^5.2^3.3^3}{2^6.3^4}=\frac{2^8.3^3}{2^6.3^4}=\frac{2^2}{3}=\frac{4}{3}\)
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 \(A=\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}\)
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.
Bài 1
\(a,\left|x\right|=-\left|-\frac{5}{7}\right|=>x\in\varnothing\)
\(b,\left|x+4,3\right|-\left|-2,8\right|=0\)
\(=>\left|x+4,3\right|-2,8=0\)
\(=>\left|x+4,3\right|=0+2,8=2,8\)
\(=>x+4,3=\pm2,8\)
\(=>\hept{\begin{cases}x+4,3=2,8\\x+4,3=-2,8\end{cases}=>\hept{\begin{cases}x=-1,5\\x=-7,1\end{cases}}}\)
\(c,\left|x\right|+x=\frac{2}{3}\)
\(=>\hept{\begin{cases}x+x=\frac{2}{3}\\-x+x=\frac{2}{3}\end{cases}}=>\hept{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}\)
a) \(\frac{\left(17\frac{2}{9}-15\frac{2}{15}\right):5\frac{2}{9}}{\left(18+3,75\right):0,25}.25\%=\frac{\left(17+\frac{2}{9}-15-\frac{2}{15}\right):\frac{47}{9}}{\left(18+3,75\right).4}.\frac{1}{4}\)
\(=\frac{\left(2+\frac{4}{45}\right).\frac{9}{47}}{\left(18+3,75\right).16}=\frac{\frac{94}{45}.\frac{9}{47}}{288+60}=\frac{\frac{2}{5}}{348}=\frac{2}{5}.\frac{1}{348}=\frac{174}{5}=34,8\)
b) \(\frac{\frac{5}{3}-\frac{5}{7}+\frac{5}{9}}{\frac{10}{3}-\frac{10}{7}+\frac{10}{9}}=\frac{5\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}{10\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}=\frac{1}{2}\)
b) \(\frac{\frac{5}{3}-\frac{5}{7}+\frac{5}{9}}{\frac{10}{3}-\frac{10}{7}+\frac{10}{9}}=\frac{5.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}{10.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}=\frac{5}{10}=\frac{1}{2}\)
a/ \(\left(\frac{-2}{3}\right)^4:24=\frac{16}{81}:24=\frac{2}{243}\)
b/ \(\left(\frac{3}{4}\right)^3.4^4=\frac{27}{64}.256=108\)
c/ \(\frac{3.0,8^5}{2,4^4}=\frac{3.0,32768}{33,1776}=\frac{0,98304}{33,1776}=\frac{4}{135}\)
d/ \(\frac{3^3-0,9^5}{2,7^4}=\frac{27-0,59049}{53,1441}=\frac{26,40951}{53,1441}=0,4969415231\)
e/\(\left(\frac{-7}{2}\right)^2+\left(\frac{-3}{4}\right)^3.64-\left(\frac{-61}{5}\right)\)
\(=\frac{49}{4}+\frac{-27}{64}.64+\frac{61}{5}\)
\(=12,25-27+12,2\)
\(=-2,55\)
f/ \(\frac{2^4.2^6}{\left(2^5\right)^2}-\frac{2^5.15^3}{6^3.10^2}=\frac{2^{10}}{2^{10}}-\frac{2^5.5^3.3^3}{2^3.3^3.5^2.2^2}\)
\(=1-\frac{2^5.5^3.3^3}{2^5.3^3.5^2}=1-\frac{5}{1}=-4\)
\(\)
chúc bạn học tốt
\(A=\frac{25^3.5^5}{6.5^{10}}=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}=\frac{5^6.5^5}{6.5^{10}}=\frac{5^{11}}{6.5^{10}}=\frac{5}{6}\)
\(B=\frac{2^5.6^3}{8^2.9^2}=\frac{2^5.2^3.3^3}{\left(2^3\right)^2.\left(3^2\right)^2}=\frac{2^8.3^3}{2^6.3^4}=\frac{4}{3}\)