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\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)
\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)
\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)
\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)
\(=0,2-\dfrac{2}{3}\)
\(=-\dfrac{7}{15}\)
\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)
\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)
\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)
\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)
\(=\dfrac{13}{26}\)
\(=\dfrac{1}{2}\)
#\(Toru\)
\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)
\(\left(\frac{2}{3}-\frac{4}{7}\right):\frac{5}{9}+\left(-\frac{8}{7}+\frac{1}{3}\right):\frac{5}{9}\)
\(=\left[\left(\frac{2}{3}-\frac{4}{7}\right)+\left(-\frac{8}{7}+\frac{1}{3}\right)\right]:\frac{5}{9}\)
\(=\left(\frac{2}{3}-\frac{4}{7}-\frac{8}{7}+\frac{1}{3}\right)\cdot\frac{9}{5}\)
\(=\left(1-\frac{12}{7}\right)\cdot\frac{9}{5}\)
\(=-\frac{5}{7}\cdot\frac{9}{5}\)
\(-\frac{9}{7}\)
\(\left(\frac{2}{3}-\frac{4}{7}\right):\frac{5}{9}+\left(-\frac{8}{7}+\frac{1}{3}\right):\frac{5}{9}\)
\(=\left(\frac{14}{21}-\frac{12}{21}\right):\frac{5}{9}+\left(-\frac{24}{21}+\frac{7}{21}\right):\frac{5}{9}\)
\(=\frac{2}{21}:\frac{5}{9}+\frac{-17}{21}:\frac{5}{9}\)
\(=\left(\frac{2}{21}+\frac{-17}{21}\right):\frac{5}{9}\)
\(=\frac{-15}{21}:\frac{5}{9}\)
\(=\frac{-15}{21}.\frac{9}{5}\)
\(=\frac{-9}{7}\)
\(=\frac{99}{35}\)
a) \(\frac{4}{3}-\frac{2}{5}\)
\(=\frac{20}{15}-\frac{6}{15}=\frac{14}{15}\)
b) \(\left|-\frac{1}{10}\right|-\left(-\frac{1}{3}\right)^2\div\frac{5}{9}\)
\(=\frac{1}{10}-\frac{1}{9}\cdot\frac{9}{5}\)
\(=\frac{1}{10}-\frac{1}{5}=\frac{1}{10}-\frac{2}{10}\)
\(=-\frac{1}{10}\)
c) Đề bài có vấn đề!!!
d) \(\left(-0,2\right)^2\cdot5-8^2\cdot\frac{9^4}{3^7}\cdot4^3\)
\(=0,04\cdot5-64\cdot\frac{\left(3^2\right)^4}{3^7}\cdot64\)
\(=0,2-4096\cdot\frac{3^8}{3^7}=0,2-4096\cdot3\)
\(=0,2-12288=-128878\)
Lười làm qá, hì:
Hướng dẫn thôi nha.
B1: Phá bỏ ngoặc của các phép tính.
B2: Ghép những số nguyên vào vs nhau, phân số vào vs nhau
B3: Giao hoán những phân số có cùng mẫu để cộng vào, ở đây chỉ nói cộng vì trừ lp 7 là cộng vs số đối mà
B4: Tính hết ra là xong
B = (8+6-3) - (9/4-5/4-2/4) + (2/7-3/7-9/7)
B = 11 - 1/2 -10/7
B = 21/2 - 10/7
B = 127/14
a: \(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)=\dfrac{7}{2}\cdot\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\cdot\dfrac{-42}{13}=\dfrac{-147}{13}\)
b: \(=-12+\dfrac{8}{9}-\dfrac{5}{18}=\dfrac{-216}{18}+\dfrac{16}{18}-\dfrac{5}{18}=\dfrac{-205}{18}\)
c: \(=\dfrac{45}{4}-\dfrac{19}{7}-\dfrac{21}{4}=6-\dfrac{19}{7}=\dfrac{23}{7}\)
d: \(=\dfrac{-1}{4}\left(\dfrac{152}{11}+\dfrac{68}{11}\right)=\dfrac{-1}{4}\cdot20=-5\)
Mình ko ghi lại đề nha
=>(-1/5)2x5-(26x38/37x26)
=1/5-3
=-14/5