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Câu 1 xem kỉ đề
\(B,\frac{49^6.5-7^{11}}{\left(-7\right)^{10}.5-2.49^5}=\frac{7^{12}.5-7^{11}}{7^{10}.5-2.7^{10}}=\frac{7^{11}.\left(7.5-1\right)}{7^{10}.\left(5-2\right)}=\frac{7.34}{3}=\frac{238}{3}\)
a) A=212.35-\(\frac{2^{12}.3^6}{2^{12}}\)+93+84.35
=212.35-36+36+212.35
=213.35
b)B=496.5-5.\(\frac{7^{11}}{\left(-7\right)^{10}}-2.49^5\)
=496.5-7.5-2.495
=712.5-7.5-2.710
mk ko viết lại đề
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)
\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)
Vậy A= \(\frac{1}{2}\)
\(\frac{2^{12}.3^5-4^6.3^6}{2^{12}.9^3+8^4.3^5}\)
\(=\frac{2^{12}.3^5-4^6.3.3^5}{2^{12}.3.3^5+4^6.3^5}\)
\(=\frac{1-3}{3+1}\)
\(=\frac{-1}{2}\)
P=212.35-46.36/212.93+84.35
P=212.35-(22)6.36/212.(33)3+(23)4.35
P=212.35-212.36/212.39+212.35
P=212.(35-36)/212.(39+35)
P=35-36/39+35
\(\frac{27^6.81^8.3^9}{3^7.3^6.3^{15}}=\frac{\left(3^3\right)^6.\left(3^4\right)^8.3^9}{3^7.3^6.3^{15}}=\frac{3^{18}.3^{32}.3^9}{3^7.3^6.3^{15}}\)
= \(\frac{3^{59}}{3^{28}}=3^{59}:3^{28}=3^{31}\)
Lời giải:
\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.234.8^2}=\frac{(3^2)^4.(3^3)^5.3^6.3^4}{3^8.(3^4)^4.2.3^2.13.(2^3)^2}\)
\(=\frac{3^8.3^{15}.3^6.3^4}{3^8.3^{16}.2.3^2.13.2^6}=\frac{3^{33}}{3^{26}.2^7.13}=\frac{3^7}{2^7.13}\)
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
\(m=\dfrac{2^7\cdot3^5+2^4\cdot3^9}{2^6\cdot3^5+2^3\cdot3^9}\)
\(m=\dfrac{2^4\cdot3^5\cdot\left(2^3+3^4\right)}{2^3\cdot3^5\cdot\left(2^3+3^4\right)}\)
\(m=\dfrac{2^4\cdot3^5}{2^3\cdot3^5}\)
\(m=\dfrac{2^4}{2^3}\)
\(m=2^{4-3}\)
\(m=2\)