a) Đặt: A = \(\frac{5.7^{12}}{7^{10}.5^{10}}=\frac{5.7^2.7^{10}}{7^{10}.5^{10}}=\frac{7^2}{5^9}\)

b) Đặt: B = \(\frac{5^8.7^{12}-5^9.7^{11}}{7^{11}.5^{10}+7^{12}.5^{11}}=\frac{5^8.7.7^{11}-5.5^8.7^{11}}{7^{11}.5^{10}+7.7^{11}.5.5^{10}}\\ =\frac{5^8.7^{11}.\left(7-5\right)}{7^{11}.5^{10}.\left(1+7.5\right)}=\frac{2}{5^2.36}=\frac{2}{25.36}=\frac{1}{450}\)

a)\(\frac{5.7^{12}}{7^{10}.5^{10}}=\frac{5.7^{10}.7^2}{7^{10}.5.5^9}=\frac{7^2}{5^9}\)

\(\dfrac{\left(-6\right).11}{\left(-11\right).\left(-8\right)}=\dfrac{\left(-3\right).2.11}{11.2.4}=-\dfrac{3}{4}\)

\(\dfrac{21.\left(-5\right)}{25.\left(-7\right)}=\dfrac{\left(-3\right).\left(-7\right).\left(-5\right)}{\left(-5\right).\left(-5\right).\left(-7\right)}=\dfrac{3}{5}\)

\(\dfrac{32.9.11}{12.24.22}=\dfrac{4.4.2.3.3.11}{3.4.4.2.3.11.2}=\dfrac{4^2.2.3^2.11}{3^2.4^2.2^2.11}=\dfrac{1}{2}\)

\(\dfrac{\left(-6\right).11}{\left(-11\right).\left(-8\right)}\)=\(\dfrac{-3}{4}\)

\(\dfrac{21.\left(-5\right)}{25.\left(-7\right)}\)=\(\dfrac{2}{5}\)

\(\dfrac{32.9.11}{12.24.22}\)=\(\dfrac{1}{2}\)

D=115/132/103/132=115/103

\(D=\frac{\frac{88}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}\)

\(D=\frac{\frac{115}{132}}{\frac{103}{132}}\)

\(D=\frac{115}{103}\)

D = \(\frac{115}{103}\)

\(a,=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)

\(b,=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}=-6\sqrt{3}\)

\(c,=3\sqrt{3}+7\sqrt{3}-9\sqrt{3}+11\sqrt{3}=12\sqrt{3}\)

a) Ta có: \(-\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{5}\sqrt{125}\)

\(=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{1}{5}\cdot5\sqrt{5}\)

\(=-17\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)

b) Ta có: \(2\sqrt{3}-\sqrt{75}+2\sqrt{12}-\sqrt{147}\)

\(=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}\)

\(=-6\sqrt{3}\)

\(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}\)

=\(\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.7+9\right)}\)

= \(\frac{8}{35+9}\)

= \(\frac{8}{44}\)

= \(\frac{2}{11}\)

\(\frac{5^{11}.7^{12}+5^{11}.7^{10}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}7^{10}\left(7^2+1\right)}{5^{11}7^{11}\left(5.7+9\right)}=\frac{50}{7.44}=\frac{25}{154}\)

\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}=\dfrac{2.6}{3.5}=\dfrac{4}{5}\)

Giải giúp em với mn

\(A=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+6^9.6.2^2.5}{\left(2^3\right)^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\dfrac{2^{11}.3^{10}\left(2^1+2^1.5\right)}{2^{11}.3^{10}\left(2^1.3^2-1.3^1\right)}=\dfrac{2+10}{2.9-1.3}=\dfrac{12}{15}=\dfrac{4}{5}\)

=45 nha

olm duyet di

= 45

tích nha!!!..^.^

bằng \(\frac{2}{11}\)bn nhé

dell biết giải mà đăng hoài

\(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}.\left(7+1\right)}{5^{11}.7^{11}\left(5.7+9\right)}=\frac{8}{35+9}=\frac{8}{44}=\frac{2}{11}\)

marion loz gì đó... bạn nói cái đách lone gì vậy?

\(\frac{5^{11}\cdot7^{12}+5^{11}\cdot7^{11}}{5^{12}\cdot7^{11}+9\cdot5^{11}\cdot7^{11}}\)

=\(\frac{5^{11}\cdot\left(7^{12}+7^{11}\right)}{5^{11}\left(5\cdot7^{11}+9\cdot7^{11}\right)}\)

=\(\frac{7^{12}+7^{11}}{5\cdot7^{11}+9\cdot7^{11}}\)

=\(\frac{7^{11}\left(7+1\right)}{7^{11}\left(5+9\right)}\)

=\(\frac{8}{14}=\frac{4}{7}\)

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