kệ cha nhà bây

a)   378

b)   3

c)    2

d)    2

e)    \(\frac{8748}{1715}\)

Mình thấy bài e) bạn có ghi thiếu ko vậy.81^2 x;: hay là cộng trừ vậy?

\(a,\) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\frac{2^{12}.3^{10}+\left(2.3\right)^9.2^3.3.5}{2^{12}.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{\left(2^{12}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(=\frac{\left(2^{12}.3^{10}\right).6}{\left(2^{11}.3^{11}\right).5}\)
\(=\frac{2.6}{3.5}\)
\(=\frac{2.2}{5}\)
\(=\frac{4}{5}\)
\(b,\) \(\frac{2^{15}.9^4}{6^3.8^3}\)
\(=\frac{2^{15}.3^8}{2^3.3^3.2^9}\)
\(=\frac{2^{15}.3^8}{2^{12}.3^3}\)
\(=2^3.3^5\)
\(=8.243\)
\(=1944\)
Chúc bạn học tốt ^^


 

a) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+6^9.120}{\left(2^3\right)^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+6^9.120}{2^{12}.3^{12}-6^{11}}=\frac{6^{10}.4+6^{10}.20}{6^{12}-6^{11}}=\frac{6^{10}.\left(4+20\right)}{6^{11}.\left(6-1\right)}=\frac{6^{11}.4}{6^{11}.5}=\frac{4}{5}\)

b) \(\frac{2^{15}.9^4}{6^3.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^3.3^3.2^9}=\frac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5=1944\)

c) \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(4.2\right)^{10}+4^{10}}{\left(2^3\right)^4+4^6.4^5}=\frac{4^{10}.2^{10}+4^{10}}{2^{12}+4^6.4^5}=\frac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.2^{10}}=\frac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(1+2^{10}\right)}=\frac{4^{10}}{4^6}=4^4=256\)

a/\(378\)

b/\(3\)

a: \(\frac{\left(-5\right)^{60}\cdot30^5}{15^5\cdot5^{61}}\)

\(=\frac{5^{60}}{5^{61}}\cdot\left(\frac{30}{15}\right)^5\)

\(=\frac{2^5}{5}=\frac{32}{5}\)

b: \(\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot\left(-1\right)\cdot3^{14}}=\frac{-3^{15}}{3^{14}}\cdot\frac15=-\frac35\)

c: \(4^6\cdot9^5+6^9\cdot120\)

\(=2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5\)

\(=2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5=2^{12}\cdot3^{10}\cdot6=2^{13}\cdot3^{11}\)

\(8^4\cdot3^{12}-6^{11}\)

\(=2^{12}\cdot3^{12}-2^{11}\cdot3^{11}\)

\(=2^{11}\cdot3^{11}\left(2\cdot3-1\right)=2^{11}\cdot3^{11}\cdot5\)

Ta có: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=\frac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot5}=\frac{2^2}{5}=\frac45\)

d: \(4^2\cdot25^2+32\cdot125\)

\(=2^4\cdot5^4+2^5\cdot5^3\)

\(=2^4\cdot5^3\left(5+2\right)=2^4\cdot5^3\cdot7\)

\(\frac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\)

\(=\frac{2^4\cdot5^3\cdot7}{2^3\cdot5^2}=2\cdot5\cdot7=10\cdot7=70\)

GIÚP MÌNH VỚI MN ƠI

\(C=\dfrac{6^3+3\cdot6^2+3^3}{13}=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}=27\)

a. - \(\frac{-1}{2}\)

b. \(\frac{238}{3}\)

Kotori Minami bạn có thể giải chi tiết ra duoc ko