\(\frac{10^3+2.5^3+5^3}{55}-\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)\(=\frac{2^3.5^3+2.5^3+5^3}{5.11}-\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\frac{5^3\left(2^3+2+1\right)}{5.11}-\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{5^2.11}{11}-\frac{2.6}{3.5}=25-\frac{4}{5}=\frac{121}{5}\)

Sửa đề : \(10.\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)

\(10\cdot\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=10\cdot\frac{2^{12}\cdot3^{10}+2\cdot9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=10\cdot\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)

\(=10\cdot\frac{2\cdot6}{3\cdot5}=8\)

Bài làm:

\(10.\frac{4^6.9^5+6^9.120}{8^4.3^{12}.6^{11}}\)

\(=\frac{2.5.2^{12}.3^{10}+2^9.3^9.2.5.2^2.3}{2^{12}.3^{12}.2^{11}.3^{11}}\)

\(=\frac{2^{13}.3^{10}.5+2^{12}.3^{10}.5}{2^{23}.3^{23}}\)

\(=\frac{2^{12}.3^{10}.5\left(2+1\right)}{2^{23}.3^{23}}\)

\(=\frac{2^{12}.3^{11}.5}{2^{23}.3^{23}}=\frac{5}{2^{11}.3^{12}}\)

sợ sai

Mình làm nhầm chỗ 2⁹.3⁹.2³.3.5 phải bằng 2¹².3¹⁰.5. Thông cảm nha bạn 

\(\frac{4^6.9^5-6^9.120}{8^4.3^{12}-6^{11}}\)

\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)

\(=\frac{2.6}{3.5}\)

\(=\frac{4}{5}\)

Phải biến đổi

thì bạn giải tớ với

Biến đổi ra đi , ví dụ : 4⁶=(2²)⁶=2^2x6=2¹²

^{Còn cách trình bày thì viết kiểu phân số , đánh trên máy tính ko quen nhé , mong bạn thông cảm}

\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)= \(\frac{\left(2^2\right)^6.\left(3^2\right)^5+6^9.6.20}{\left(2^3\right)^4.3^{12}-6^{11}}\)= \(\frac{2^{12}.3^{10}+6^{10}.20}{2^{12}.3^{12}-6^{11}}\)

\(\frac{2^{10}.2^2.3^{10}+6^{10}.20}{6^{12}-6^{11}}\)= \(\frac{6^{10}\left(2^2+20\right)}{6^{10}\left(6^2-6\right)}\)= \(\frac{24}{30}\)= \(\frac{4}{5}\)

\(\frac{4}{5}\)

những cơ số lớn thì bạn cứ quy về cơ số nhỏ để tìm nhân tử chung.

vd: 4^3=2^6

9^5=3^10...

\(a,\) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\frac{2^{12}.3^{10}+\left(2.3\right)^9.2^3.3.5}{2^{12}.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{\left(2^{12}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(=\frac{\left(2^{12}.3^{10}\right).6}{\left(2^{11}.3^{11}\right).5}\)
\(=\frac{2.6}{3.5}\)
\(=\frac{2.2}{5}\)
\(=\frac{4}{5}\)
\(b,\) \(\frac{2^{15}.9^4}{6^3.8^3}\)
\(=\frac{2^{15}.3^8}{2^3.3^3.2^9}\)
\(=\frac{2^{15}.3^8}{2^{12}.3^3}\)
\(=2^3.3^5\)
\(=8.243\)
\(=1944\)
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) \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(4.2\right)^{10}+4^{10}}{\left(2^3\right)^4+4^6.4^5}=\frac{4^{10}.2^{10}+4^{10}}{2^{12}+4^6.4^5}=\frac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.2^{10}}=\frac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(1+2^{10}\right)}=\frac{4^{10}}{4^6}=4^4=256\)

........... là gì vậy

trrertre

=5^1+3+5+7+.....+49

Số các số hạng từ 1 đến 49 là (49-1):1+1=50

Tổng các số từ 1 đên 49 là      (49+1).50:2=1250

   =5^1250