\(\frac{3^4\cdot5-3^6}{3^4\cdot13+3^4}\)

\(=\frac{3^4\cdot\left(5-3^2\right)}{3^4\cdot\left(13+1\right)}\)

\(=\frac{5-9}{13+1}\)

\(=-\frac{4}{14}=-\frac{2}{7}\)

\(\frac{3^4.5-3^6}{3^4.13+3^4}=\frac{3^4.5-3^4.9}{3^4\left(13+1\right)}=\frac{3^4\left(5-9\right)}{3^4.14}=\frac{3^4.\left(-4\right)}{3^4.14}=\frac{-4}{14}=\frac{-2}{7}\)

\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

\(\Rightarrow\frac{2^{12}.\left(13+65\right)}{2^{10}.104}+\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)

\(\Rightarrow\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.2^4}{3^9.2^4}\)

\(=\frac{2^2.3}{4}+3\)

\(=3+3=6\)

1). Sx:-3/4 ; 1/-4 ; 2/5 ; 4/9

2). \(\frac{2.5.13}{26.35}+\frac{24.5-24}{4-28}+\frac{123.6+123.4}{3-126}=\frac{2.5.13}{2.13.5.7}+\frac{24.\left(5-1\right)}{4.\left(1-7\right)}+\frac{123.\left(6+4\right)}{-123}=\frac{1}{7}+4-10=\frac{1}{7}+\frac{42}{7}=\frac{43}{7}\) 

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\frac{100}{100}-\frac{1}{x+1}=\frac{99}{100}\)

\(\frac{1}{x+1}=\frac{1}{100}\)

\(\Rightarrow x+1=100\)

\(x=99\)

x=99 nha ban ! ai k minh se tk lai !

\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(=\frac{1}{1.3}-\frac{1}{11.13}\)

\(=\frac{1}{3}-\frac{1}{143}\)

\(=\frac{140}{429}\)

a,\(\frac{4.7}{9.32}=\frac{7}{9.8}=\frac{7}{72}\)

a) \(\frac{4.7}{9.32}\)=\(\frac{28}{288}\)=\(\frac{7}{72}\)

b)\(\frac{3.21}{14.15}\)=\(\frac{63}{210}\)=\(\frac{3}{10}\)

c)\(\frac{2.5.13}{26.35}\)=\(\frac{130}{910}\)=\(\frac{1}{7}\)

d)\(\frac{9.6-9.3}{18}\)=\(\frac{27}{18}\)=\(\frac{3}{2}\)

e)\(\frac{17.5-17}{3-20}\)=\(\frac{68}{-17}\)=\(-4\)

f)\(\frac{49+7.49}{49}\)=\(\frac{392}{49}\)=\(8\)

\(a,\frac{4.7}{9.32}=\frac{4.7}{9.4.8}=\frac{7}{9.8}=\frac{7}{72}\)

\(b,\frac{3.21}{14.15}=\frac{3.3.7}{2.7.3.5}=\frac{3}{2.5}=\frac{3}{10}\)

\(c,\frac{2.5.13}{26.35}=\frac{2.5.13}{13.2.5.7}=\frac{1}{7}\)

\(d,\frac{9.6-9.3}{18}=\frac{9.\left(6-3\right)}{18}=\frac{9.3}{18}=\frac{9.3}{2.9}=\frac{3}{2}\)

\(e,\)\(\frac{17.5-17}{3-20}=\frac{17.\left(5-1\right)}{-17}=\frac{17.4}{-17}=\frac{-17.\left(-4\right)}{-17}=-4\)

\(f,\frac{49+7.49}{49}=\frac{49.\left(1+7\right)}{49}=\frac{49.8}{49}=8\)

ta có: \(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(2^5-3\right)}=\frac{8}{2^5-3}=\frac{8}{29}=\frac{104}{377}\)

\(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}=\frac{3^{10}.5}{3^8.13}=\frac{3^2.5}{13}=\frac{45}{13}=\frac{1305}{377}\)

\(\Rightarrow\frac{104}{377}< \frac{1305}{377}\Rightarrow\frac{2^5.7+2^5}{2^5.2^5-2^5.3}< \frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)

Ta cứ tính ra tử số và mỗi số của từng phân số ra nhé Jerry Gaming:

\(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}\)= \(\frac{2^5.\left(7+1\right)}{2^5.\left(2^5-3\right)}=\frac{2^5.8}{2^5.\left(32-3\right)}=\frac{32.8}{2^5.29}=\frac{32.8}{32.29}=\frac{8}{29}\)

\(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)= \(\frac{3^4.5.3^6}{3^8.13}=\frac{3^{10}.5}{3^8.13}=\frac{3^2.5}{13}=\frac{9.5}{13}=\frac{45}{13}\)

\(\frac{8}{29}\)và \(\frac{45}{13}\)MSC: 377

Ta có:

\(\frac{8}{29}=\frac{8.13}{29.13}=\frac{104}{377}\)

\(\frac{45}{13}=\frac{45.29}{13.29}=\frac{1305}{377}\)

Vậy quy đồng \(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}\)và \(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)ta được \(\frac{104}{377}\)và \(\frac{1305}{377}\)

Chúc bạn học tốt!

\(=\frac{2^3.3^3.5^3.2^3.7}{3.2^4.5^3.2.7}=\frac{2^6.5^3.3^3}{3.2^5.5^3}=2.3^2=2.9=18\)

=\(\frac{2^3\cdot3^3\cdot5^3\cdot2^3\cdot7}{3\cdot2^4\cdot5^3\cdot2\cdot7}=\frac{2^{3_{ }}\cdot3^3\cdot2^3}{3\cdot2^4\cdot2}=\frac{1728}{96}=\frac{18}{1}=18\)

2 câu trên câu nào cũng đúng. nhưng làm theo cách 2 thì nhanh hơn