$\begin{array}{l}a)\\\text{- Ta có :}\\\quad\dfrac{4.5+4.11}{8.7+4.3}\\=\dfrac{4.(5+11)}{4.14+4.3}\\=\dfrac{4.16}{4.(14+3)}\\=\dfrac{16}{14+3}\\=\dfrac{16}{17}\\\quad\dfrac{(-15).8+10.7}{5.6+20.3}\\=\dfrac{-3.5.8+5.2.7}{5.6+5.4.3}\\=\dfrac{-24.5+5.14}{5.6+5.12}\\=\dfrac{5.(-24+14)}{5.(6+12)}\\=\dfrac{5.(-10)}{5.18}\\=\dfrac{5.2.(-5)}{5.2.9}\\=\dfrac{-5}9\\\quad\dfrac{2^4.5^2.7}{2^3.5.7^2.11}\\=\dfrac{2.5}{7.11}\\=\dfrac{10}{77}\\\text{- Ta lại có : $\rm BCNN(17,9,77)= 11781$}\\\to\begin{cases}\dfrac{16}{17}=\dfrac{16.693}{17.693}=\dfrac{11088}{11781}\\\dfrac{-5}9=\dfrac{-5.1309}{9.1309}=\dfrac{-6545}{11781}\\\dfrac{10}{17}=\dfrac{10.153}{17.153}=\dfrac{1530}{11781}\end{cases}\\\,\\b)\\\text{- Ta có :}\\\quad\dfrac{-5^2-5.3^2}{5^2+5^2.3^2}\\=\dfrac{-(5^2+5.3^2)}{5^2.(1+3^2)}\\=\dfrac{-[5.(5+3^2)]}{5^2.10}\\=\dfrac{-5.14}{5^2.10}\\=\dfrac{-14.5}{5^2.10}\\=\dfrac{-14}{5.10}\\=\dfrac{-14}{50}\\=\dfrac{-7}{25}\\\quad\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\\=\dfrac{(2^2)^6.(3^2)^5+(2.3)^9.2^3.3.5}{(2^3)^4.3^{12}-(2.3)^{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}.(2.3-1)}\\=\dfrac{2^{12}.3^{10}.(1+5)}{2^{11}.3^{11}.5}\\=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\\=\dfrac{2.6}{3.5}\\=\dfrac{12}{15}\\=\dfrac45\\\quad\dfrac{2929-101}{2.1919+404}\\=\dfrac{29.101-101}{2.19.101+4.101}\\=\dfrac{101.(29-1)}{38.101+4.101}\\=\dfrac{101.28}{101.(38+4)}\\=\dfrac{101.28}{101.42}\\=\dfrac{28}{42}\\=\dfrac23\\\text{- Ta lại có : $\rm BCNN(25,5,3)=75$}\\\to\begin{cases}\dfrac{-7}{25}=\dfrac{-7.3}{25.3}=\dfrac{-21}{75}\\\dfrac45=\dfrac{4.15}{5.15}=\dfrac{60}{75}\\\dfrac23=\dfrac{2.25}{3.25}=\dfrac{50}{75}\end{cases} \end{array}$
