Đáp án + Giải thích các bước giải:
Ta có :
`B=\frac{3^{9}-2^{3}.3^{7}+2^{10}.3^{2}-2^{13}}{3^{10}-2^{2}.3^{7}+2^{10}.3^{3}-2^{12}}`
`→B=\frac{(3^{9}-2^{3}.3^{7})+(2^{10}.3^{2}-2^{13})}{(3^{10}-2^{2}.3^{7})+(2^{10}.3^{3}-2^{12})}`
`→B=\frac{3^{7}(3^{2}-2^{3})+2^{10}(3^{2}-2^{3})}{3^{7}(3^{3}-2^{2})+2^{10}(3^{3}-2^{2})}`
`→B=\frac{(3^{2}-2^{3})(3^{7}+2^{10})}{(3^{3}-2^{2})(3^{7}+2^{10})}`
`→B=\frac{3^{2}-2^{3}}{3^{3}-2^{2}}`
`→B=\frac{9-8}{27-4}`
`→B=\frac{1}{23}`
Vậy `B=\frac{1}{23}`
