lấy vở bồi dưỡng toán ra xem ^^ ko có thì thôi^^

câu 1 ; câu 2 tính ra rồi so sánh

giải:

a) A = \(\frac{10}{27}\) +\(\frac{9}{16}\)+\(\frac{11}{34}\)=1,256 

=> A < 2

b) B = \(\frac{1}{12}\)+\(\frac{1}{13}\)+\(\frac{1}{14}\)+....+\(\frac{1}{22}\)=0,477  ; \(\frac{1}{2}\)=0,5

=> ko tke chúng tỏ vì B < 0,5

hok tốt nha !!!

ta có: \(\frac{10}{27}< \frac{10}{30}=\frac{1}{3}\)

\(\frac{9}{16}< \frac{9}{9}=1\)

\(\frac{11}{34}< \frac{11}{22}=\frac{1}{2}\)

=>A<\(\frac{1}{3}+1+\frac{1}{2}\)<2

vậy A<2

10/27 < 10/15, 9/16 < 9/15, 11/34 < 11/15

=>10/27 + 9/16 + 11/34 < 10/15 + 9/15 + 11/15

=>A < 30/15

=>A < 2 (đpcm)

\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)

\(=\frac{5\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\left(1-\frac{1}{11}+\frac{1}{121}\right)}\)

\(=\frac{5}{8}:\frac{15}{16}\)

\(=\frac{2}{3}\)

2/3

Tk mk nha

\(\frac{5\times\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\times\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}\div\frac{15\times\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\times\left(1-\frac{1}{11}+\frac{1}{121}\right)}=\frac{5}{8}\div\frac{15}{16}=\frac{2}{3}\)

F=5-5x(1/3+1/9-1/27) /8-8x(1/3+1/9-1/27)

: 15-15x(1/11+1/121) /16-16x(1/11+1/121)

=5-5x1/8-8x1

: 15-15x1/16-16x1

=0:0=0

chắc vậy!