Chứng tỏ rằng:A=\(\frac{10}{27}\)+\(\frac{9}{16}\) +\(\frac{11}{34}\) <2
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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)
\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)
\(=\frac{5\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\left(1-\frac{1}{11}+\frac{1}{121}\right)}\)
\(=\frac{5}{8}:\frac{15}{16}\)
\(=\frac{2}{3}\)
\(\frac{5\times\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\times\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}\div\frac{15\times\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\times\left(1-\frac{1}{11}+\frac{1}{121}\right)}=\frac{5}{8}\div\frac{15}{16}=\frac{2}{3}\)
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!