so sánh a , M = 33 . 10^3 / 23 . 5 . 10^3 + 7000 và N = 3774/5214 b , B = 4/7 + 5 + 3/7^2 + 5/7^3 + 6/7^4 và Q = 5/7^3 + 6/7^2 + 5/7^4 + 4/7 + 5

0 bình luận về “so sánh a , M = 33 . 10^3 / 23 . 5 . 10^3 + 7000 và N = 3774/5214 b , B = 4/7 + 5 + 3/7^2 + 5/7^3 + 6/7^4 và Q = 5/7^3 + 6/7^2 + 5/7^4 + 4/7 + 5”

1. Giải thích các bước giải:

   a.Ta có:

   $M=\dfrac{33.10^3}{23.5.10^3+7000}$

   $\to M=\dfrac{33.10^3}{115.10^3+7.10^3}$

   $\to M=\dfrac{33.10^3}{(115+7).10^3}$

   $\to M=\dfrac{33.10^3}{122.10^3}$

   $\to M=\dfrac{33}{122}<\dfrac{33}{66}=\dfrac12$

   Ta có:

   $N=\dfrac{3773}{5214}>\dfrac{2607}{5214}=\dfrac12$

   $\to M<\dfrac12<N$

   $\to M<N$

   b.Ta có:

   $B=\dfrac47+5+\dfrac3{7^2}+\dfrac5{7^3}+\dfrac6{7^4}$

   $\to B=\dfrac6{7^4}+\dfrac5{7^3}+\dfrac3{7^2}+\dfrac47+5$

   $Q=\dfrac{5}{7^3}+\dfrac{6}{7^2}+\dfrac{5}{7^4}+\dfrac47+5$

   $\to Q=\dfrac{5}{7^4}+\dfrac{5}{7^3}+\dfrac{6}{7^2}+\dfrac47+5$

   $\to B-Q=\dfrac{1}{7^4}-\dfrac{3}{7^2}$

   $\to B-Q<\dfrac{1}{7^2}-\dfrac{3}{7^2}$

   $\to B-Q<\dfrac{3}{7^2}-\dfrac{3}{7^2}$

   $\to B-Q<0$

   $\to B<Q$

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