Question

So Sánh M và N

\(M=\frac{19^{30}+5}{19^{31}+5}\)và \(N=\frac{19^{31}+5}{19^{32}+5}\)

Answer 1

\(M=\frac{19^{30}+5}{19^{31}+5}\)

\(19M=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5}{19^{31}+5}+\frac{90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)

\(N=\frac{19^{31}+5}{19^{32}+5}\)

\(19N=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5}{19^{32}+5}+\frac{90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)

chung tử rồi so sánh mẫu đi

Answer 2

#)Giải :

\(M=\frac{19^{30}+5}{19^{31}+5}\Rightarrow19M=\frac{19\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)

\(N=\frac{19^{31}+5}{19^{32}+5}\Rightarrow19N=\frac{19\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)

Vì \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\Rightarrow19M>19N\Rightarrow M>N\)

              #~Will~be~Pens~#

Answer 3

Ta có : \(N=\frac{19^{31}+5}{19^{32}+5}< 1\)

Áp dụng công thức \(\forall a,b,m\in N;b,m\inℕ^∗\)

\(\Rightarrow\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\)

Ta có :

\(N=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\cdot\left(19^{30}+5\right)}{19\cdot\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=M\)

Vậy N < M

Answer 4

Ta có :

\(M=\frac{19^{30}+5}{19^{31}+5}=\frac{\left(19^{31}+5\right):19+90:19}{19^{31}+5}\)

\(=1:\frac{19+90:19}{19^{31}+5}=1.\frac{19^{31}+5}{19+90:19}\)

\(N=\frac{19^{31}+5}{19^{32}+5}=\frac{\left(19^{32}+5\right):19+90:19}{19^{32}+5}\)

\(=1:\frac{19+90:19}{19^{32}+5}=1.\frac{19^{32}+5}{19+90:19}\)

Mà \(\frac{19^{31}+5}{19+90:19}< \frac{19^{32}+5}{19+90:19}\)

\(\Leftrightarrow M< N\)