Hãy so sánh phân số sau \(A=\frac{19^{30}+4}{19^{31}+4}vàB=\frac{19^{31}+4}{19^{32}+4}\)
So sánh
\(\frac{19^{30}+5}{^{19^{31}}-5}va\frac{19^{31}+5}{19^{32}+5}\)
to lam ko biết là đúng hay sai đây đấy
bỏ hai số 5 nằm ở 2 mẫu số
ta có biểu thức 1
(19^30+5).(19^32)/19^31.19^32
= (19^30+5).(19^31.19)/19^31.19^32
biểu thức 2
(19^31+5).19^31/19^31.19^32
=(19^30+5).(19.19^31)/19^31.19^32
suy ra bằng nhau
So sánh A và B
\(A=\frac{19^{30}+5}{19^{31}+5}\)
\(B=\frac{19^{31}+5}{19^{32}+5}\)
Xét B = \(\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+14}{19^{32}+5+14}=\frac{19^{31}.19}{19^{32}.19}=\frac{19\left(19^{30}+1\right)}{19\left(19^{31}+1\right)}=\frac{19^{30}+1}{19^{31}+1}< \frac{19^{30}+5}{19^{31}+5}=A\)Vậy A > B
Xét B = \(\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+19}{19^{32}+19}=\frac{19\left(19^{30}+1\right)}{19\left(19^{31}+1\right)}=\frac{19^{30}+1}{19^{31}+1}< \frac{19^{30}+5}{19^{31}+5}=A\)
Vậy A > B
So Sánh M và N
\(M=\frac{19^{30}+5}{19^{31}+5}\)và \(N=\frac{19^{31}+5}{19^{32}+5}\)
\(M=\frac{19^{30}+5}{19^{31}+5}\)
\(19M=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5}{19^{31}+5}+\frac{90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)
\(N=\frac{19^{31}+5}{19^{32}+5}\)
\(19N=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5}{19^{32}+5}+\frac{90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)
chung tử rồi so sánh mẫu đi
#)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~#
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
So sánh A và B:
A=\(\frac{19^{30}+5}{19^{31}+5}\) B=\(\frac{19^{31}+5}{19^{32}+5}\)
\(19A=\frac{19^{31}+95}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)
\(19B=\frac{19^{32}+95}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)
Ta thấy \(19A>19B\) nên A > B
Ta có \(A=\frac{19^{30}+5}{19^{31}+5}\)
Suy ra \(19A=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5}{19^{31}+5}+\frac{90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)
Ta có \(B=\frac{19^{31}+5}{19^{32}+5}\)
Suy ra \(19B=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5}{19^{32}+5}+\frac{90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)
Vì \(19^{31}+5< 19^{32}+5\Rightarrow\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\)
Do đó \(19A>19B\Rightarrow A>B\)
Vậy A > B
\(19A=\frac{19\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)
\(19B=\frac{19\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)
Do \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\)
Nên \(1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\)
Hay 19A>19B
Suy ra A>B
Vậy A>B
So sánh M và N biết:
M = \(\frac{19^{30}+5}{19^{31}+5}\); N = \(\frac{19^{31}+5}{19^{32}+5}\)
\(19M=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)
\(19N=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)
Vì \(19^{31}+5< 19^{32}+5\) nên \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\) \(\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\)
Do đó \(M>N\)
Ta có :
\(N=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19.\left(19^{30}+5\right)}{19.\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=M\)
=> N < M
So sánh \(y=\frac{19^{30}+5}{19^{31}^{ }+5}\)và\(y=\frac{19^{31}+5}{19^{32}+5}\)
Bài 1: So sánh lũy thừa
a) 125^80 và 25^125
b) 31^11 và 17^14
c) \(A=\frac{19^{30}+5}{19^{31}+5}vàB=\frac{19^{31+5}}{19^{32}+5}\)
d)\(A=\frac{2^{18}-3}{2^{20}-3}vàB=\frac{2^{20-3}}{2^{22}-3}\)
e) \(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}vàB=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
Bài 2: Cho \(A=1+2+2^2+...+2^{30}\)
Viết A+1 dưới dạng lũy thừa
2) A=1+2+22+...+230=>2A=2+22+23+...+231
=>2A-A=A=(2+22+...+231)-(1+2+22+...+230)=231-1
=>A+1=(231-1)+1=231-(1-1)=231-0=231
lm xog chc'..............................................ặc ặc
so sánh A và B
\(A=\frac{19^{30}+5}{19^{31+5}}\) B=\(\frac{19^{31}+5}{19^{32}+5}\)
Cho hỏi, mẫu ở A là: "1931+5" hay là "1931 + 5"?
so sánh : M = \(\frac{19^{30}+5}{19^{31}+5}\) N = \(\frac{19^{31}+5}{19^{32}+5}\)
M= \(\frac{1}{19}\)
N= \(\frac{1}{19}\)
=> M=N