Chưng minh rằng:\(\frac{23}{34}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{70}<\frac{4}{3}\)
cho abc chia hết cho 23 .chưng minh rằng 30a+3b-2cchia hết cho 23
ta có: 23a + 23b chia hết cho 23
=> 7a + 3b + 16a + 20b chia hết cho 23
=> 7a + 3b + 4(4a + 5b) chia hết cho 23
do 7a + 3b chia hết cho 23 nên 4(4a + 5b) chia hết cho 23
mà 4 không chia hết cho 23 nên 4a + 5b phải chia hết cho 23
Chứng minh rằng :\(\frac{23}{34}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{70}<\frac{4}{3}\)
Chứng minh rằng 23! , 29 + 34! chia hết 23
Chưng minh rằng :\(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{100^2}<1\)
Chưng minh rằng:
\(\frac{2002}{\sqrt{2003}}+\frac{2003}{\sqrt{2002}}>\sqrt{2002}+\sqrt{2003}\)
Chưng minh rằng:\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{100}}>10\)
Ta có:
\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(......................\)
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
Cộng theo vế ta có: \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{99}{10}\)
\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{99}{10}+\frac{1}{10}=\frac{100}{10}=10\)
cho x;y>0 chưng minh rằng \(\frac{x}{y}+\frac{y}{x}\ge2\)
áp dụng bđt cosi với 2 số x,y>0 ta có: \(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\)=> đpcm
Có \(\frac{x}{y}+\frac{y}{x}=\frac{x^2+y^2}{xy}\ge\frac{1+1}{1}=2\)2
Chưng Minh:\(\frac{1}{21}\)+\(\frac{1}{22}\)+\(\frac{1}{23}\)+....+\(\frac{1}{49}\)+\(\frac{1}{50}\)<\(1\frac{1}{12}\)
(1/21+1/22+...+1/30)+(1/31+...+1/40)+(1/41+...+1/50)
(1/21+1/22+...+1/30)<1/20+..+1/20=1/20*10=1/2
(1/31+...+1/40)<1/30+..+1/30=1/30*10=1/3
(1/41+...+1/50)<1/40+...+1/40=1/40*10=1/4
Suy ra day so <1/2+1/3+1/4=13/12=1/1/12=>dpcm
k cho minh nhe
Chứng tỏ rằng :\(\frac{23}{34}<\frac{1}{31}+\frac{1}{32}+...+\frac{1}{70}<\frac{4}{3}\)