\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
Vậy \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\left(đpcm\right)\)
CMR:
\(\frac{1}{1.2.3}+\frac{1}{2.3.\text{4}}+\frac{1}{3.\text{4}.5}+...+\frac{1}{98.99.100}=\frac{\text{4}9\text{4}9}{19800}\)
( Câu siêu khó dành cho học sinh đội tuyển Toán )
( Giúp với )
Bài 10 :
b) \(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}\)
Cho \(\frac{x^{\text{4}}}{a}+\frac{y^{\text{4}}}{b}=\frac{1}{a+b};x^2+y^2=1\)
Chứng minh rằng:\(\frac{x^{200\text{4}}}{a^{1002}}+\frac{y^{200\text{4}}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)
\(\text{Chứng minh rằng : }\)\(\forall n\in Z\left(n\ne0,n\ne-1\right)\)\(\text{thì }\)\(Q=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n\left(n+1\right)}\)
\(\text{Không phải là số nguyên}\)
chứng tỏ rằng : \(\frac{\text{1}}{2}\) + \(\frac{\text{1}}{3}\) + \(\frac{\text{1}}{4}\) + ..... + \(\frac{\text{1}}{63}\) > 2
4.\(\left(\frac{1}{4}\right)^2+25.\text{[}\text{ }\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\text{]}:\left(\frac{3}{2}\right)^3\)
\(\frac{\text{1}}{5x8}\) + \(\frac{\text{1}}{8x\text{1}\text{1}}\)+ \(\frac{\text{1}}{\text{1}\text{1}x\text{1}4}\) + ..... + \(\frac{\text{1}}{x\left(x+3\right)}\)= \(\frac{\text{1}0\text{1}}{\text{1}540}\)
1.chứng minh rằng : \(\frac{1}{2}!+\frac{2}{3}!+\frac{3}{4}!+...+\frac{99}{100}!< 1\)
2. Chứng minh rằng :\(\frac{1.2-1}{2}+\frac{2.3-1}{3}+\frac{3.4-1}{4}+...+\frac{99.100-1}{100}< 2\)
\(\frac{\text{3}}{\text{4}}+\frac{\text{1}}{\text{4}}:x=\frac{\text{2}}{5}\)
biet x thuoc Q
cac ban giup mik nhe cam on nhieu :3
