Ta có: \(\frac{1}{1^2}=\frac{1}{1\cdot1};\frac{1}{2^2}<\frac{1}{1\cdot2};...;\frac{1}{50^2}<\frac{1}{49\cdot50}\)

=>\(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}<1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}=1+1-\frac{1}{50}=2-\frac{1}{50}=1,98\)

hay A<1,98 mà 1,98<2 nên A<2

Vậy A<2

M=1+1/2^2+1/3^2+1/4^2+...+1/10^2>1+1/2*3+1/3*4+1/4^5+...+1/10*11

                             M>1+1/2-1/3+1/4-1/4+1/5-...-1/11

                            M>1+1/2-1/11

                              M>1+9/22

                               M>31/22

                                vì 31/22>4/3 nên M>4/3

khúc đằng trước hỉu j chết liền

lên olm đăg thử đi 

\(\dfrac{1}{k^2}<\dfrac{1}{k(k-1)}=\dfrac{1}{k-1}-\dfrac{1}{k}\)

Ap dung:

\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\ldots+\dfrac{1}{n^2}<1+\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\ldots+\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)=2-\dfrac{1}{n}<2\)

Không chép lại đề nhé

Ta có:

P=\(\frac{50-49}{49}+\frac{50-48}{48}+...+\frac{50-2}{2}+\frac{50-1}{1}\)

P=\(\frac{50}{49}-\frac{49}{49}+\frac{50}{48}-\frac{48}{48}+...+\frac{50}{2}-\frac{2}{2}+\frac{50}{1}-\frac{1}{1}\)

P=\(\left(\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\right)+\frac{50}{1}-\left(\frac{49}{49}+\frac{48}{48}+...+\frac{2}{2}+\frac{1}{1}\right)\)

P=\(50\cdot\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)+50-49\)                 (chỗ này gộp nha)

P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{48}+\frac{1}{49}\right)+1\)

P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)+\frac{50}{50}\)

P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)\)

=>P=50S

=>\(\frac{S}{P}=\frac{S}{50S}=\frac{1}{50}\)

Vừa nãy mình nói nhầm, Sorry.

Tích nha

Ta có : \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{100^2}<\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{100}<1\)

Mà \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}<1\) nên A không phải số tự nhiên

nhin la biet ko phai so tu nhien

Câu 1 :\(P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{100}=\frac{1}{100}\)

like mình làm hết

bình phương 2 vế của 1/a + 1/b +1/c =2 ta đk:

1/a^2 +1/b^2 + 1/c^2 + 2 x (a+b+c) / abc =4

1/a^2 + 1/b^2 + 1/c^2 +2 =4

=> 1/a^2 + 1/b^2 + 1/c^2 =2

 olm đã có