>11 điểm thì các bạn ấy mới nhận 3 T.I.C.K bạn ạ

Câu trả lời A > B

Cách chứng minh đề sai : Số số phân số là 

(10101-3):5+1=\(\frac{10103}{5}\)

không biết KQ  OK

Câu hỏi của Lê Tiến Cường - Toán lớp 6 - Học toán với OnlineMath

\(A=\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{10101}{10100}=\frac{2+1}{2}+\frac{6+1}{6}+\frac{12+1}{12}+...+\frac{10100+1}{10100}\)

\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{10100}\right)\)

\(A=\left(1+\frac{1}{1\times2}\right)+\left(1+\frac{1}{2\times3}\right)+\left(1+\frac{1}{3\times4}\right)+...+\left(1+\frac{1}{100\times101}\right)\)

\(A=\left(1+1+1+....+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{100\times101}\right)\)

\(A=100+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\right)\)

\(A=100+1-\frac{1}{101}=101-\frac{1}{101}< 101=B\)

\(\Rightarrow A< B\)

So easy

a: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{100\cdot101}\)

=1-1/2+1/2-1/3+...+1/100-1/101

=1-1/101=100/101

b: \(A=1+\dfrac{1}{2}+1+\dfrac{1}{6}+1+\dfrac{1}{12}+...+1+\dfrac{1}{10100}\)

\(=100+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)

\(=101-\dfrac{1}{101}< 101\)

ta có: 2 = 1 x 2

6 = 2 x 3

12 = 3 x 4

...

10100 = 100 x 101

=> Số số hạng của dãy 2;6;12;...;10100 là: ( 101 -1) : 1 = 100 ( số hạng)

ta có: \(A=\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{10101}{10100}\) 

\(A=1+\frac{1}{2}+1+\frac{1}{6}+1+\frac{1}{12}+...+1+\frac{1}{10100}\)

\(A=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\right)\) ( có 100 số 1)

\(A=100+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)

\(A=100+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(A=100+\left(1-\frac{1}{101}\right)\)

\(A=100+1-\frac{1}{101}=101-\frac{1}{101}< 101\)

=> A < B

a) \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{100x101}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

a: \(\frac{10^{99}+5}{10^{99}-8}=\frac{10^{99}-8+13}{10^{99}-8}=1+\frac{13}{10^{99}-8}\)

\(\frac{10^{100}+6}{10^{100}-7}=\frac{10^{100}-7+13}{10^{100}-7}=1+\frac{13}{10^{100}-7}\)

Ta có: \(10^{99}-8-\left(10^{100}-7\right)\)

\(=10^{99}-10^{100}-8+7=10^{99}\left(1-10\right)-1=-9\cdot10^{99}-1<0\)

=>\(10^{99}-8<10^{100}-7\)

=>\(\frac{13}{10^{99}-8}>\frac{13}{10^{100}-7}\)

=>\(\frac{13}{10^{99}-8}+1>\frac{13}{10^{100}-7}+1\)

=>\(\frac{10^{99}+5}{10^{99}-8}>\frac{10^{100}+6}{10^{100}-7}\)

b: Đặt \(A=\frac{10^{1010}+1}{10^{1011}+1};B=\frac{10^{1011}-4}{10^{1012}-4}\)

Ta có: \(10A=\frac{10^{1011}+10}{10^{1011}+1}=1+\frac{9}{10^{1011}+1}\)

\(10B=\frac{10^{1012}-40}{10^{1012}-4}=1-\frac{36}{10^{1012}-4}\)

mà \(\frac{9}{10^{1011}+1}>\frac{-36}{10^{1012}-4}\)

nên 10A>10B

=>A>B

\(7A=7+7^2+....+7^{101}\)

\(7A-A=\left(7-7\right)+\left(7^2-7^2\right)+......+\left(7^{100}-7^{100}\right)+7^{101}-1\)

\(A=\frac{7^{101}-1}{6}\)

Vậy Biểu thức A = B = \(\frac{7^{101}-1}{6}\)

no

A = 101 x 50

B = 50 x 49 + 53 x 50 = 50 x (49+53) = 50 x 102

Có 101<102 => 50 x 101 < 50 x 102

=> A<B