:))))

?

?

uhm :)))

sao mn lại lm vậy? Ấn nút report thì có j kinh tởm chứ? T lm suốt thấy bth mà

Tung nhỉ? :)))

hi

vt truyện sẽ về 2 con rắn độc trog lp

ơ, @🏅Saitrocchoida💫ッ ko bị bay acc à?

ok

Ta có: \(A=\frac89+\frac{24}{25}+\frac{48}{49}+\cdots+\frac{10200}{10201}\)

\(A=\frac{3^2-1}{3^2}+\frac{5^2-1}{5^2}+\frac{7^2-1}{7^2}+\cdots+\frac{101^2-1}{101^2}\)

\(A=1-\frac{1}{3^2}+1-\frac{1}{5^2}+1-\frac{1}{7^2}+\cdots+1-\frac{1}{101^2}\)

\(A=\left(1+1+1+\cdots+1\right)-\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots+\frac{1}{101^2}\right)\)

\(A=100-\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots+\frac{1}{101^2}\right)\)

Đặt \(B=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots+\frac{1}{101^2}\)

\(B<\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\cdots+\frac{1}{100\cdot102}\)

\(B<\frac12\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+\cdots+\frac{2}{100\cdot102}\right)\)

\(B<\frac12\cdot\left(\frac12-\frac14+\frac14-\frac16+\frac16-\frac18+\cdots+\frac{1}{100}-\frac{1}{102}\right)\)

\(B<\frac12\cdot\left(\frac12-\frac{1}{102}\right)\)

\(B<\frac14-\frac{1}{204}\)

\(\Rightarrow B<\frac14\)

⇒ B < 0,25

⇒ A > 100 - 0,25

⇒ A > 99,75

Vậy A > 99,75 (đpcm)