\(\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+\frac{64}{63}+\frac{100}{99}\\ =\frac{2.2}{1.3}+\frac{4.4}{3.5}+\frac{6.6}{5.7}+\frac{8.8}{7.9}+\frac{10.10}{9.11}\)

\(\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+\frac{64}{65}+\frac{100}{99}\)

\(1+\frac{1}{3}+1+\frac{1}{15}+1+\frac{1}{35}+1+\frac{1}{65}+1+\frac{1}{99}\)

\(\left(1+1+1+1+1\right)+\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{65}+\frac{1}{99}\right)\)

\(\frac{60}{11}\)

\(\frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\) 

bằng nhau

https://olm.vn/hoi-dap/detail/219583530854.html?pos=498103189513

tham khảo bài này nhé mà hình như là đúng đề bài

\(\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)

\(=5\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(=5\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=5\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=5.\frac{49}{100}\)

\(=\frac{49}{20}\)

bạn copy phép tính đánh vào goolge

bấm enter người ta khác cho biết

mk toàn làm thế thôi

lười mà

21,22 + 9,072 x 10 + 24,72 : 12

= 21,22 + 90,72 + 2,06

= 114

Hok tốt

Từ \(1=a+b+c\Rightarrow1=\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right).\)(bất đẳng thức bunhiacopxki)

\(\Leftrightarrow a^2+b^2+c^2\ge\frac{1}{3}\)(*)

Ta có  : \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)(1)

Dễ thấy \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{b}{b}+\frac{c}{a}+\frac{a}{c}\)

\(\ge3+2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{c}{b}\frac{b}{c}}+2\sqrt{\frac{a}{c}\frac{c}{a}}=3+2+2+2=9\)(bất đẳng thức cô si)

\(Hay:\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\left(do:a+b+c=1\right)\)(2)

Từ (1) và (2) suy ra \(9^2\le\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

\(\Rightarrow\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge27\)(**)

Ta có \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\)

\(=a^2+2+\frac{1}{a^2}+b^2+2+\frac{1}{b^2}+c^2+2+\frac{1}{c^2}\)

\(=\left(a^2+b^2+c^2\right)+\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+6\)

\(\ge\frac{1}{3}+27+6=33+\frac{1}{3}>33\)(theo (*) và (**) )