vậy \(x=2\)\(x=2\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{9}\)\(\left(2\dfrac{2}{9}-x\right)\)=\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)

\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{8\times9}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

thế học sinh lớp 6 có giải đc hông đẻ tui còn giải hộ

\(\dfrac{1}{6}x+\dfrac{1}{12}x+\dfrac{1}{20}x+...+\dfrac{1}{2450}x=1\)

\(x\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2450}\right)\)=1

\(x\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{49\times50}\right)\)=1

\(x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\)

\(x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\)

\(x\times\)\(\dfrac{12}{25}=1\)

\(\Rightarrow x=1\div\dfrac{12}{25}\)

\(x=1\times\dfrac{25}{12}=\dfrac{25}{12}\)

vậy \(x=\dfrac{25}{12}\)

A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)

2A=2(\(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\))

2A= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

2A-A=1-\(\dfrac{1}{2^{10}}\)=\(\dfrac{1023}{1024}\)

A=\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)

A=\(\dfrac{1}{1\times1}+\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+...+\dfrac{1}{100\times100}\)

<\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{100\times101}\)

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

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

vì \(\dfrac{100}{101}< 1\Rightarrow A< 1\)

a) Số học sinh lớp 6A là :

\(12\div30\%=40\)(học sinh)

b) Số học sinh khá lớp đó là :

40\(\times\)55%=22(học sinh)

bạn tự đáp số nhé !!!

cho mk với!!! mk đang cần