\(\frac{5}{9}-\frac{7}{13}-\frac{5}{9}.\frac{3}{13}+\frac{-5}{9}.\frac{6}{9}\)

= \(\frac{5}{9}-\frac{7}{13}-\frac{5}{3}.\frac{1}{13}+\frac{-5}{3}.\frac{2}{9}\)

=\(\frac{5}{9}-\frac{7}{13}-\frac{5}{39}+\frac{-10}{27}\)

=\(\frac{5}{9}+\frac{-7}{13}+\frac{-5}{39}+\frac{-10}{27}\)

=\(\frac{-13}{27}\)

Chúc bạn học tốt nhea ♥♥♥

\(\frac{5}{9}-\frac{7}{13}-\frac{5}{9}.\frac{3}{13}+\frac{-5}{9}.\frac{61}{9}\)

\(=\frac{5}{9}\left(1-\frac{3}{13}\right)-\frac{7}{13}-\frac{5}{9}.\frac{61}{9}\)

\(=\frac{5}{9}.\frac{10}{13}-\frac{5}{9}.\frac{61}{9}-\frac{7}{13}\)

\(=\frac{5}{9}\left(\frac{10}{13}-\frac{61}{9}\right)-\frac{7}{13}\)\(=-\frac{314}{81}\)

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

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

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

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\) ta có : 

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\right)\)

\(A=1-\frac{1}{2^{2018}}< 1\)

Vậy \(A< 1\) ( đpcm ) 

Chúc bạn học tốt ~ 

Đặt A=\(\left(-1\right)\cdot\left(-1\right)^2\cdot\left(-1\right)^3\cdot\left(-1\right)^4.....\left(-1\right)^{2000}\cdot\left(-1\right)^{2001}\)

\(A=\left(-1\right)\cdot1\cdot\left(-1\right)\cdot1\cdot\cdot\cdot\cdot\cdot\cdot1\cdot\left(-1\right)\)

\(A=-1\)

\(\left(-1\right)\cdot\left(-1^2\right)\cdot\left(-1^3\right)\cdot\left(-1^4\right)....\left(-1\right)^{2000}\cdot\left(-1\right)^{2001}\)

\(=\left(-1\right)\cdot1\cdot\left(-1\right)\cdot1....1\cdot\left(-1\right)\)

\(=\left(-1\right)\)

a) = 3/3 x ( -24/54 +45/54 ) : 7/12

   = 1 x 21/54 x 12/7

   = 18/27 

( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )

a, -7/3 × 4/9 + -7/3 × 5/9                                 b thì mình chưa làm ra

= -7/3 × (4/9 + 5/9)

= -7/3 × 1

= -7/3

\(\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)

\(=>\frac{1}{2}x-1+\frac{3}{5}\left(x-2\right)=2\)

\(=>\frac{1}{2}\left(x-2\right)+\frac{3}{5}\left(x-2\right)=2\)

\(=>\left(x-2\right)\left(\frac{1}{2}+\frac{3}{5}\right)=2\)

\(=>\left(x-2\right)\frac{11}{10}=2\)

\(=>x-2=\frac{20}{11}\)

\(=>x=\frac{20}{11}+2\)

\(=>x=\frac{42}{11}\)

Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có : 

\(A< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

Vậy \(A< \frac{1}{2}\) ( đpcm ) 

Chúc bạn học tốt ~ 

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