\(\frac{7}{9}\cdot\frac{8}{5}-\frac{7}{9}\cdot\frac{3}{5}\)

\(=\frac{7}{9}\left(\frac{8}{5}-\frac{3}{5}\right)\)

\(=\frac{7}{9}\cdot\frac{5}{5}\)

\(=\frac{7}{9}\cdot1=\frac{7}{9}\)

ra \(\frac{7}{9}\)nha

7/9 * 8/5 - 7/9 * 3/5 

= 7/9 * ( 8/5 - 3/5 ) 

= 7/9 * 1 

= 7/9 

^^ Ủng hộ nha!!!

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)

\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{2015}:2\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2013}{4030}\)

\(\frac{1}{x+1}=\frac{1}{2015}\)

=>x+1=2015

=>x=2014

=2014 k mk đi

\(x=2014\)

hình như là sai đề

Áp dụng tc của dãy tỉ số bằng nhau ta có

\(\frac{x_1-1}{10}=.....=\frac{x_{10}-10}{1}=\frac{\left(x_1+x_2+....+x_{10}\right)-\left(1+2+3+...+10\right)}{1+2+3+...+10}\)

\(=\frac{45}{55}=\frac{9}{11}\)

Giải ra ta được

\(x_1=\frac{101}{11}\)

\(x_2=\frac{103}{11}\)

........

\(x_{10}=\frac{119}{11}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:: \(\frac{x_1-1}{10}=\frac{x_2-2}{9}=...=\frac{x_9-9}{2}=\frac{x_{10}-10}{1}=\frac{x_1-1+x_2-2+...+x_9-9+x_{10}-10}{10+9+...+2+1}\)

\(=\frac{\left(x_1+x_2+..+x_9+x_{10}\right)-\left(1+2+...+9+10\right)}{10+9+...+2+1}=\frac{100-55}{55}=\frac{9}{11}\)

=> \(\frac{x_1-1}{10}=\frac{9}{11}\Leftrightarrow x_1-1=\frac{10\cdot9}{11}\Leftrightarrow x_1=\frac{90}{11}+1=\frac{101}{11}\)

     \(\frac{x_2-2}{9}=\frac{9}{11}\Leftrightarrow x_2-2=\frac{9\cdot9}{11}\Leftrightarrow x_2=\frac{81}{11}+2=\frac{103}{11}\)

Tương tự ta cũng có:

     \(x_3=\frac{105}{11}\)

    \(x_4=\frac{107}{11}\)

    \(x_5=\frac{109}{11}\)

     \(x_6=\frac{111}{11}\)

    \(x_7=\frac{113}{11}\)

    \(x_8=\frac{115}{11}\)

    \(x_9=\frac{117}{11}\)

   \(x_{10}=\frac{119}{11}\)

\(-\frac{1}{2}+\frac{7}{6}+\left(-\frac{7}{8}\right)=-\frac{1}{2}+\frac{7}{6}-\frac{7}{8}\)\(\frac{7}{8}=\frac{-12}{24}+\frac{28}{24}-\frac{21}{24}\)=\(\frac{-12+28-21}{24}=\frac{-5}{24}\)

LHL mik đi, mik tl rồi, bảo đảm đúng

\(-\frac{1}{2}+\frac{7}{6}+\left(-\frac{7}{8}\right)=\frac{-4}{8}+\frac{7}{6}+\left(\frac{-7}{8}\right)=\frac{-4}{8}+\frac{-7}{8}+\frac{7}{6}=\frac{-11}{8}+\frac{7}{6}=\frac{-33}{24}+\frac{28}{24}=\frac{-5}{24}\)

a, B=[(x+3)/(x-3)+(2x^2-6)/(9-x^2)+x/(x+3)]:[(6x-12)/(2x^2-18)]

=[(x+3)/(x-3)+ -(2x^2-6)/(x^2-9)+x/(x+3)]:[(6x-12)/(2x^2-18)]

=[(x+3)/(x-3)+ -(2x^2-6)/(x-3)(x+3)+x/(x+3)]:[(6x-12)/2(x-3)(x+3)]

={[(x+3)^2-2x^2+6+x(x-3)]/(x-3)(x+3)}:[6(x-2)/2(x-3)(x+3)]

=(x^2+6x+9-2x^2+6+x^2-3x)/(x-3)(x+3): 6(x-2)/2(x-3)(x+3)

=3x+15/(x-3)(x+3): 6(x-2)/2(x-3)(x+3)

=3(x+5)/(x-3)(x+3): 6(x-2)/2(x-3)(x+3

=3(x+5)/(x-3)(x+3).2(x-3)(x+3)/6(x-2)

=3(x+5).6/(x-2)

=6(x+5)/6(x-2)

=x+5/x-2

b,Ta thay : x=1

=>x+5/x-2=1+5/1-2=-6

Ta thay : x=-3

=>x+5/x-2=-3+5/-3-2=-2/5

c, Ta co : x+5/x-2=0

x+5=(x-2).0

x+5=0

x=-5

Vậy : x=-5

ta có \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}=\frac{3}{5}\)

và \(\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}+\frac{5}{8}-\frac{5}{6}}=\frac{2\left(\frac{1}{2.2}-\frac{1}{3.2}+\frac{1}{4.2}\right)}{5\left(\frac{1}{4}+\frac{1}{8}-\frac{1}{6}\right)}=\frac{2\left(\frac{1}{4}+\frac{1}{8}-\frac{1}{6}\right)}{5\left(\frac{1}{4}+\frac{1}{8}-\frac{1}{6}\right)}=\frac{2}{5}\)

Vậy \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}+\frac{5}{8}-\frac{5}{6}}=\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)

ĐS: 1

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