Ta có : a//b 

=>góc A₃+góc B₂=180^o(2 góc trong cùng phía)

Mà góc B₂=góc B₄ nên: góc A₃+góc B₄=180^o

Mặt khác góc A₃- góc B₄=80 (giả thiết)

Cộng vế theo vế 2 đẳng thức trên ta được : 2 góc A₃=260^o

=>góc A₃=130^o => - góc A₁=130^o(đối đỉnh) 

                                  -góc A₂=180^o-góc A₃=180^o-130^o=50^o(kề bù )

                                  -góc A₄=góc A₂=50^o(đối đỉnh )

=> góc B₄=-80^o+góc A₃=130^o-80^o=50^o

                               => -góc B₂=góc B₄ =50^o(đối đỉnh)

                                    -góc B₃=180^o-góc B₄=180^o-50^o=130^o(kề bù)

                                   -góc B₁=góc B₃=130^o(đối đỉnh)

\(G=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2=\left(\frac{\left(n+1\right).n}{2}\right)^2=\frac{\left(n+1\right)^2.n^2}{4}\)

kẻ Oz // Nb // Ma

*Oz//Nb

=>góc bNy = góc zOy (2 góc đồng vị)

Mà góc bNy=35^o nên: góc zOy=35^o

*Oz//Ma

=>góc xMa = góc xOz( 2 góc đông vị)

Mà góc xMa=50^o nên: góc xOz=50^o

Suy ra: góc xOy= goc zOy+ xOz=35^o+50^o=85^o

\(\frac{2x+7}{4}=\frac{3-5y}{7}=\frac{2x-5y}{9}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{2x+7}{4}=\frac{3-5y}{7}=\frac{2x-5y}{9}=\frac{\left(2x+7\right)+\left(3-5y\right)-\left(2x-5y\right)}{4+7-9}\)

\(=\frac{2x+7+3-5y-2x+5y}{2}=\frac{10}{2}=5\)

Suy ra:\(\frac{2x+7}{4}=5\Rightarrow2x+7=20\Rightarrow x=\frac{13}{2}\)

\(\frac{3-5y}{7}=5\Rightarrow3-5y=35\Rightarrow x=-\frac{32}{5}\)

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (x khác 0; khác -3)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

<=>\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

<=>\(\frac{1}{x+3}=\frac{1}{308}\)

=>x+3=308

<=>x=305 (nhận)

Vậy x=305

\(I=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right)\left(2n+3\right)}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right)\left(2n+3\right)}\right)\)

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

\(=\frac{2n+3}{2n+3}-\frac{1}{2n+3}=\frac{2n+2}{2n+3}\)

*\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left[\frac{1}{14}+\frac{1}{7}-\left(-\frac{3}{35}\right)\right].\frac{4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right).\frac{4}{3}}=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{4}{3}}=\frac{\frac{-1}{12}}{\frac{14}{35}}=-\frac{1}{12}.\frac{35}{14}=\frac{-35}{168}\)

*\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{63}{10}.12-21.\frac{18}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{378}{5}-\frac{378}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)

thấy ngưởi trên ko

ae đọc hỉu ko

* B=6a+2b=13a-5b-7a+7b=(13a-5b)-7.(a-b)=A-7.(a-b)

Vì A chia hết cho 7 ; 7.(a-b) chia hết cho 7 nên:

B chia hết cho 7

*A=13a-5b=6a+2b+7a-7b=B+7.(a-b)

Vì B chia hết cho 7; 7(a-b) chia hết cho 7

Nên: A chia hết cho 7