a, Đi qua gốc tọa độ

\(\Rightarrow O\left(0;0\right)\) 

Thay \(x=0;y=0\)

\(\Rightarrow0=\left(2k-1\right).0+k\\ \Rightarrow k=0\left(t/m\right)\)

\(b,\)  Cắt trục hoành tại điểm có hoành độ bằng 3 

\(\Rightarrow\) đi qua điểm \(\left(3;0\right)\)

Thay \(x=3;y=0\)

\(\Rightarrow\left(2k-1\right).3+k=0\\\Rightarrow6k-3+k=0\\ \Rightarrow7k=3\\ \Rightarrow k=\dfrac{3}{7}\left(t/m\right)\)

\(c,\) Song song với đường thẳng \(y=\dfrac{3}{5}x+4\)

\(\Rightarrow\left\{{}\begin{matrix}a=a'\\b\ne b'\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2k-1=\dfrac{3}{5}\\k\ne4\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}k=\dfrac{4}{5}\left(t/m\right)\\k\ne4\end{matrix}\right.\)

\(a,\dfrac{3x}{2x+4}=\dfrac{3x}{2\left(x+2\right)}=\dfrac{3x\left(x-2\right)}{2\left(x^2-4\right)}=\dfrac{3x^2-6x}{2\left(x^2-4\right)}\)

\(\dfrac{x+3}{x^2-4}=\dfrac{2\left(x+3\right)}{2\left(x^2-4\right)}=\dfrac{2x+6}{2\left(x^2-4\right)}\)

\(b,\dfrac{x+5}{x^2+4x+4}=\dfrac{x+5}{\left(x+2\right)^2}=\dfrac{3\left(x+5\right)}{3\left(x+2\right)^2}=\dfrac{3x+15}{3\left(x+2\right)^2}\\ \dfrac{x}{3x+6}=\dfrac{x}{3\left(x+2\right)}=\dfrac{x\left(x+2\right)}{3\left(x+2\right)^2}=\dfrac{x^2+2x}{3\left(x+2\right)^2}\)

\(a,\dfrac{4x+13}{5x^2-35x}+\dfrac{48-x}{2x\left(7-x\right)}\\ =\dfrac{4x+13}{5x\left(x-7\right)}+\dfrac{x-48}{2x\left(x-7\right)}\\ =\dfrac{2.\left(4x+13\right)+5.\left(x-48\right)}{10x\left(x-7\right)}=\dfrac{8x+26+5x-240}{10x\left(x-7\right)}\\ =\dfrac{13x-214}{10x\left(x-7\right)}\)

\(b,\dfrac{3x+1}{x^2-2x+1}+\dfrac{-1}{x+1}+\dfrac{x+3}{1-x^2}\\ =\dfrac{3x+1}{\left(x-1\right)^2}+\dfrac{-1}{x+1}+\dfrac{-x-3}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(3x+1\right)\left(x-1\right)-\left(x-1\right)^2+\left(-x-3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{3x^2+x-3x-1-x^2+2x-1-x^2-3x+x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{1}{x+1}\)

Ta có

\(\widehat{cAa}=\widehat{ADb}=105^o\)

mà hai góc vị trí đồng vị

\(=>m//n\)

\(\Rightarrow\widehat{DCt}=\widehat{CBA}=40^o\) ( hai góc đồng vị)

\(\widehat{cBA}=\widehat{dBm}=x=40^o\)

\(\left(3x+2\right)^2.5=65\\ \Rightarrow\left(3x+2\right)^2=65:5\\ \Rightarrow\left(3x+2\right)^2=13\\ \Rightarrow\left[{}\begin{matrix}3x+2=\sqrt{13}\\3x+2=-\sqrt{13}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{13}-2}{3}\\x=\dfrac{-\sqrt{13}-2}{3}\end{matrix}\right.\)

\(=23,19.35+23,19.66-23,19\\ =23,19.\left(35+66-1\right)\\ =23,19.100\\ =2319\)

\(=1,42.30.10+14,2.57+14,2.14-14,2\\ =14,2.30+14,2.57+14,2.14-14,2\\ =14,2\left(30+57+14-1\right)\\ =14,2.100\\ =1420\)

\(\widehat{A_1}+\widehat{A_2}=180^o\) ( kề bù )

\(\Rightarrow\widehat{A_1}+4\widehat{A_1}=180^o\\ \Rightarrow5\widehat{A_1}=180^o\\ \Rightarrow\widehat{A_1}=180^o:5=36^o\)

mà \(m//n\)

\(\Rightarrow\widehat{A_1}=\widehat{B_1}=36^o\)

\(AC=sinB.BC=sin30^o.20=10\)

=> Bạn Nam sai 

\(=\sqrt{144}.\sqrt{6,25}.\sqrt{36}=12.2,5.6=180\)