\(d,=\dfrac{17}{12}\left(\dfrac{15}{11}-\dfrac{4}{13}+\dfrac{7}{11}-\dfrac{9}{13}\right)=\dfrac{17}{12}\left(2-1\right)=\dfrac{17}{12}\)
Bài 2:
\(a,\Rightarrow x=\dfrac{7}{5}-\dfrac{3}{4}=\dfrac{13}{20}\\ b,\Rightarrow\left|x-\dfrac{2}{3}\right|=1\dfrac{3}{5}-\dfrac{3}{5}=1\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=1\\\dfrac{2}{3}-x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\\ c,\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{3}=\dfrac{2}{3}\\\dfrac{2}{3}x+\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{1}{3}\\\dfrac{2}{3}x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Bài 3:
\(a,\dfrac{a}{b}< \dfrac{a+c}{b+c}\Leftrightarrow ab+ac< ab+bc\Leftrightarrow ac< bc\Leftrightarrow a< b\left(luôn.đúng\right)\\ b,A>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\\ A< \dfrac{a+c}{a+b+c}+\dfrac{b+c}{a+b+c}+\dfrac{c+a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Leftrightarrow1< A< 2\Leftrightarrow A\notin Z\)
