HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
bài 1) a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{15}\right) \) \(\left(\dfrac{5}{42}-x\right)=\dfrac{11}{13}+\dfrac{15}{28}-\dfrac{11}{15}\) \(x=\dfrac{5}{42}-\dfrac{3541}{5460}=-\dfrac{413}{780}\) b) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\) \(\left|x+\dfrac{4}{15}\right|=-\left|2,15\right|+\left|3,75\right|=1,6\) \(\Rightarrow x+\dfrac{4}{15}=1,6\) hoặc \(x+\dfrac{4}{15}=-1,6\) \(\Rightarrow x=\dfrac{4}{3}\) hoặc \(x=-\dfrac{28}{15}\) c) \(\dfrac{5}{3}-\left|x-\dfrac{3}{2}\right|=-\dfrac{1}{2}\) \(\Rightarrow\left|x-\dfrac{3}{2}\right|=\dfrac{5}{3}+\dfrac{1}{2}=\dfrac{13}{6}\) \(\Rightarrow x-\dfrac{3}{2}=\dfrac{13}{6}\) hoặc \(x-\dfrac{3}{2}=-\dfrac{13}{6}\) \(\Rightarrow x=\dfrac{11}{3}\) hoặc \(x=-\dfrac{2}{3}\) d)\(\left(x-\dfrac{2}{3}\right).\left(2x-\dfrac{3}{2}\right)=0\) \(\Rightarrow x-\dfrac{2}{3}=0\) hoặc \(2x-\dfrac{3}{2}=0\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{3}{4}\end{matrix}\right.\) 3) a) \(\left(x^{^2}-4\right)^{^2}+\left(x+2\right)^{^2}=0\) Vì \(\left(x^{^2}-4\right)^{^2}\ge0,\left(x+2\right)^{^2}\ge0\) nên : \(\left\{{}\begin{matrix}x^{^2}-4=0\\x+2=0\end{matrix}\right.\Rightarrow x=\pm2\) b) \(\left(x-y\right)^{^2}+\left|y+2\right|=0\) Vì \(\left\{{}\begin{matrix}\left(x-y\right)^{^2}\ge0\\\left|y+2\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\y=-2\end{matrix}\right.\Rightarrow x=-2;y=-2\) c) \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\) Vì \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+\dfrac{9}{25}\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow y=-\dfrac{9}{25};x=-\dfrac{9}{25}\) d) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\left(-\dfrac{1}{4}\right)-\left|y\right|\) \(\Rightarrow\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\) Vì \(\left\{{}\begin{matrix}\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|\ge0\\\left|y\right|\ge0\end{matrix}\right.\) mà \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\) nên không tồn tại x,y thỏa mãn đề bài .