1) Tìm x, y biết : \(\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\)
Ta có :
\(\left|x-y\right|\ge0\forall x;y\)
\(\left|y+\frac{9}{25}\right|\ge0\forall y\)
\(\Rightarrow\left|x-y\right|+\left|y+\frac{9}{25}\right|\ge0\forall x,y\)
\(\Rightarrow\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\Leftrightarrow\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y+\frac{9}{25}\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\frac{9}{25}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=-\frac{9}{25}\end{matrix}\right.\)
Vậy : \(x=y=-\frac{9}{25}\)
2) Tìm x biết :
a) \(\left|x+\frac{2}{11}\right|>\left|-5,5\right|\)
\(\Rightarrow\left|x+\frac{2}{11}\right|>5,5=\frac{11}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{2}{11}>\frac{11}{2}\\x+\frac{2}{11}>-\frac{11}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\frac{11}{2}-\frac{2}{11}=\frac{117}{22}\\x>-\frac{11}{2}-\frac{2}{11}=-\frac{125}{22}\end{matrix}\right.\Rightarrow x>-\frac{125}{22}\)
Vậy : \(x>-\frac{125}{22}\)
Đúng không ta ? Mình không chắc lắm ....
2.Tìm x biết :
a) \(\left|x+\frac{2}{11}\right|>\left|-5,5\right|\)
\(\Rightarrow\left|x+\frac{11}{2}\right|>5,5\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{11}{2}>5,5\\x+\frac{11}{2}< -5,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>0\\x< 11\end{matrix}\right.\)
Vậy : \(0< x< 11\)
P/s: Xin lỗi bạn , phần a) câu 2 này mình làm lại nhiều lần quá. Lần này chắc chắn đúng nha ...
Mình làm lại phần b) Câu 2 nhé ( Mình nhận ra mình làm sai ):
a) \(\left|x+\frac{2}{11}\right|>\left|-5,5\right|\)
\(\Rightarrow\left|x+\frac{2}{11}\right|>5,5\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{2}{11}>5,5\\x+\frac{2}{11}< -5,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>0\\x< 11\end{matrix}\right.\)
Vậy : \(0< x< 11\)
2) Tìm x biết :
b) \(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Rightarrow\pm\frac{2}{5}< x-\frac{7}{5}< \pm\frac{3}{5}\)
Xét : \(\frac{2}{5}< x-\frac{7}{5}< \frac{3}{5}\)
\(\Rightarrow\frac{2}{5}+\frac{7}{5}< x< \frac{3}{5}+\frac{7}{5}\)
\(\Rightarrow\frac{9}{5}< x< 2\)
Xét : \(-\frac{2}{5}>x-\frac{7}{5}>-\frac{3}{5}\)
\(\Rightarrow-\frac{2}{5}+\frac{7}{5}>x>-\frac{3}{5}+\frac{7}{5}\)
\(\Rightarrow1>x>\frac{4}{5}\)
Vậy : \(\left[{}\begin{matrix}\frac{9}{5}< x< 2\\\frac{4}{5}< x< 1\end{matrix}\right.\)
\(2a,\left|x+\frac{11}{2}\right|>\left|-5,5\right|\)
\(\Leftrightarrow\left|x+\frac{11}{2}\right|>5,5\)
Ta có: \(\left|x+\frac{11}{2}\right|>5,5\) nên \(\left|x+\frac{11}{2}\right|>0\)
\(\Rightarrow\left|x+\frac{11}{2}\right|=x+\frac{11}{2}\)
\(pt:x+\frac{11}{2}>5,5\Leftrightarrow x>5,5-\frac{11}{2}=0\)
Vậy \(S=\left\{x/x>0\right\}\)
