Câu hỏi của quynh nhu nguyen

Question

BT3. Tìm x, biết:

a/ |x|=$\dfrac{1}{2}$ ;                         b/ |x|= 3,12;                          c/ |x|= 0;                           d/ |x|=2$\dfrac{1}{7}$ ;

BT3. Tìm x, biết:

a/|x|=2,...

Nguyễn Thanh Hằng · Accepted Answer

1, a/ $\left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\x=\dfrac{-1}{2}\end{matrix}\right.$ Vậy ............. b/ $\left|x\right|=3,12\Leftrightarrow\left[{}\begin{matrix}x=3,12\x=-3,12\end{matrix}\right.$ Vậy ........... c/ $\left|x\right|=0\Leftrightarrow x=0$ Vậy .......... d/ $\left|x\right|=2\dfrac{1}{7}$ $\Leftrightarrow\left[{}\begin{matrix}x=2\dfrac{1}{7}\x=-2\dfrac{1}{7}\end{matrix}\right.$ Vậy .............. 2, a/ $\left|x\right|=2,1\Leftrightarrow\left[{}\begin{matrix}x=2,1\x=-2,1\end{matrix}\right.$ Vậy ........... b/ $\left|x\right|=\dfrac{17}{9}$ ; $x< 0$ $\Leftrightarrow x=-\dfrac{17}{9}$ Vậy .......... c/ $\left|x\right|=1\dfrac{2}{5}\Leftrightarrow\left[{}\begin{matrix}x=1\dfrac{2}{5}\x=-1\dfrac{2}{5}\end{matrix}\right.$ Vậy ........... d/ $\left|x\right|=0,35$ ; $x>0\Leftrightarrow x=0,35$ 3, a/ $\left|x-1,7\right|=2,3$ $\Leftrightarrow\left[{}\begin{matrix}x-1,7=2,3\x-1,7=-2,3\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=4\x=-0,6\end{matrix}\right.$ Vậy ........... b/ $\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0$ $\Leftrightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}$ $\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{12}\x=-\dfrac{13}{12}\end{matrix}\right.$ Vậy ...........Câu trả lời: 1, a/ $\left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\x=\dfrac{-1}{2}\end{matrix}\right.$ Vậy ............. b/ $\left|x\right|=3,12\Leftrightarrow\left[{}\begin{matrix}x=3,12\x=-3,12\end{matrix}\right.$ Vậy ........... c/ $\left|x\right|=0\Leftrightarrow x=0$ Vậy .......... d/ $\left|x\right|=2\dfrac{1}{7}$ $\Leftrightarrow\left[{}\begin{matrix}x=2\dfrac{1}{7}\x=-2\dfrac{1}{7}\end{matrix}\right.$ Vậy .............. 2, a/ $\left|x\right|=2,1\Leftrightarrow\left[{}\begin{matrix}x=2,1\x=-2,1\end{matrix}\right.$ Vậy ........... b/ $\left|x\right|=\dfrac{17}{9}$ ; $x< 0$ $\Leftrightarrow x=-\dfrac{17}{9}$ Vậy .......... c/ $\left|x\right|=1\dfrac{2}{5}\Leftrightarrow\left[{}\begin{matrix}x=1\dfrac{2}{5}\x=-1\dfrac{2}{5}\end{matrix}\right.$ Vậy ........... d/ $\left|x\right|=0,35$ ; $x>0\Leftrightarrow x=0,35$ 3, a/ $\left|x-1,7\right|=2,3$ $\Leftrightarrow\left[{}\begin{matrix}x-1,7=2,3\x-1,7=-2,3\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=4\x=-0,6\end{matrix}\right.$ Vậy ........... b/ $\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0$ $\Leftrightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}$ $\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{12}\x=-\dfrac{13}{12}\end{matrix}\right.$ Vậy ...........

Đạt Trần · Answer

Đề dễ lắm sao ko tự làm điCâu trả lời: Đề dễ lắm sao ko tự làm đi

Nguyễn Thị Hồng Nhung · Answer

Bài1:

a/ |x|=$\dfrac{1}{2}$

=>$x=-\dfrac{1}{2}$hoặc $x=\dfrac{1}{2}$

Vậy..

Các câu sau tương tự

Bài2:

a/|x|=2,1

=>$x=-2,1$hoặc $x=2,1$

Vậy...

Bài3:

a/ | x - 1,7| =2,3

=>$x-1,7=2,3$hoặc$x-1,7=-2,3$

=>x=4 hoặc x=-0,6

Vậy ...Câu trả lời: Bài1: