Question

[ha phuong anh]

\(\left|x+\dfrac{1}{5}\right|:\dfrac{9}{10}\)

\(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|=0\)

\(\left|2x\right|-\left|3.5\right|=\left|-6.5\right|\)

\(\left|x-1.7\right|=2.3\)

Answer 1

[Đức Hiếu]

a,?????

b, Với mọi giá trị của x;y ta có:

\(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|\ge0\)

Để \(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|=0\) thì:

\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|x+y\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{1}{2}+y=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy..........

c, \(\left|2x\right|-\left|3,5\right|=\left|-6,5\right|\)

\(\Rightarrow\left|2x\right|=6,5+3,5=10\)

\(\Rightarrow\left\{{}\begin{matrix}2x=10\\2x=-10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

Vậy..........

d, \(\left|x-1,7\right|=2,3\)

\(\Rightarrow\left\{{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)

Vậy.........

Chúc bạn học tốt!!!

Answer 2

[ha phuong anh]

cám ơn p, câu a mik viết sai.