Đáp án+Giải thích các bước giải:
Bài 4:
`a) |x-1,7|=2,3`
`<=>`\(\left[ \begin{array}{l}x-1,7=2,3\\x-1,7=-2,3\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=4\\x=-0,6\end{array} \right.\)
Vậy `x∈{-0,6; 4}`
`b) |x+4/(15)|-|-3,75|=-|-2,15|`
`<=> |x+4/(15)|-3,75=-2,15`
`<=> |x+4/(15)|=1,6`
`<=>`\(\left[ \begin{array}{l}x+\dfrac{4}{15}=1,6\\x+\dfrac{4}{15}=-1,6\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{4}{3}\\x=-\dfrac{28}{15}\end{array} \right.\)
Vậy `x∈{4/3; -(28)/(15)}`
`c) |3x+4|+|3y+5|=0`
`<=>` $\left\{\begin{matrix} 3x+4=0\\ 3y+5=0 \end{matrix}\right.$
`<=>` $\left\{\begin{matrix} 3x=-4\\ 3y=-5 \end{matrix}\right.$
`<=>` $\left\{\begin{matrix} x=\dfrac{-4}{3}\\ y=\dfrac{-5}{3}\end{matrix}\right.$
Vậy `x=4/3; y=-5/3`
