Câu 25: Trang 16- sgk toán 7 tập 1
Tìm x, biết :
a. $\left | x-1,7 \right |=2,3$
b. $\left | x+\frac{3}{4} \right |-\frac{1}{3}=0$
Bài Làm:
a. $\left | x-1,7 \right |=2,3$
<=> $\left\{\begin{matrix} x-1,7=2,3 & \\ x-1,7=-2,3 & \end{matrix}\right.$
<=> $\left\{\begin{matrix} x=2,3+1,7 & \\ x=-2,3+1,7 & \end{matrix}\right.$
<=> $\left\{\begin{matrix} x=4 & \\ x=-0,6 & \end{matrix}\right.$
Vậy $\left\{\begin{matrix} x=4 & \\ x=-0,6 & \end{matrix}\right.$
b. $\left | x+\frac{3}{4} \right |-\frac{1}{3}=0$
<=> $\left | x+\frac{3}{4} \right |=\frac{1}{3}$
<=> $\left\{\begin{matrix}x+\frac{3}{4}=\frac{1}{3} & \\ x+\frac{3}{4}=-\frac{1}{3} & \end{matrix}\right.$
<=> $\left\{\begin{matrix}x=\frac{1}{3}-\frac{3}{4} & \\ x=-\frac{1}{3}-\frac{3}{4} & \end{matrix}\right.$
<=> $\left\{\begin{matrix}x=\frac{-5}{12} & \\ x=\frac{-13}{12} & \end{matrix}\right.$
Vậy $\left\{\begin{matrix}x=\frac{-5}{12} & \\ x=\frac{-13}{12} & \end{matrix}\right.$