Bài 9 trang 11 SGK Toán 9 tập 1

Tìm x biết:

a) \(\sqrt {{x^2}}  = 7\) ;                

b) \(\sqrt {{x^2}} = \left| { - 8} \right| \)

c) \(\sqrt {4{{\rm{x}}^2}}  = 6\)               

d) \(\sqrt {9{{\rm{x}}^2}} = \left| { - 12} \right|\);

a) Ta có:

\(\eqalign{
& \sqrt {{x^2}} = 7 \cr
& \Leftrightarrow \left| x \right| = 7 \cr
& \Leftrightarrow x = \pm 7 \cr} \)

Vậy \(x= \pm 7\).

b) Ta có:

\(\eqalign{
& \sqrt {{x^2}} = \left| { - 8} \right| \cr 
& \Leftrightarrow \left| x \right| = 8 \cr 
& \Leftrightarrow x = \pm 8 \cr} \)

Vậy \(x= \pm 8 \).

c) Ta có:

\(\eqalign{
& \sqrt {4{x^2}} = 6 \cr
& \Leftrightarrow \sqrt {{2^2}.{x^2}} = 6 \cr
& \Leftrightarrow \sqrt {{{\left( {2x} \right)}^2}} = 6 \cr
& \Leftrightarrow \left| {2x} \right| = 6 \cr
& \Leftrightarrow 2x = \pm 6 \cr
& \Leftrightarrow x = \pm 3 \cr} \)

Vậy \(x= \pm 3 \).

d) Ta có:

\(\eqalign{
& \sqrt {9{x^2}} = \left| { - 12} \right| \cr
& \Leftrightarrow \sqrt {{3^2}.{x^2}} = 12 \cr
& \Leftrightarrow \sqrt {{{\left( {3x} \right)}^2}} = 12 \cr
& \Leftrightarrow \left| {3x} \right| = 12 \cr
& \Leftrightarrow 3x = \pm 12 \cr
& \Leftrightarrow x = \pm 4 \cr} \).

Vậy \(x= \pm 4 \).