Tìm \(x\) biết:

    a) \[\sqrt {{x^2}}  = 5\];                                                    b) \[\sqrt {25{x^2}}  = 10\];

    c) \[\sqrt {4{x^2} - 28x + 49}  = 7\];                                       d) \[\sqrt {x - 10\sqrt x  + 25}  = 3\].

a) Ta có \[4{x^2} - 64 = 0\]

\(4{x^2} = 64\)

\({x^2} = 16\)

\(x = 4\) hoặc \(x =  - 4\).

Vậy \(S = \left\{ { - 4;4} \right\}\).    

b) Ta có \[\sqrt {{x^4}}  - 7 = 0\]

\(\sqrt {{{\left( {{x^2}} \right)}^2}}  = 7\)

\(\left| {{x^2}} \right| = 7\)

\({x^2} = 7\)

\(x = \sqrt 7 \) hoặc \(x =  - \sqrt 7 \).

 Vậy \(S = \left\{ { - \sqrt 7 ;\sqrt 7 } \right\}\).    

c) Ta có \[\sqrt {9{x^2}}  = 2x + 1\]

\(\sqrt {{{\left( {3x} \right)}^2}}  = 2x + 1\)

\(\left| {3x} \right| = 2x + 1\)

TH1: \(\left\{ {\begin{array}{*{20}{c}}{3x \ge 0}\\{3x = 2x + 1}\end{array}} \right.\) nên \(\left\{ {\begin{array}{*{20}{c}}{x \ge 0}\\{x = 1}\end{array}} \right.\), do đó \(x = 1\).

TH2: \(\left\{ {\begin{array}{*{20}{c}}{3x < 0}\\{ - 3x = 2x + 1}\end{array}} \right.\) nên \(\left\{ {\begin{array}{*{20}{c}}{x < 0}\\{x =  - \frac{1}{5}}\end{array}} \right.\), do đó \(x =  - \frac{1}{5}\).

Vậy \(S = \left\{ { - \frac{1}{5};1} \right\}\).

d) Ta có \[\sqrt {{x^2} - 4x + 4}  - \sqrt {{x^2} + 4x + 4}  = 0\]

\(\sqrt {{x^2} - 4x + 4}  = \sqrt {{x^2} + 4x + 4} \)\(\sqrt {{{\left( {x - 2} \right)}^2}}  = \sqrt {{{\left( {x + 2} \right)}^2}} \)

\(\left| {x - 2} \right| = \left| {x + 2} \right|\)

\(x - 2 = x + 2\) hoặc \(x - 2 =  - \left( {x + 2} \right)\)

\(0x =  - 4\) (loại) hoặc \(2x = 0\)

\(x = 0\).

Vậy \(S = \left\{ 0 \right\}\).